Solucionario Mecanica Vectorial para ingenieros Estatica

Transcripción

CHAPTER 3
PROBLEM
3.1
A foot valve for a pneumatic system is hinged at B.
80
that
mm
a - 28°, determine the moment of the
B by
Point
resolving the force into
Knowing
6-N force about
horizontal and vertical
1
components.
SOLUTION
Note
= a- 20° = 28° -20° = 8°
that
Fx = 6 N)cos 8° = 5.8443 N
Fv =(16N)sin8° = 2.2268N
and
1
(1
x = (0.
Also
1
7 m) cos 20°
= 0. 1 59748 m
^£*.
*
\©*
—
tL> Kl
U^r^-^T^P,,
/|t^^û^
d
^r^^^^-,
>n^y c
y = (0.17 m)sin 20° = 0.058143 m.
Noting that the direction of the moment of each force component about
B is counterclockwise,
MB =xFy +yF
x
= (0.1 59748 m)(2.2268N)
+(0.058143 m)(l 5.8443 N)
= 1.277 N-m
or
©
M B =1.277N-m^)4
PROPRIETARY MATERIAL.
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153
PROBLEM
A
3.2
foot valve for a pneumatic system
that
a
= 28°,
determine the
hinged at B.
is
moment of
Point B by resolving the force into
Knowing
6-N force about
components along ABC and
the
1
in a direction perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into
components P and Q, where
g = (16 N) sin 28°
= 7.5115 N
°^7^
M B = rm Q
Then
= (0.17m)(7.5115N)
= 1.277N-m
or
PROPRIETARY MATERIAL. ©
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teachers and educators permitted by McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual,
reproduced or distributed
distribution to
you are using
M B = 1.277 N-m^^
it
in
without permission.
154
j
-100
mm-
200
mm
PROBLEM
i
200
'">
mm
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
same moment about D.
creates the
in..,
3.3
25"
SOLUTION
(a)
F
=(300N)cos25°
v
= 27.1.89 N
O.2.
Fy =(300 N) sin
= 126.785 N
F = (27
.89 N)i
1
+ (1 26.785 N)
= ZM = -(0.1m)i-(0.2m)j
r
M D =rxF
M D = HO.
PC
25°
1
m)i - (0.2 m)j]
x [(271 .89 N)i + (1.26.785
N)j]
= -(12.6785 N m)k + (54.378 N m)k
= (41.700 N-m)k
•
•
0>T-**>
M
(b)
The
smallest force
=41.7N-nO^
Q at B must be perpendicular to
DB at 45°^£L
M D =Q(DB)
41 .700
Q = 147.4 N ^L 45° <
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student using this Manual,
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N m = (2(0.28284 m)
in
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155
-200
•-lOOnim-*
mm
PROBLEM3.4
*.
A
125
300-N
force
is
applied at
moment of the 300-N
il
A
as shown. Determine (a) the
force about D, (b) the magnitude and
sense of the horizontal force applied at
200 mn.
same moment about D, (c) the smallest
creates the same moment about D.
•
C
C
that creates the
force applied at
C that
H
25°
SOLUTION
(a)
See Problem 3.3 for the figure and analysis leading
to the determination
of Md
M
=41.7N-m^H
Cl^n
0>\7.Siy\
c
(b)
Since
=
at.
C is horizontal C = Ci
r
= DC = (0.2 m)i - (0. 25 m) j
1
M D =rxCi = C(0.l25m)k
4l.7N-m = (0.l25m)(C)
C = 334N <
C = 333.60 N
(c)
The
smallest force
C must be perpendicular to DC; thus,
it
forms
a
with the vertical
m
0.2 m
0.125
tan6^
a = 32.0°
M D = C(£>C); DC = V(
-
2
m
= 0.23585
41.70
2
)
+ (°- 1 25
Nm = C(0.23585m)
C = 176.8 N^L 5HX)°<
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)'
m
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m
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156
PROBLEM
3.5
An
P
8-1 b force
when a
is
equal
is
to.
applied to a shift lever. Determine the
25 9
moment of V about
B
.
SOLUTION
Px = (8
First note
lb) cos 25°
= 7.2505
/^
=(8
lb
lb) sin
= 3.3809
25°
i*22.
lb
Noting that the direction of the moment of each force component about
•**.
B is
clockwise, have
= -(8in.)(3.3809
1b)
- (22 in.)(7.2505
= -186.6
lb)
lb -in.
or
distribution to
Mj =186.6 lb -in. J) ^
in
157
PROBLEM
3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force
22
P that has a 21 0-lb
•
in.
clockwise
moment about B.
in.
SOLUTION
For
P to be minimum
it
must be perpendicular to the
line joining Points A.
and B. Thus,
a=
e
^T^s*. K
/
8 in
,
22
in.
i
= 19.98°
MB =dP^
and
p
n
d = rAIB
Where
ZZ
i«.
fl
= ^m.y+(22m.y
= 23.409
$u
JB w
in.
_210Ib-in.
Then
'
min
" 23.409
-8.97
in.
Pmin =8.97
lb
lb
^
19.98°
<
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158
PROBLEM
3.7
An
P
1 1
-lb force
22
The moment of P about
Determine the value of a.
applied to a shift lever.
is
and has a magnitude of 250 lb
•
in.
B is clockwise
in.
SOLUTION
By
M
definition
B
m
=r Psm.6
= a + (9Q°-tf>)
where
and
<p
- tan"_i
8
in.
19.9831'
22
in.
2
also
r f/fl =V(8in.) +(22in.)
= 23.409
Then
250lb-in
in.
= (23.409in.)(lllb)
xsin(tf
or
sin («
2
+ 90° -19.9831°)
+ 70.01 69°) = 0.97088
or
a + 70.0 169° = 76. 1391°
and
a+ 70.0169° = 103.861°
a = 6.12°
33.8°
<
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it
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159
PROBLEM
3.8
known
that a vertical force of 200 lb is required to remove the nail
from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
It is
at
C
the force
P
same moment about B if a ~
same moment about B.
that creates the
smallest force
P
.1
0°,
(c) the
that creates the
r4 n
V—A
in.
SOLUTION
(a)
MB =raB FN
We have
(4 in.)(200 lb)
800
lb -in.
or
A-
m.
(/;)
M
By definition
B
~rA/B P sin
6»
= 10° + (180°~70°)
MB =H00\b-m.)<
= 120°
Then
800
lb in.
•
= (18
in.)
x Psin 1 20°
or
(c)
For
P to be minimum,
Points
A and B.
or
must be perpendicular to the
<
line joining
P must be directed as shown.
800
cl
= f A IB
lb in.
= (18
•
in.)
î„=44.4
1b
Pm
Pmitl =44.41b^l20
^
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to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
distribution
it
lb
A*W™.
Thus
or
Thus,
P = 51.3
you are using
it
in
without permission.
160
m
PROBLEM
A
3.9
winch puller
length
d
force into
AB
is
used to straighten a fence post. Knowing that the tension in cable
m, determine the moment about D of the force exerted by the cable
horizontal and vertical components applied (a) at Point C, (b) at Point E.
is
1
,90
0.875
0.2
at
BC is 1040 N and
C by resolving that
m
m
SOLUTION
(a)
Slope of line
0.875
EC =
1.90
Then
*abx
-
m
m + 0.2 m
12
,~ (Tab)
12
(1040 N)
0,1«y
13
960
N
*W=-0040N)
and
= 400
N
MD - T
Then
ABx (0.875
m)~TABy (0.2
m)
= (960 N)(0.875 m) - (400 N)(0.2 m)
= 760 N
(b)
We have
>
or M. /J
•
=760N-m
)^
MD ~r m (y) + TABx (x)
= (960 N)(0) + (400 N)(I .90 m)
= 760 N
•
or
reproduced
distribution to
v
7>
= 760N-m )<«
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or.
you are using
M
distributed in
it
without permission.
161
PROBLEM
It is
known
3.10
that a force with a
d- 2,80 m, determine
required
moment of 960 N
the tension that
•
m about D
must be developed
is
required to straighten the fence post
in the cable
of winch puller
AB
CD.
If
to create the
moment about Point D.
0.875
0.2
m
j.i»
SOLUTION
o.
a? 5î
\*.
£^
»
'My
0.875
EC =
Slope of line
2.80
7'
Then
My
r
and
OiZ^
z-acw
*AB
/ffl)>
m
m
+ 0.2
7
m
24
24
T
25
1
/)B
rAH
25
We have
960N-m=
24
—
7^(0) +— 7^(2.80*)
7
7^= 1224 N
©
or
r^=1224N ^
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PROPRIETARY MATERIAL;
162
PROBLEM
It is
3.11
known that a
moment about
moment of 960 N
force with a
capacity of winch puller
AB
is
m about D is required to straighten the fence post CD. If the
•
2400 N, determine the minimum value of distance d
to create the specified
Point D.
0.875 in
fe
0.2 in
SOLUTION
o.mtn
czom
The minimum value of d can be found based on
the equation relating the
moment of the
force
TAB
about D:
MD ={TABmK ),(d)
MD =960N-m
where
(^flmax )y
0.875m
.
.
Now
= TAIHmx sin & = ( 2400 N)sfo
sin
#
^(t/
960
+ 0.20) 2 +(0.875) 2 m
0.875
N m = 2400 N
(d)
•
+ 0.20) 2 +(0.875) 2
^
or
2
(J + 0.20) + (0.875)
or
or
+ 0.20) 2 + (0.875) 2 = 2.
3 .7852</
2
2
!
875 d
= 4.7852rf 2
- 0.40c/ - .8056 =
Using the quadratic equation, the minimum values of d are 0.51719
Since only the positive value applies here,
d — 0.5
1
719
m and -.41151 m.
m
or
©
d~5
1
7
mm ^
PROPRIETARY MATERIAL
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reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limited
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163
5.3
PROBLEM
in.
The
12.0 in.
lift
tailgate
of a car
is
supported by the hydraulic
exerts a 125-lb force directed along
and socket
2.33
3.12
at B,
determine the
its
moment of the
BC. If the
on the ball
lift
centerline
force about A.
in.
mmM
1
>iii.
i
SOLUTION
2
First note
dcB
==
=
Then
cos
7(12.0
2.33
9-
\«5»,a>
im.
\Z.&
2.33
*cb =
>N.
in.
in.
12.2241
and
2
in.
12.2241
sin
+ (2.33 in.)
12.224 lin.
12.0
9=
in.)
ttvi.
in.
FCB cos 9\ -FCB $m9l
1251b
[(12.0 in.)i- (2.33 in.) j]
12.2241
in.
Now
M.A = VB/A X ^CB
where
XBIA = (15.3in.)i-- (12.0
=
M
Then
in.
(15.3 in.) i--(14.33
+ 2.33 in.) j
in.) j
ru-nimn*
—
1251b
12.2241
(12.01
— 2.331)
in.
(1393.87 lb in.)k
(116.156 lb -ft)k
or
M„ = 116.2 lb- ft *)<
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164
-
1
—
20.5
PROBLEM
in.
h
3.13
4.38 in
i
The
*Ti
tailgate
of a car
is
supported by the hydraulic
exerts a 125-lb force directed along
socket at B, determine the
7.62
its
moment of the
lift
centerline
EC.
on the
If the
ball
lift
and
force about A.
t§|
1.7.2 hi.
SOLUTION
First
dCB
note
= 7(17.2 in.)
:
=
18.8123
2
+ (7.62
2
in.)
in.
Z.O.'S «w.
17.2 in.
Then
cos
-
18.8123
sin
6-
7.62
t:v*
=
Now
XKvZ
>KJ.
in.
S
fcd = ( F
>Ni.
in.
18.8123
and
4.?.e.
in.
cos 0)\ - (FCB sin
(,7 2i "-)i
'
n.z
.«»».
6>) j
+ (7 62i
where
r^ = (20.5 in.)i- (4.38
Then
M, = [(20.5 in.)i - (4.38 in.) j] x
'
^
in.)j
1251b
(1
.
t
18.8 123
7.21
- 7.62 j)
in.
(1538.53 lb -in.)k
(128.2 lb -ft)k
or
M^ =128.2 lb- ft ^H
MW
IROlRlhTARl MAlhRlAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved.
M> ,«,#/ «///,&
,»«> 6* rfwpfe^
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used bevond the. limited
d.stnbuUoiMo teachers and educators permitted by McGraw-Hill
for their individual course, preparation. ^
you are using
it
without permission.
165
PROBLEM
120mm
3.14
mechanic uses a piece of pipe AB as a lever when tightening an
When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
A
alternator belt.
about bolt
65
C if its line of action passes through O.
mm
SOLUTION
M c =rwc xFj,
We have
4
Noting the direction of the moment of each force component about
j,
C is
clockwise.
Mc =xF
By
+yFBx
mm = 55 mm
y - 72 mm + 90 mm - 62 mm
x - 1 20 mm - 65
Where
1
65
F
and
lix
-(485N) = 325N
2
7(65) +(72)
72
Fa
.(485
2
V(65)
iW c
2
+
N)- 360 N
3
(72)
= (55 mm)(360 N) + (1 62)(325 N)
= 72450 N- mm
= 72.450 N-m
or
a student using this Manual,
teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are
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M c =72.5 N-m J) <
it
in
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166
PROBLEM
Form the
3.15
vector products
B
x
C and B'
* C, where
B = B\ and use the results
obtained to prove the identity
sin a cos/?
= -sin (a + P) + -sin (a - p).
SOLUTION
N(>te:
B = £(cos/?i + sin/?j)
B' = 5(cos/?i-sin/?j)
C - C(cos a + sin a j)
i
By definition:
|BxC| = flCsin(a-jff)
0,)
|B'xC| = 5Csin(flf + y?)
Now
(2)
B x C = Z?(cos /?i + sin y9j) x C(cos tfi + sin orj)
= BC(cos
/?sin
« - sin /?cos «)k
B'x C - /?(cos /?i - sin
and
- £C(cos yffsin
Equating the magnitudes of
/?j)x C(cos ai
ar
-f
(3)
sin
#j)
+ sin /?cos ar)k
(4)
BxC from Equations (I ) and (3) yields:
BCs'm(a -p)~ BC(cos ps'm
Similarly, equating the magnitudes
of
B'xC
from Equations
a - sin pcos a)
(2)
and
(5)
(4) yields;
BCsm(a + p) = BC(cos ps'm a + sln pcos a)
Adding Equations
(5)
and
sin(a - p) + s'm(a + p) - 2 cos /?sin
a
or sin
you are using
« cos /? - - sin(ar + /?) + -sin(flf - /?) ^
AAW
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(6)
(6) gives:
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167
PROBLEM
3.16
A line passes through the Points (20 m, 16 m) and (-1 m, ~4 m). Determine the perpendicular distance rffrom
the tine to the origin O of the system of coordinates.
SOLUTION
E>
2
dAB = V[20m- (-1 m)] + [1 6
- 29
Assume that a force F,
directed ixomA to B.
(Zom,
K-rti^
m - (-4 m)f
m
or magnitude F(N), acts at Point A and
is
-*x
¥~FX m
Then,,
*B~ rA
Where
'All
d
.41!
=
—
<21i
V
\r
xF\ = dF
-
29
By
MQ =
definition
A
+ 20j)
r,=-(lm)i-(4m)j
Where
M
Then
=[-(-1 m)i-(4
F
m)j]x—-[(21
29
a
-(20)k
m
m)i + (20 m)j]
+ (84)k]
~F|k N-m
29
64
Finally
F = </(F)
29
.
rf-
—
m
29
64
</
= 2.21m
^
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limited
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the
using this Manual,
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student
©
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168
PROBLEM
The
vectors
3.17
P and
(tf)P--7i +
Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
3j-3kandQ = 2i + 2j + 5k,(6)P = 6i-5j-2kandQ = -2i + 5j~k.
SOLUTION
(a)
We have
v4
where
P = -7i + 3j-3k
= |PxQf
Q = 2i + 2j + 5k
J
k
-7
3
-3
2
2
5
>
PxQ
Then
= [(15 + 6)i + (-6 + 35)j + (-14-6)k]
= (21)1 + (29) j(-20)k
^ = V(20) 2 +(29) 2 +(-20) 2
(b)
We have
A = \PxQ\
where
P = 6i~5j-2k
or
4 = 41.0
<4
or
4 = 26.9
^
Q = ~2i + 5in.j~lk
PxQ
Then
i
i
6
-5
-2
5
fc
= [(5 + 1 0)i + (4 + 6)j + (30-1 0)k]
«(15)i
+ (10)j + (20)k
A=j(i5)-+(\oy + (2oy
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169
PROBLEM
A
3.18
plane contains the vectors
to,
respectively, (a)
i
A and
B. Determine the unit vector normal to the plane
5k, (b) 3i - 3j + 2k and -2i + 6j - 4k.
when
A and B are equal
+ 2j - 5k and 4i - 7j -
SOLUTION
{a)
AxB
We have
|AxB|
A
where
B
+ 2j-5k
:li
=
4i
- 7 j - 5k
J
*
1
+2
-5
4
-7
-5
i
AxB
Then
|AxB|
and
(-1
~ 35)i + (20 + 5)j + (-7 - 8)k
15(31
-1J -Ik)
2
15V(-3)
+H)
2
2
+(-l) -I5>/ri
15(-3f-lJ-lk)
X--
or
X
(-3i-j~k)
4
+ 3k)
<
15VH
(b)
AxB
We have
X-
where
A
|AxB|
B
3i-3j + 2k
:-2i
+ 6j-4k
I
AxB
Then
k
J
3-3
-2
-4
6
(12-12)i
(8j
|AxB|
and
2
+ (-4 + l2)j + (18-6)k
+ 12k)
2
4^(2)
+(37 = 4^13
4(2j + 3k)
distribution to
(2j
X,
©
2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
you are using
or
V^
4^13
it
in
without permission.
170
PROBLEM
3.19
Determine the moment about the origin O of the force F = 4i + 5j - 3 k that acts at a Point A. Assume that the
positioji vector of A is {a) r - 2i - 3j + 4k, (£) r - 2i + 2.5j - 1 .5k, (c) r - 2i + 5j + 6k.
SOLUTION
M,
(a)
k
J
'
2-3
4
4
-3
5
(9-20)i + (16 +
M,
(h)
i
j
k
2
2.5
-1.5
4
5
(-7.5
M,
(<0
'
J
2
5
4
5
(-1 5
6)j
+ (I0 + 12)k
-:s
+ 7.5)i + (-6 + 6) j + (10-1 0)k
- 30)i + (24 + 6)j + (10- 20)k
Note: The answer to Part b could have been anticipated since the elements of the
determinant are proportional.
M,
M,
last,
-45i + 30j-10k
«
A
two rows of the
to teachers and educators permitted by McGraw-Hillfor their individual course preparation, if you are a student using this Manual,
distribution
M a = -lli + 22j + 22k ^
you are using
it
in
without permission.
171
PROBLEM
3.20
of the force F == -2i + 3j + 5k that acts
Determine the moment about the origin
position vector of A is (a) r = = i + j + k,(6)r = = 2i + 3j- 5k, (c)r == -4i + 6j + 10k.
at a
Point A.
Assume
that the
SOLUTION
M
(a)
(}
=
i
j
k
1
1
1
-2
3
5
M
= (5-3)i + (-2-5)j + (3 + 2)k
i
M
(h)
- 2
-2
j
k
3
-f
3
5
J
k
Ma
(c)
- -4
6
10
-2
3
5
M
= (30 - 30)i + (-20 + 20) j + (-.12 + 1 2)k
Note: The answer to Part c could have been anticipated since the elements of the
A
M o =30i + I2k A
= (15 + 15)i + (10-10)j + (6 + 6)k
1
= 2i-7j + 5k
last
=
<
two rows of the
determinant are proportional.
©
distribution to teachers and educators permitted. by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
in
172
PROBLEM
-VI
200
3.21
N
A
r
200-N force
the
applied as
is
moment of the
shown
to the bracket
ABC. Determine
force about A.
SOLUTION
We have
M,
where
rCIA = (0.06 m)i
=
%.i
x *c
+ (0.075 m)j
*c = ~(200 N)cos 30°j + (200 N)sin 30°k
i
Then
MA
=
-.
200 0.06
&
j
0.075
-cos 30°
=
200[(0.075sin 30°)i
sin 30°
- (0.06sin
or M,,
30°) j
~ (0.06 cos 30°)k]
= (7.50 N m)i- (6.00 N-m)j-(l 0.39 N-m)k
<4
PROPRIETARY MATERIAL <Q 201 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation.
Ifyou are a student using this Manual,
173
PROBLEM
'/
3.22
^-^
£
Before the trunk of a large tree
attached as shown.
Knowing
is felled,
cables
AB
that the tensions in cables
and
BC
are
AB
and
BC
N and 660 N, respectively, determine the moment about
of the resultant force exerted on the tree by the cables at B.
are 555
.
.'vjB
O
4.25 in
^J""
/
0.75
,.-"r
-^1
in/\
"I
.T
SOLUTION
We have
where
M,
1
ff/O
r
l3/Q
(7m)j
Tdff
'yf/f
X F#
+ Tec
~ ,VBA' AB
-(0.75m)i-(7m)J + (6m)k
(555
2
(.75)
l
J!C
2
+(7) +(6)
2
N)
m
*-BC*BC
(4.25m)i-(7m)j + (lm)k
-(660 N)
2
2
^(4.25) +(7) +(l)
2
m
F = [-(45.00 N)i - (420.0 N) j
-f-
i?
(360.0 N)k]
+[(340.0 N)i - (560.0 N)j + (80.00 M)k]
= (295.0 N)i - (980.0 N)j + (440.0 N)k
i
and
M,
k
J
Nm
7
295
(3080
980
N
•
440
m)i
- (2070 N m)k
•
©
or
M,} = (3080 N-m)i- (2070 N-m)k <
174
PROBLEM
3.23
f
i
in
^5
The 6-m boom
the free end
B
has a fixed end
of the boom
If the tension in the cable is 2.5
the force exerted
/*.
to a Point
by the cable
A steel cable is stretched from
C located on the vertical wall.
kN, determine the moment about A of
at B.
SOLUTION
2
2
:
First note
^c-VK>)
= 7.6 m
Then
+ 2.4j-4k)
V=-?4^(-«
7.6
We have
Â ~ r
/i//l
where
l
B/A
+(2.4) +(-4)
X TBC
(6 m)i
M;4 ^(6m.)ix^4r^-(~6» + 2.4j-4k)
Then
7.6
or
distribution to
=
(7.89
kN-m)j + (4.74 kN-m)k
<
you are using
M //=
it
in
without permission.
175
PROBLEM
3.24
A wooden board AB,
C- 48itii
3{ji)i..
which
a small roof, exerts at
used as a temporary prop to support
Point A of the roof a 57-lb force directed
along BA. Determine the
:'6-iftl
is
moment about
C of that force.
SOLUTION
We have
where
rA/c
and
\<
BA
= (48 in.)i - (6 in.)j + (36 in.)k
— xM p M
-(5in.)i
+ (90in.)j-(30in.)k
(57 lb)
2
2
V(5) +(90) + (30)
2
in.
= -(31b)i + (541b)j-(181b)k
Mc
i
J
k
48
6
36
3
54
18
-(1 836 lb
•
lb -in.
in.)i
+ (756 lb
or
in.)j
+ (2574 lb
M c = -(1 53
.0 lb
•
in.)
•
ft)i
+ (63.0 lb
•
ft)j
+ (2 1 5 lb
•
ft)k
A
any form or by any means, without the. prior written permission of the publisher, or used beyond the limited
distribution
•
you are using
it
in
without permission.
176
PROBLEM
3.25
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
«'
/.)
0.6 id
0.6
m "X
SOLUTION
(a)
M^r
We have
,xTM
iy
where
**/•/,(=
Tw: =
-
(2-3
m)j
*"DE^f.)E
+ (3.3in)J-(3m)k
(0.6m)l
2
2
V(0.6) +(3.3) +(3)
2
m
= (1 08 N)i + (594 N)j - (540 N)k
i
k
J
M,,=
Nm
2.3
-540
594
108
= -(1 242 N m)i - (248.4 N m)k
-
•
M
or
(b)
We
Â=r
have
G//l
where
/i
=-(1242N-m)i-(248N-m)k
<
xTCG
r(,,=(2.7m)i
+ (2.3m)j
T*CO — *"CG'CG
1
T
~~
=
-(.6m)i + (3.3m)j-(3m)k
2
2
V(.6) +(3.3) +(3)
2
(810N)
m
= -(108 N)i + (594 N)j - (540 N)k
MA =
k
i
J
2.7
2.3
-108
594
= -(1 242 N
•
N-m
-540
m)i
or
+ (1458 N m)j + (1 852 N m)k
•
M A = -(1242 N
©
•
•
m)i
+ (1 458 N m)j + (1 852 N m)k
•
•
A
you are using
it
without permission.
177
PROBLEM
A
3.26
smaJI boat hangs from two davits, one of which
The
figure.
tension in line
ABAD
is
82
C of the resultant force R,, exerted,
about
is
shown
in the
Determine the moment
the
davit at A.
on
lb.
SOLUTION
R.=2R
We have
.
(/)
where
,„~-(82
and
AD
I
AD
+ F, D
1b)j
AD
—
AD
= (82
6i~7.75j-3k
lb)
10.25
F, /3 =(48lb)i-(62 1b)j-(24 1b)k
R A = 2¥AB + FAD = (48 lb)i - (226
Thus
Also
l
A/C
(7.75ft)j
1
Using Eq.
Mc
(3.21):
48
(492
- (24
lb)k
+ (3ft)k
J
k
7.75
3
-226
-24
lb ft)i
•
lb)j
+ (1 44
lb
•
ft) j
- (372
M c = (492
©
lb ft)k
•
lb ft)i
•
+ (1 44.0
lb ft)j
- (372
lb ft)k
•
<
178
PROBLEM
In
3.27
Problem 3.22, determine the perpendicular distance from
O to cable AB.
Point
PROBLEM
3.22 Before the trunk of a large tree
is felled,
AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
cables
respectively, determine the
force exerted on the tree
moment about
by the cables
O of the resultant
at B.
SOLUTION
We have
IM
TBA d
d = perpendicular distance from O to line AB.
where
Now
M,
and
rBIO
={lm)\
*BA
~~
r
B/0
1
'"BA
BA'
X
*BA
AB
(0.75m)i-(7m)j + (6m)k
(555
2
2
+(7) +(6)
'(0.75)
-(45.0 N)i
- (420 N)j + (360 N)k
k
j
i
M,
N-m
7
-45
-420
(2520.0
N
|M
and
N)
m
2
•
360
m)i
1
+ (315.00 N m)k
•
= V( 252o -°) 2 + (31 5.00) ;
= 2539.6 N-m
2539.6
N-m = (555 N)rf
d = 4.5759 m
or
PROPRIETARY MATERIAL. © 2010
d = 4.58 m
<
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distribution to
or
in
179
PROBLEM
In
3,28
O to cable BC.
Point
PROBLEM
3.22 Before the trunk of a large tree
A B and BC are
cables
tensions in cables
AB
and
respectively, determine the
force exerted
on
the tree
BC
are 555
moment about
by the cables
is felled,
Knowing
attached as shown.
N
that the
and 660 N,
O of the resultant
at. #.
SOLUTION
We have
[M,
Tscd
d — perpendicular distance from O
where
M,
l
B/0
P
'BC
rB/0
to line
BC.
X ^liC
7mj
—
1
T
™ *"BC*
HC
(4.25m)i-(7m)j + (lm)k
2
2
V(4.25) +(7) +(l)
(340 N)i
I
(660 N)
m
- (560 N)j + (80 N)k
k
J
M,
2
7
340
(560
-560
N
•
m)i - (2380
|M
and
80
!
N m)k
•
= 7(560) 2 +(2380)
= 2445.0
2445.0
N-m
N-m = (660 N)rf
d = 3.7045 m
or
d~ 3.70 m
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you are using
it
without permission.
180
PROBLEM
A
."(i
Problem 3.24, determine the perpendicular distance from Point
drawn through Points A and B.
In
•IS in.
in.
3.29
D
to a line
Cj-iii.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-lb
force directed along BA. Determine the moment about C of that
force.
.90
()(>
B
~~*S
in,:
in.
~~^~£
SOLUTION
IMJ^rf
We have
d — perpendicular distance from D
where
M. l}
AB.
^rAlii x¥]iA
r,/D =-(6in.)j
^liA
to line
+ (36in.)k
~ ^BA^BA
(-(5in.)i
+ (90in.)j-(30
in.)k)
(57 lb)
2
V(5) +(90)
-(31b)i
I
M,
;)
-3
2
+ (30)
in.
+ (541b)j-(181b)k
J
k
-6
36
54
-18
lb -in.
-(1 836.00 lb in.)i
•
and
2
- (1 08.000
IM
lb in.) j
836.00)
•
2
- (1 8.0000 lb
+(108.000)
2
•
in.)k
+ (18.0000)'
= 1839.26 lb -in.
1839.26 lb -in =(57
</
lb)e/
= 32.268 in.
PROPRIETARY MATERIAL. &
</
= 32.3in.
<
distribution
or
you are using
it
in
without permission.
181
PROBLEM
3.30
Problem 3.24, determine the perpendicular distance from Point
line drawn through Points A and B.
In
PROBLEM
3.24
A
wooden board AB, which
prop to support a small roof, exerts at Point
directed along
(if)
C to a
A
is used as a temporary
of the roof a 57-3 b force
BA. Determine the moment about
C of that force.
in.
B^-<
SOLUTION
4
We have
\M c \~FBA d
3fc..
d - perpendicular distance from C
where
AB.
M
^^\
^%f
M. c
^rA/c x¥BA
rAIC
- (48 in.)i - (6 in,)j + (36 in.)k
fylA
to line
(~(5
D
/
\
~ ^BAÎIA
_
.
^-.
>l
// S^
f*
in.)i
+ (90 in.) j - (30 in.)k)
2
2
7(5) +(90) +(30)
fe*
2
in.
= ™(3 lb)i + (54 lb) j - (1 8 lb)k
J
k
M c = 48
-6
36
-3
54
-18
i
lb -in.
= -(1 836.001b -i n.)i - (756.00 lb
•
in.)j
+ (2574.0 lb
•
in.)k
2
2
2
|M c = V(l 836.00) + (756.00) + (2574.0)
and
|
-3250.8 lb -in.
3250.8 lb
in.
= 57
lb
d- 57.032 in.
or
d=
57.0
in.
^
2010 The McGraw-Hill Companies, Inc. All rights reserved. ;Vt> />flrf o//Afc Manual may be displayed,
you are using
it
without permission.
182
—
PROBLEM
In
3.31
DE of cable DEF.
Point A to portion
PROBLEM
comers
3.25
C and D,
Determine the
The ramp ABCD is supported by cables at
The tension in each of the cables is 810 N.
moment about A of the force exerted by (a)
cable at D, (b) the cable at C.
0.6
the
.
m
SOLUTION
We have
TDE d
IM
d
where
perpendicular distance from A to line DE,
M
r„
HIA
'DE
(2.3
m)j
~ A"DE*-DE
(0.6m)i + (3.3m)j-(3m)k
2
V(0.6)
(108 N)i
i
O'fcn,
+ (594 N)j - (540 N)k
b
J
M
N-m
2.3
108
~(1
(8I0N)
+ (3.3) 2 +(3) 2 m
594
242.00
540
N
and
•
m)i - (248.00
|MJ
N m)k
•
=- /(1242.00) 2 +(248.00) :
N
= 1266.52 N-m
1266.52
Nm =
(8 10
N)d
d = .56360
1
m
PROPRIETARY MATERIAL. © 2010 The
or
d = 1.564 m
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183
PROBLEM
In
3.32
C and G.
Point A to a line drawn through Points
PROBLEM 3.25 The ramp ABCD is supported by cables at
corners C and D. The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (h) the cable at C.
0.6
m
SOLUTION
We have
d - perpendicular distance from A
where
M
rG/A
IVu/A
(2.7m)i + (2.3m)j
l
X
to line
CG.
Ice
CG
-(0.6.m)i + (3.3m)j-(3
2
+(3.3)
A/(0.6)
2
+ (3)
2
m)k
(810 N)
m
-(108N)i + (594N)j-(540N)k
M
k
i
J
2.7
2.3
-108
594
-(1242.00
N-m
-540
N
•
m)i + (1458.00
N
•
m)j +(1 852.00
N m)k
and
= 2664.3 N-m
2664.3
N-m = (8 ION) d
d = 3.2893
m
distribution to
d = 3.29 m 4
teachers and educators permitted by McGraw-Hill Jbr their individual course preparation. Ifyou are a student using this Manual,
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or
it
in
without permission.
184
PROBLEM
.In
3.33
C to portion AD of the line ABAD.
Point
PROBLEM 3.26 A small
boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
R4 exerted on the davit at A.
SOLUTION
First
compute the moment about
From Problem
C of the force FDA
exerted by the line on D:
3.26:
= -(48
lb)i
+ (62
lb) j
+ (24
lb)k
lb)i
+ (62
lb)j
M c =rm; xF^
= +(6
x[-(48
ft)i
= -(1441b-ft)j + (372
M c =V0 44
2
>
= 398.90
Then
M c = VM d
Since
F,w
= 82
+(372)
lb
•
lb)k]
1b-ft)k
2
ft
1b
398.90
82
lb -ft
d = 4.86
ft
<
1b
distribution
+ (24
yon are using
it
in
without permission.
185
*-*
PROBLEM
3.34
Determine the value of a
16
that
minimizes the
C to a section
of pipeline that passes through Points A and B.
perpendicular distance from Point
ft
i
SOLUTION
Assuming a
force
F acts
along AB,
\M. c
MrA/c xF\^F(d)
d - perpendicular distance from
Where
W
C to line AB
F = %AB F
(24ft)i + (24ft)j-(28)k
•F
2
2
7(24) +(24) +(18)
(6)i
AK;
ft) j
i
J
k
3
-10
10a
6
.[(10
ft
+ (6)j-(7)k
= (3 ft)i - (1
Mc
2
- {a -
1
ft)k
6-7
+ 6fl)i + (81-6.ai)j + 78k]
11
Since
^/c
xr
r^xF
or
2
|
=
W
+ 6a) 2 +(81- 6a) 1 + (78) 2 = d*
(1
12!
d / j2
Setting -j-{d
)-
to find
a
to
minimize
c/
1
[2(6)(l
+ 6a) + 2(-6)(8 - 6a)] =
1
121
a
Solving
= 5.92 ft
or
- 5.92
ft
^
© 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo par/ o//Aw Minna/ »m>' fc displayed
distribution
o
you are using
it
in
without permission.
186
PROBLEM
3.35
Given the vectors P =3i --J +
S, and Q S.
P
2k,Q
==
4i
+ 5j-3k, and S = -2i + 3j - k, compute
the scalar products
P Q,
•
•
SOLUTION
P-Q = (31-1j + 2k)-(4i-5J-3k)
= (3)(4) + (-l)(~5) + (2)(-3)
=1
P S = (3i - lj + 2k)
•
or
•
(-2i
P
Q=
I
A
+ 3j - Ik)
= (3)(-2) + H)(3) + (2)H)
= -U.
or
P-S = -U
<
Q-S-(4i-5j~3k)-(-2i + 3j-l.k)
= (4K-2) + (5)(3) + (-3X-l)
= 10
or
to teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,
distribution
Q-S = 10 <
you are using
it
in
without permission.
187
PROBLEM
Form
3.36
B C and B' C, where B =
prove the identity
the scalar products
results obtained to
cos a cos
•
•
B',
and use the
P ~ — cos (a + /?) +— cos (a - ft).
SOLUTION
y
B-C = £Ccos(a~/?)
By definition
B = /?[(cos/?)i + (sin/?)j]
where
C = C[(cos a)\ + (sin a)j]
(B cos /?)(Ccos a) + (B sin /?)(Csin a) cos /?cos
or
By definition
B'
•
a+ sin /?sin a = cos(a - j3)
(B cos p) (C cos
+ (~B sin
cos /?cos
or
Adding Equations
or)
(1)
and
(1)
C - BC cos (or + /?)
B' = [(cos
where
BCcos(a~ (3)
fi)i
~ (sin /?)j]
/?)(C sin
or)
= BC cos (a + /?)
« - sin ft sin a = cos {a + /?)
(2)
(2),
2 cos /?cos
a = cos (a-fi) + cos (a + /?)
or
cosacosfi = ~-cos(a + fi) + ~cos(a~ ft)
A
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188
PROBLEM
3,37
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First
AB = AB{$m 37°j - cos 37°k)
note
CD = CD(- cos 40° cos 55°j + sin 40°j- cos 40° sin 55°k)
fc.b\
Now
AB- CD = (AB)(CD) cos
or
AB(sm
37°j - cos 37°k)
•
CD(-cos 40° cos 55°i + sin 40°j - cos 40°sin 55°k)
~(AB)(CD) cos
or
cose?
= (sin 37°)(sin 40°) + (~cos
37°)(--cos 4()°sin 55°)
= 0.88799
or
(9
= 27.4°
A
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189
:
PROBLEM
3.38
Section AB of a pipeline
lies in the yz plane and forms an angle
37°
with the z axis. Branch lines CD and EF join AB as
of
shown. Determine the angle formed by pipes AB and EF.
SOLUTION
AB
First note
= /f£?(sm37
j-cos37
k)
__.
= £F(cos 32° cos 45°i + sin. 32° j - cos 32° sin 45°k)
F
1
£
l/^ (€t\ *W
Now
A
AB-EF--= (AB)(EF)cosO
AB(sm
or
17° j
- cos 37°k) £F(cos 32° cos 45° j + sin 32° j - cos 32° sin 45°k)
-
= (AB)(EF)cos0
cos
or
= (sin
37°)(s.in
32°) + (-cos 37°)(-cos 32° sin 45°)
= 0.79782
or
©
(9
= 37.1° ^
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it
without permission.
1
90
PROBLEM
3.39
Consider the volleyball net shown.
Determine the angle formed by guy
wires AB and
AC,
SOLUTION
First
note
AB - V(-6.5) 2 + (-8) 2 + (2) 2 =
AC^yj(0f+(~S) 2 + (6) 2 =]0
and
AB = -(6.5
ft)i
- (8
ft)j
+ (2
1
0.5
ft
ft
ft)k
7c = -(8ft)j + (6ft)k
By definition
AB-AC = (ABXAC)cos0
or
(-6,51
- 8 j + 2k) (-8 j + 6k) = (1 0.5)0 0) cos
•
-
(-6.5)(0) + (-8)(-8) + (2)(6) - 1 05 cos 6
or
cos
= 0.72381
or
= 43.6°
<
191
PROBLEM
3.40
Consider the volleyball net shown.
Determine the angle formed by guy
wires
AC and AD.
sit
SOLUTION
^C = V(0) 2
First note
-10
2
+(~8) +(6)
2
ft
AD = J(4) 2 +(-&)2 + (l?
=9
IC = ~(8ft)j + (6ft)k
75 = (4ft)j-(8ft)j + (lft)k
and
By
ft
AC AD = (AC)(AD)cos0
definition
(-8 j + 6k) (4i - 8 j + k) =
or
•
(0)(4)
t
+ (-8)(-8) + (6)(1) = 90 cos 6
or
# = 38.9°
<
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distribution to
0)(9) cos
cos0 = 0.77778
or
you are using
(1
it
in
without permission.
192
PROBLEM
3.41
2.-1
AC is 1260 N, determine
between cable AC and the boom AB, (b) the
projection on AB of the force exerted by cable AC at Points.
Knowing
1.2 in
that the tension in cable
(a) the angle
-f
2.6 in
2.4
m
SOLUTION
(a)
First
note
AC~-=^j(~2A)
2
+ (0.8) 2
+(1.2)
2
= 2.8m
AB = yj(-2 A) 2 +(-].&?
+(Q)
2
= 3.0m
^C = -(2.4 m)i + (0.8
and
I/?
= -(2.4m)i-(K8m)j
AC -AB = (AC){AB) cos
By definition
or
(-2.41
or
+ 0.8 j +
1
.2k) (-2.4i
•
-
1
.8 j)
= (2.8)(30) x cos
(-24X-2.4) + (0.8X-1 .8) + (1 .2X0) - 8.4cos
cos# = 0.51429
or
(b)
m) j + (1 .2 m)k
or
= 59.O C
We have
= 7^ c cos<9
= (1260N)(0.51429)
or
C^cU=648N
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193
PROBLEM
2.4
3.42
m
Knowing
AD
and the boom AB, (b) the projection on
of the force exerted by cable AD at Point A.
angle between cable
AB
AD is 405 N, determine (a) the
SOLUTION
{a)
AD = >/(-2.4) 2 + (i ,2) 2 + (-2.4) 2
First note
= 3.6m
AB = V(-2.4) 2 + (-1 .8) 2 + (0) 2
-3.0m
AD = -(2.4 m)i + (1 .2 m)j - (2.4 m)k
AB = -(2.4m)i~(1.8m)j
and
ADAB = {AD)(AB)cos0
By definition,
(-2.4i
+
1
.2j
(-2.4)(-2.4)
- 2.4k)
4- (1
(-2.4i
.2)(-I.8)
-
.1
.8j)
= (3.6)(3.0)cos#
+ (-2.4X0) = 10.8 cos
cos $ — —
3
(b)
(* AD'AB
~
^^70.5°
<
*AD Âli
'
= TAD cos&
= (405N)^
Pad)ab = 135.0
N <
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194
PROBLEM
3.43
Slider P can move along rod OA. An
elastic cord
is
attached to the slider and to the vertical member
BC. Knowing
that the distance from
to P is 6 in. and that the tension
PC
O
cord
is
3
in the
determine (a) the angle between the elastic cord
lb,
and the rod OA, (b) the
by cord PC at Point P.
projection
on
OA
of the force exerted
SOLUTION
^=
First note
2
OA
»
Then
2
+(12) +(-6)
>/(12)
1
2
=18in.
,
= i(2i + 2j~k)
Now
OP = 6
The coordinates of Point
P are (4
4
in.,
PC = (5
so that
and
(a)
in.,
in,
-2
in.)i
or
in.)
+ (1
PC = 7(5) 2 +
We have
OP = -(04)
=>
in..)j
.1
2
(1 1)
+ (14
in.)k
+ (14) 2 = V342
in.
~PCX OA ^(PC)cosO
(5i
+ llj + 14k)--(2i + 2 i-k) = 7342cos^
(
or
]
cos
[(5)(2)
+ (ll)(2) + (14)(~l)j
3V342
0.32444
= 71.1°
^
or (Tpc)oa =0-973 lb
^
or
(b)
We have
= {Tpc 'K ]>c )-X OA
PC
ft
pc
OA
— Tpr COS &
= (3
1b)(0.32444)
=^^S~S5£=SS£=f£S
195
PROBLEM
P
Slider
can
3.44
move
along rod OA.
An
elastic cord
attached to the slider and to the vertical
Determine the distance from
rod
OA
O
to
P
for
PC
is
member BC.
which cord
PC and
are perpendicular.
SOLUTION
2
2
(M^/(12) +(12) +(-6)- =18
First note
~
X OA
CH
Then
OA
OA
—
(2i
Let the coordinates of Point
P be
in.
1
(12i
+ 12j-6k)
18
+ 2j-k)
(x in., j> in., z in.).
Then
PC = [(9 - x)in.]i + (15- y)in.] j + [(12- z)in.]k
Also,
OP = rfoô, = -^(21 + 2J-k)
and
OP ~ (x in.)i + (>'
2
The requirement
that
CM
in.)j
+ (z
in.)k
2
dOP
and PC" be perpendicular implies
that
^PC-0
-(2j
or
+ 2j-k)-[(9-.v)i + (15-y)j + (12-2)k] =
(2)|9--^>| + (2) 'l5-|rfw
or
|
+
H)
12
-rf0P
or
W
Jnp = 12.00
in.
^
displayed,
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permission of the publisher, or used beyond the limited
reproduced or distributed in anvform or by any means, without the prior written
course preparation. Ifyou are a student using tins Manual,
yon are using
it
without permission.
196
PROBLEM
3.45
Determine the volume of the parallelepiped of Fig. 3.25 when
(a) P = 4i - 3j + 2k,
= -2i - 5j + k, and S = 7i + j - k,
(/;) P = 5i - j + 6k, Q = 2i + 3j + k and S = -3i - 2j + 4k.
Q
5
SOLUTION
Volume of a
parallelepiped
O)
is
found using the mixed
VoI
triple product.
= P-(QxS)
4-3
2
-2
-5
1
7
1
3
in.
-1
(20-21.-4 + 70 + 6-4)
67
or
Volume =67.0
4
Volunie = 111.0
A
Vol=P-(QxS)
(b)
5
-1
6
2
3
I
-3
-2
4
(60 +
3
in.
3-24 + 54 + 8+10)
111
or
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197
PROBLEM
3.46
Given the vectors
P - 4i - 2 j + 3k,Q = 2i + 4j - 5k,
and S
= SJ - j + 2k, determine the value of Sx
for
which
the three vectors are coplanar.
SOLUTION
If P,
Q, and S are coplanar, then P must be perpendicular
to
(Qx S).
P-(QxS) = ()
(or, the
volume of a parallelepiped defined by P, Q, and S
4-2
Then
Sx
-1
zero).
3
-5-0
2
-6-20 + 8-125;. =0
S =7
<
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any form or bv any means, without the prior written permission of the publisher, or used beyond the limited
teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a
distribution to
4
32 + lO.S:
or
you are using
2
is
it
©
in
without permission.
198
]
PROBLEM
3.47
The 0.61x].00-m
along side AB and
hook
frictionless
the force exerted
1
ABCD
of a storage
'bin
held open by looping cord
hinged
is
DEC over a
at E. If the tension in the cord is
moment about each of the
determine the
0.1
lid
is
by the cord
at
66 N,
coordinate axes of
D.
in...
SOLUTION
First
note
z
= Vi°- 61 ) 2 -(0.Jl) 2
Then
dm =
=
and
1
0.11m
m
0.60
2
2
V(0.3) +(0.6) +(-0.6r
0.9
m
66
N
(0.3i
l)E
+ 0.6j-0.6k)
0.9
= 22[(lN)i + (2N)j-(2N)k]
Now
MA = r
where
r/)//(
D/A
xTDf:
= (0.ll m)j + (0.60m)k
J
k
0.11
0.60
2
-2
i
M
Then
/(
= 220
1
=
~ 1 .20)i + 0.60 j - 0. 1 k
N m)i + (1 3.20 N m) j - (2.42 N m)k
22[(-0.22
= - (3
1
.24
1
•
•
•
-3
PROPRIETARY MATERIAL ©
1
.2
N
•
m,
M
= 1 3.20 N m,
•
v
A/ 2
= -2.42 N m
-
A
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it
1
in
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199
PROBLEM
3.48
lid ABCD of a storage bin is hinged
and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
The 0.61xl.00-m
along side
AB
the force exerted
by
the cord at C.
SOLUTION
x6\y-(o.uy
First note
0.60
m
Then
= l.lm
66
and
l
N
(-0.7i
CE
+ 0.6j-0.6k)
1.1
= 6[-(7N)i + (6N)j-(6N)k]
Now
M A ^vm x\E
where
r£M
=
(0.3m)I + (0.71m)J
i
M A -6 0.3
Then
-7
=
k
J
0.71
-6
6
6[-4.26i +
.8 j
1
+ (1 .8 + 4.97)k]
= - (25.56 N m)i + (1 0.80 N m)j + (40.62
•
•
M
- -25.6 N m,
•
x
M
N m)k
•
= 1 0.80 N m, Afz = 40.6 N
•
y
•
m
-4
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are a student using this Manual,
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©
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it
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200
PROBLEM
3,49
To
lift a heavy crate, a man uses a block
and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the
y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N
and -460 N m, determine the distance a.
•
m
•
SOLUTION
^ = (2.2m)i-(3.2 m)j-(am)k
First note
Now
™*D ~ VAID X
where
(2.2m}i + (1.6m)j
'AID
T,BA
T
A
I!
Then
'«/j
d ISA
T,BA
M,
(2.2i-3.2j-ak)(N)
J
2.2
1.6
2.2
-3.2
dBA
T,BA
{- .6a \
1
d
M
Thus
v
k
i
-a
+ 22a\ + [(2.2)(~3 .2) - (1 .6)(2.2)]k}
t.
=2.2-^-a
(N m)
•
BA
M,
-I0.56-
7 *4
(N m)
•
d HA
Then forming the
ratio
it
M.
N-m
-460 N-m
120
2 2 7Z
dB,t"(N-m)
-
or
-10.56-^-
(N-m)
a=
1
.252
m
^j
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part of ,his Manual may be displayed
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for their individual course preparation. Ifyou are a student using this Manual
201
PROBLEM
3.50
lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the man applies a 195-N
force to end A of the rope and that the moment of that force about they
axis is 132 N m, determine the distance a.
To
•
SOLUTION
2
dBA - J(22f+(-3.2) + (-af
First note
= Vl5.08 + « 2 m
195
N
and
Ts ,=^^- -(2.2i-3.2j-ok)
Now
My ~\ -(rA/D xTw
where
Vf/0
!
(2.2m)i
)
+ (1.6m)j
1
M,
Then
195
2.2
1.6
2.2
-3.2
195
(2.2a)
-a
(N m)
•
d HA
Substituting for
My and d
HA
t
132
N-m
195
(2.2a)
J\Jm+a*
0.30769^1 5.08 + a* - a
or
Squaring both sides of the equation
0.094675(15.08 + a
2
)
= a2
or
you are using
m A
displayed,
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using this Manual,
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distribution to
a = 1.256
it
in
without permission.
202
PROBLEM
A
3.51
small boat hangs from two davits, one of which
the figure.
It is
resultant force
279
lb
•
ft
known
R
t
exerted
in absolute value.
tension in line
is
shown
in
moment about the z axis of the
on the davit at A must not exceed
that the
ABAD when
Determine the largest allowable
x ~ 6 ft.
SOLUTION
R
First note
Also note
that only
TAD will
21^ +TAD
contribute to the
moment about the z axis.
Now
= 10.25
Then,
'AD
T
ft
AD
AD
r
(6i-7.75j-3k)
10.25
Now
M..
where
y
AIC
(7.75ft)j
+ (3ft)k
1
Then
7\
for
279
71.
7,75
3
-7.75
-3
10.25
6
~^H-(1)(7.75)(6)|
A ;I
10.25'
or
7"
=61.5
lb
<
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it
in
without permission.
203
PROBLEM
3.52
For the davit of Problem 3.51, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From
the solution
of Problem 3.51,
TAD is now
AD
l
AI)
AD
60
lb
(xi-7.75j-3k)
2
2
yj X +(-lJS) +(-lf
Then
M
=k
•
(rA/c
xT,(0 ) becomes
1
60
7.75
3
-7.75
-3
279
2
V*
2
+(-7.75) +(-3)'
jc
60
279.
(D(7.75)(x)
2
six
279>/x
0.6Vjc
Squaring both
2
+ 69^0625 = 465.x
2
+ 69.0625
=x
0.36x
sides:
+69.0625
2
+ 24.8625 = r
1
x =38.848
distribution to
= 6.23
ft
4
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x
it
in
without permission.
204
PROBLEM
To
3,53
loosen a frozen valve, a force
F of magnitude 70 lb is
Knowing that 6 = 25°,
applied to the handle of the valve.
Mx
-6.1 lb
•
M
and
ft,
z
~ -43 lb
ft,
•
determine
(j)
and
d.
SOLUTION
We have
rivi
(
where
VAIO
'
F(cos #eos fi - sin $\ + cos #sin
F
70
Pol-
F
+ (llin.)j-(rf)k
-(4in.)i
-
1b,
9 = 25°
(70 lb)[(0.9063 1 cos 0)i - 0.42262 j + (0.9063 1 sin 0)k]
j
k
11
-J
i
ML:
-4
(701b)
-0.90631 cos
(70 lb)[(9.9694sin
-0.42262
M.
(70 lb)(9.9694sin
M
(70 lb)(-0.9063 Irfcos
M.
(70
y
From Equation
(3)
^ = cos
lb)(l .69048
0.90631 sin
- 0.42262 d)m. = -(61
(j)
+ 3.6252 sin
24.636'
+ 3.6252sin
<f>)
lb ft)(I2 in./ft)
•
d
(2)
•
ft(l
2
in./ft)
(3)
or
(p
1022.90
=
34.577
29.583
in.
(1)
in.
697.86
From Equation (1)
<p) j
in.
- 9.9694cos ^ in. - -43 lb
U 634.33
in.
- 0.42262^)1 + (-0.9063 Wcos
+ (1.69048 -9.9694 cos 0)k]
and
<J)k)
or
c/
= 24.6°
<
= 34.6in. ^
PROPRIETARY MATERIAL. €5 2010 The McGraw-Hill Companies, inc. All righis reserved. No part of this Manual may be displayed,
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it
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205
—
PROBLEM
3.54
When
F
a force
is
applied to the handle of the valve shown,
moments about the x and z axes
and M.z - -81 lb ft. For
F about the y axis.
•
i I
in.
are, respectively,
d~21 in.,
M
determine the
x
~ -11 lb
moment
its
•
ft
My of
SOLUTION
w xF = M
We have
ZM
Where
r^=-(4in.)i + (lim.)j-(27in.)k
:
r
(
F = F(cos 0cos <p\ - sin. $\ + cos
i
M a =F
sin flk)
k
J
-4
1
cos 0cos
F [(11 cos
-27
.1
-sin 6
(p
sin
0-27 sin
+ (4 sin
M
M
and
x
J?
M
z
Now, Equation (1)
cos #sin
0)i
cos 0cos
1
cos
^)k](lb
= F(l. 1 cos (9 sin 0-27 sin 0)(lb
•
0)j
•
in.)
0)
in.)
= F(-21 cos 0cos0 + 4cos 0sin 0) (lb
= F(4 sin <9 -
M^
<f>11
and Equation (3)
1
6* cos
F
1
.1
cos
in.
cos #sin
+ (-27 cos Bcos <p+ 4 cos #sin
-
lb
cos (j)) (lb
•
•
(2)
in.)
(3)
in.)
+ 27sin#
0- — 4sin
n{
—
(4)
-
(5)
F
Substituting Equations (4) and (5) into Equation (2),
M, =/N-27
F
+4
(
111
M
*-
+ 27sin0
F
(27M Z +4.M V )
M.
or
1
4sin0-
11
&
206
PROBLEM
Noting
that the ratios
yp
and
3.54 (Continued)
± are the ratios of lengths, have
M =— (-81
v
-
lb -ft)
+—(-77 lb -ft)
11
11
= 226.82 lb
•
or
ft
©
M
v
-=
-227
lb
-
ft
A
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207
PROBLEM
3.55
0.35 in
The frame
ACD
is
hinged
at
A and D and
cable that passes through a ring at B and
is
is
supported by a
attached to hooks
G
and H. Knowing that the tension in the cable is 450 N,
moment about the diagonal AD of the force
exerted on the frame by portion BH of the cable.
at
determine the
0.75
m
*
0.75 in
SOLUTION
M AD ~ "'AD
'
\ VBIA
X
*Bll)
X D ---(4i-3k)
Where
„,
(0.5
'a/a
m)i
2
2
d m - vv 0.375) 2 +(0.75) +(-0.75)
and
= 1.125
450
Then
T.SH
m
N
— (0.375i + 0.751J - 0,75k)
1.1.25
(1
50 N)i + (300 N) j - (300 N)k
-3
4
|
Hu> =
Finally
0.5
150
!
300
-300;
[(-3X0.5X300)]
or
©
MAD
90.0
N-m
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it
without permission.
208
PROBLEM
0.35
3.56
m
Problem 3.55, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
In
0.75
m
SOLUTION
mad=âd-(*biaXTbg)
Where
M/)
r»
MA
=-(4i-3k)
(0.5
m)j
BG - V(-°- 5 ) + (0.925)
and
= 1.125
T
Then
liC!
2
+ (-0.4)
2
ra
450 N
=-(-0.5i + 0.925j-0.4k)
J
'
1.125
= -(200 N)i + (370 N) j - (1 60 N)k
-3
4
u
0.5
»'i -200
Finally
370
-160
[(-3X0.5X370)]
M AD
-Ul.ON-m
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preparation. Ifyou are a student using this Manual,
©
209
PROBLEM
The
at
0.7
3.57
triangular plate
B
D and
and
is
ABC
supported by ball-and-socket joints
is
held in the position
shown by
cables
AE and
CF. If the force exerted by cable AE at A is 55 N, determine the
moment of that force about the line joining Points and B.
m
D
.O.G.Ti]
SOLUTION
First
note
4E
= V(°- 9 ) + (~0-6) + (0-2)
2
=1.1
m
«=-yp(0.9i-0.6J + 0.2k)
Then
= 5[(9N)i-(6N)j + (2N)k]
DB - y[(\ 2f + (-0.35) 2 + (0) 2
Also
1.25m
X OB
Then
DB
DB
(1.2I-0.35J)
1.25
1
- 7 j)
(241
25
M DB ~ '"DB
"*DB
Now
'\ V
AID
X T Ui )
TO/} =-(0.1m)j + (0.2m)k
where
24
M»-^(5)
-7
!
Then
9
-0.1
0.2
-6
2
H1.8- 12.6 + 28.8)
or
M
Dfl
=2.28N-m ^
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it
without permission.
210
PROBLEM
The
3.58
triangular plate
ABC is
D and is held
supported by ball-and-socket joints
shown by cables AE and
CF. If the force exerted by cable CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
at
B and
in the position
SOLUTION
dCF
First note
Then
\:f
DB
Also
Then
^DB
=
--
2
V(0.6) +(-0.9)
2
+ (- 0.2) 2 =
1.1
m
-_33 N (0.6i-0.9j + 0.2k)
'-
:
=
3[(6N)i~(9N)j-(2N)k]
=
V0-2) +(-0.35) +(0)
2
~
1
_
'
DB
DB
~~
.25
=—
2
2
m
(I.2i- 0.35 j)
J
1.25
=
_L(24i-7j)
V
J;
25
Mm
Now
where
yan
-A'OB
:
'( r
C/D
Mm
*cp)
-(0.2m)j-(0.4m)k
24
Then
X
= -(3)
25
6
-7
0.2
-0.4
-9
-2
= —(-9.6 + 16.8-86.4)
25
or
©
Mm ^~9.50N-m <
any farm or by any means, without the prior written permission of the publisher, or used beyond the limited
you are using
it
in
without permission.
211
PROBLEM
A
3.59
regular tetrahedron has six edges of length a.
shown along edge BC. Determine
about edge OA.
directed as
l>
the
A
force
P
is
moment of P
>-
SOLUTION
We have
MoA=*OA<*aoX*)
where
From
triangle
OBC
{OA) x
2
(OA) z =(OA) x tan
f
W
\
i
n/3,/
vv1
2
(OA) =(OA)l +(OA)l
Since
2
+{OAz f
(
a
[
+ {OA)l +
or
2>/3
a
Y
2
a
Then
4,0
a
.
h
2>/3
2.
J.
1
- = l ,+ VlJ+ ivr k
x
and
12
,
2
=
p=
,
( a sin30°)i-(
a co S 30°)k
(/))
«
rc/0
=
P
(|
2
_
^
=oi
1
M
V3
2
2V3
Kf)
(A!
I
-73"
]
T
'
2
v
-
3
0X->/3) =
aP
Moa =
aP_
72"
y
©
20 JO The McGraw-Hill Companies, Inc. All rights reserved. JVo /wirt o///h,s Manual may be displayed,
212
PROBLEM
A
3.60
Show that two
and BC, are perpendicular to each
other, (b) Use this property and the result obtained in Problem 3.59
to determine the perpendicular distance between edges OA and BC.
regular tetrahedron has six edges of length a. (a)
opposite edges, such as
OA
SOLUTION
(a)
For edge
OA
to
be perpendicular to edge BC\
OA-BC = Q
where
From
triangle
OBC
(OA) x
>
(
(OA) z =(OA) x tm\30° =
i
\
-
V3
04 = I-|I + (CM)J +
2^3
J
a
k
\2Sj
BC = (asm 30°) - (a cos 30°) k
and
i
2
2
a
Then
2
c
a
.
\
(i-V3k)~ =
i+(.oa) v i+
^4 + (O^) (0)-~ =
or
}v
4
OA-BC^0
so that
(6)
<9/i is
Have A^ fW =
From the
Pt/,
results
with
perpendicular to #C.
P acting along BC and d the perpendicular distance from OA to
5C.
of Problem 3.57
M0A
Pa
ft?
4i
~Pd
or
4i
d = —T^
72
A
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in
213
:
y
PROBLEM
y^ji
\
A
45 in.
sign erected
on uneven ground
guyed by cables
4 V
11
si
8»v
determine
?;"" .--..
jm^M^^^W%
EF
by cable
force exerted
_^#g&
.^-^.,-f
/,:
96 ii r.
3.61
the
is
and EG. If the
£7** at
moment
E is 46
of that
about the line join ing Poi nts
A and
lb,
force
D.
=
;|*:
i|
fPI
1
\
r
'/
.
Ml
L#j
>'
M^:
^îÎbr >>V
47
-
in
l:-;iy';!t'^
-
.--•'•
••
.
-'•
,"-
-/••-
' :>-^
'"
':
:
:
'
<*><"^
'
.
"
.
"^v
"
~-J
\^Xl7in.&%8>
iftô^
.--^
i
•
v-ii:—;T*V
->'''
T.
•
-'*'"- I'-i i"»?
»*"
^' fT
/^
i
"''
1
'
"x*''
\
SOLUTION
First note that
EC ~-V(48) 2 + (36) 2 - 60
then(fx48,96, J> 36)
or
(36
in.,
96
in.
and
that
J§ = -g = |. The coordinates
27
in,).
Then
in,,
of Point
E are
2
2
2
^,=>/H5) + H10) +(30)
= 115
in.
T^==^(-l5I-110J + 30k)
Then
=
2[-(3 lb)i
- (22
lb)j
+ (6
lb)k]
AD = 7C48) 2 + (-12) 2 + (36) 2
Also
= 3.2726
in.
_
Then
=
—!—
(48i-12j + 36k)
12^26
=ir
(4M+3k)
MAD =-Kiiy(-r
Now
EI,4
XTEF)
©
214
PROBLEM
m = (36
where
r
M
Then
in.)i
(Continued)
3.61
+ (96
in.)j
+ (27
4
-I
3
36
96
27
-3
-22
6
in.)k
1
(2)
2
(2304
+ 81-2376 + 864 + 216 + 2376)
or
lb -in. <«
© 2010 The McGraw-Hill Companies, inc. All rights reserved. JVo /*»/ o//A& M»ma/ may be displayed,
distribution to
M^, =1359
in
215
PROBLEM
>J
A
sign erected on uneven ground is
guyed by cables EF and EG. If the
force exerted by cable EG at E is 54 lb,
\-1 > in.
^0^
,,^'y^^
,,
%
*-
"'
|V
"&ftK?'?^^T?î*K V^lS
i p":
X^^^x
"\
J7in
36iri7
;
-
\^
moment
of that force
about the line joining Points A and D.
in.
?
'
;
:
determine the
^M^M^M:-}-^:-^
Is.
in.
3.62
]ff ;1
-
:
>^'
\|
*
12
®^,
1
SOLUTION
First note that
then
(Jx 48,
BC =V( 48 > 2
96,
=
f <36)
+ (36)
or
(36
2
=- 60
in
,
96
in.
and
that
~~~—
27
in.).
Then
in.,
BC
60
~.
4
The coordinates of Point E are
2
2
2
4*;=Vai) +(-88) +(-44)
= 99
in.
T*;= ^(lli-88j -44k)
Then
= 6[(llb)i-(8 1b)j-(4
^/> - s/(48)
Also
2
= 12>/26
Then
AD
1.b)k]
+ (~1 2) 2 + (36) 2
in.
AD
=
—!=(48i~I2j +
J
36k)
12^26
= ~=-(4i-j + 3k)
V26
MAD ^k AD <rm xTEG
Now
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)
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21fi
PROBLEM
where
rBA
= (36
in.)i
M AD
Then
3.62 (Continued)
+ (96
in.)j
+ (27
4
-I
3
(6) 36
96
27
I
-8
-4
26
_6_ (-1536
-27
in.)k
»864 -.288 -144 + 864)
V26
or
M, n
=~23501b-in.
<
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reproduced or distributed hi any form or by any means, without the prior written permission of the publisher, or used beyond the limited
©
you are using
it
without permission.
217
)
PROBLEM
Two
forces
action of
F|
F2
is
,
3.63
and
F,
in
equal to the
space have the same magnitude F, Prove that the
moment of F2 about
the line of action
of
moment of
F,
about the line of
F,
SOLUTION
¥
First note that
Let
x
Mj = moment of F2 about the
action
of
Now, by
line
~f\Ay
F2
and
M,
of action of
~F2
and
a\
M
2
2
= moment of F,
about the line of
M,
itfi=4-(rBM
definition
xF2 )
= A\-( :wa x A2 )f2
i
Fi=F2 =F
Since
m
and
x
-r»
BM
Mi=^r(»*^xi )F
M =X -{-* xA )F
2
2
h
Using Equation (3.39)
i?BIA
x
X X 2 ) = ^2 HfiM X ^1
M
so that
BIA
2
2
=A
}
m xZ )F
-(r
2
©
M =M A
[2
2i
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218
PROBLEM
0.35
3.64
m
In
Problem 3.55, determine the perpendicular distance between
BH of the cable and the diagonal AD.
portion
0.875
^m
m
PROBLEM
//
0.925
D
3.55 The frame ACD is hinged at A and
and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
m
cable is 450 N, determine the moment about the diagonal AD
of the force exerted on the frame by portion BH of the cable.
0.75 in
L _y_
0.75
0.5
n>*K-^^
0.5
1
m
C,<
m "A,
SOLUTION
From the
solution to
W =450N
Problem 3.55:
7)
TBH = (1 50 N)i + (300 N) j - (300 N)k
|A/^! = 90,0N-m
X , D =-(4i.-3k)
Based on the discussion of Section
contribute to the
moment of Tbh about
Now
3.1
1,
follows that only the perpendicular component of
TBH
will
AD.
line
\'lill )para!lei
it
~~
'"'AD
*IIH
= (1 501 + 300 j - 300k) I(4i - 3k)
•
= ^l'050)(4) + (-300)(-3)]
= 300
T*liH
Also
so that
Since
v
BH
'perpendicular
% AD and (TM )pcrpcndicuiQr are
or
90.0
=
V
N
*/>•//
/parallel
+ ( '«// ^perpendicular
= V(450) 2 - (300) 2 = 335.41
perpendicular,
it
follows that
M AD
~ "('BH Jpcipendiculftr
N-m
= rf(335.41N)
d
= 0.26833
m
distribution to
d = 0.268
mA
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N
it
in
without permission.
219
PROBLEM
0.35
3.65
in
In
BG of the cable and the diagonal AD,
portion
0.875
(;:v.l
m
PROBLEM
3.56 In Problem 3.55, determine the
about the diagonal
portion
AD
moment
of the force exerted on the frame by
BG of the cable.
SOLUTION
From
the solution to
T}iG - 450 N
Problem 3.56:
Tbg =-(200M)i + (370N)j-(160N)k
\M AD = \WH-m
\
^D--(4i-3k)
Based on the discussion of Section 3.11,
contribute to the
moment of Tbg about
Now
line
(Tbg ) parallel
~
*
it
TS o
will
AD.
BO
'
^ 41)
- (-2001 + 370j - 1 60k)
-
~(4i
- 3k)
= |[(-200X4) + (-I60X-3)]
= -64 N
Also
*BG
~
'
"*~
*/J6" /parallel
V
*BG ) perpendicular
2
2
(TBG ) pcrp enc.icuiar = V(450) - (~64) = 445.43
so that
Since
X AD and (Tso )p
.
C1
p ,ndicular
are perpendicular,
it
N
follows that
™AD ~ "('ag) perpendicular
llJN-m = J(445.43N)
or
</
= 0.24920 m
d = 0.249
©
m A
you are using
it
without permission.
220
PROBLEM
In
3.66
D and B.
cable AE and the line joining Points
PROBLEM
ABC is supported by
and D and is held in the position
shown by cables AE and CF. If the force exerted by cable AE
at A is 55 N, determine the moment of that force about the line
joining Points
and B,
3,57 The triangular plate
ball-and-socket joints at
B
D
SOLUTION
From
the solution to
T„ =55N
Problem 3.57
?
T^=5[(9N)i-(6N)j + (2N)k]
|M D/i = 2.28N-m
|
X /«-'— (24i-7j)
contribute to the
moment of TA e about
Now
line
(*AE )parallel
it
TAE
will
DB.
~ ÂE ^ Dli
*
5(9i-6j +
[(9)(24)
2k)-— (24i~7j)
+ (-6)(-7)]
5
=
Also
*AE
so that
Since
A,
(''./: '),,,
perpend
neodfcul*
icuiar
51.
6N
~ ( *Ae) parallel + ( *AEJ perpendicular
- J(5S) 2
ra and (T (/i )pcrpendicular are perpendicular,
it
+(51.6)
2
- 1 9.0379 N
follows that
MDB ^"(TÔperpendieiilar
or
2.28
N-m =
</(!.
9.0379 N)
d~ 0.1 19761
^
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distribution
J = 0.1198 m
you are using
it
in
without permission.
Ill
PROBLEM
3.67
In Problem 3.58, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM
3.58
The
ball-and-socket joints at
triangular plate
B and
D
and
ABC
is
is
supported by
held in the position,
cables AE and CF. If the force exerted by cable CF at
33 N, determine the moment of that force about the line
joining Points D and B.
shown by
C
is
:0.3iJi
SOLUTION
From
the solution to
Problem 3.58
Tcr
= 33N
Tc/,=3[(6N)i-(9N)j-(2N)k]
jA* D«|
= 9.50N-m
Xm =
contribute to the
moment of Tcr about
Now
line
( i-CF )parallcl ~"
(241
~ 7,)
25
it
T Cp
will
DB,
*CF * o«
"
= 3(61-9j-2k)~ (241 - 7 j)
= ^[(6)(24) + (-9)(-7)]
- 24.84 N
Tc — (TcF )
AlSO
y;-
s° that
jwral le!
+ ( *-C/' ) perpendicular
= V(33) 2 - (24.84) 2
(Ta.Wendicuiar
= 2.1 .725 N
Since
X DB and (TCF ) pcrpendicillar are perpendicular,
\M DB
I
it
follows that
"V'CY'Vperpcndic
perpendicular
9,50N-m = </x21.725N
or
or
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you are using
d~ 0.437 m A
it
in
without permission.
Ill
PROBLEM
In
3.68
Problem 3.61, determine the perpendicular
distance between cable
Points
%
EF and the line joining
A and D.
PROBLEM
3.61 A sign erected on uneven
force exerted by cable EF at E is 46 lb,
determine the moment of that force about
the line joining Points A. and D.
ill
ground
is
SOLUTION
From the
solution to Problem 3.61
7V„=46
El-
1b
T^-2H3 1b)i-(22 1b)j + (6 1b)k]
\M
-1359
1Di
lb in.
1b-in.
1
X AD
contribute to the
(4i-j + 3k)
it
Now
(TeF /parallel
'
,
2
[(~3)(4) +
'26
Also
i EF
so that
(^perpendicular
\ AD and
will
~ ÊF ^ AD
2(-3i-22j + 6k)
Since
TEF
moment of TEF about line AD.
OW
= 10.9825
lb
- ( l EF ) para
||
c]
(4i-j + 3k)
(-22)H) + (6)(3)j
+ (l EF )perpemiicular
= V(46) 2 - (10.9825) 2 = 44.670
^âre perpendicular,
it
lb
follows that
MAD - d{T
El, ) pcrpeiidicular
1359
or
lb in.
•
= rfx 44.670
lb
or
t/
= 30.4in.
<
PROPRIETARY MATERIAL. €) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
you are using
it
without permission.
223
PROBLEM
3.69
Problem 3.62, determine the perpendicular
EG and the line joining
Points A and D.
In
distance between cable
PROBLEM
3.62 A sign erected on uneven
force exerted by cable EG at E is 54 lb,
determine the moment of that force about, the
line joining Points A and D.
ground
is
SOLUTION
From the
solution to
TB0 = 54
Problem 3.62
TfiC
lb
=6[(llb)i-(81b)i-(41b)k]
|M /JD = 23501b-in.
|
^ =4^(41
-j + 3k)
V26
Based on the discussion of Section
contribute to the
it
line
AD.
moment of Teg about
N»w
3.11,
(TEG ) para „e
,
TgG
will
= TEG X AD
- 6(i - 8j - 4k) -fL( 4i - j + 3k)
V26
•
= ~™L[(I)(4) + (-8X-D + (-4X3)] (TBG ) perpendiciliar = T£G = 54
Thus,
Since
% A0 and (T£6
pe[pcildicu tar
are perpendicular,
1
or
2350
™ AD
\
lb in.
•
it
lb
follows that
~ "\*EG /perpendicular
= dx 54
lb
or
rf
= 43.5
in.
4
PROPRIETARY MATERIAL €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
distribution to teachers and educators permitted by McGraw-Hillfar their individual course preparation. Ifyou are a student using this Manual,
224
PROBLEM
3.70
Two
parallel 60-N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical components
and adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the
the
moments of the two
two
forces, (c)
by summing
forces about Points.
SOLUTION
(a)
SM 8
We have
where
</,
:
-t/,C v +rf2 C,,=M
= (0.360 m) sin 55°
= 0.29489 m
d2 =(0.360 m) sin 55°
= 0.20649
m
C, =(60 N) cos 20°
- 56.382
C,,
N
=(60 N) sin 20°
= 20.521 N
M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.52
--=-(12.3893
(b)
1
N)k
N-m)k
£M A
We have
:
m x FB + r
£(r x F) = r
t
CIA
M = 12.39 N-m
or
M = 12.39 N-m
M
x Fc =
i
M = (0.520 m)(60 N)
sin 55°
-cos 20°
-sin 20°
J
+(0.800 m)(60N) cos 55°
7.8956
k
sin 55°
cos 20°
(1
k
J
cos 55°
I
sin
20°
N m - 30.285 N m)k
-(12.3892
•
•
N-m)k
distribution to
or
N [(0.360 m)sin(55° - 20°)](-k)
= -(12.3893 N-m)k
you are using
M = 12.39 N-mJ^
M = Fd(-k)
We have
= 60
(c)
or
it
in
without permission.
225
21.
PROBLEM
II >
1211./
1
'
1
1
lb
16 in.
A
3.71
couples. Determine (a) the
A"
two 21 -lb
ft
acted upon by two
couple
formed by the
of the
plate in the shape of a parallelogram
forces,
(/?)
the perpendicular distance between the 12-lb
forces if the resultant of the
if the resultant
couple
moment
is
72
is
two couples
lb- in.
is
zero, (c) the value
clockwise and dfc 42
of
a
in.
SOLUTION
(a)
c
M =d^
We have
x
d ~ 16
where
in.
x
Fj=211b
M, =(16in.)(21
-336
\Z»V?
(/?)
1
2.
(c)
336
lb in.
•
-^(1 2
lolll
-72
lb -in.
sin
and
distribution to
M, -336
lb -in.
H
=
1
= 336
</,
in.
<
2
lb -in.
a~ 0.80952
a = 54.049°
= 28.0
- (42
in.)(sin a){\
2
or
lb)
a = 54.0° ^
© 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No pari of this Manual may be displayed,
you are using
lb)
M =M +M
We have
or
or
lb -in.
M,+M. 2 =0
We have
or
Ho
lb)
it
in
without permission.
226
PROBLEM
100 imn
140
3,72
A couple M
of magnitude 18 N-m is applied to the
handle of a screwdriver to tighten a screw into a
block of wood. Determine the magnitudes of the
KiOi
mm
160
mm
two smallest horizontal forces
M
if
that are equivalent to
they are applied (a) at corners
corners
B and
A.
and D,
(b)
-at
C, (c) anywhere on the block.
2-10 mil
V^
SOLUTION
P
D
(a)
1
n
18N-m = P(.24m)
or
P = 75.0N
A
P
M = Pd
We have
or
Pmin =75.0N <
dBC ^yj(BE) 2 +{ECf
(A)
= V(-24m) 2 +(.08m) 2
= 0.25298
m
M = Pd
We have
18N-m = P(0.25298m)
P-71J52N
).24m)
0.4
o
r%
^
M=
We have
2
P = 45.0N
A
2
+ (0.32 m)
m
PdAC
P = 45.0N
or
any form or by any means, without the prior written permission of the publisher, or used beyond the
limited
teachers and educators permitted by McGraw-Hill
far their individual course preparation. Ifyou are a student using this Manual,
distribution to
<
18N-m = />(0.4m)
you are using
P = 71.2N
dAC =yl(ADf+(DCf
(c)
,32- .tv,
or
it
in
without permission.
227
PROBLEM
:
25
-in.-diameter pegs are attached to a board as shown. Two
strings are passed around the pegs and pulled with the forces
Four
.',:.)
3.73
ih
I!>
1
Determine the resultant couple acting on the board.
pegs should it pass
(/;) If only one string is used, around which
pulled
to create the same
it
be
and in what directions should
indicated, (a)
6 in.
couple with the
minimum
tension in the string? (c)
What
is
the
value of that minimum, tension?
L
25
Sin.-
lii
SOLUTION
+)
(a)
M = (35 lb)(7
-245
in.)
lb -in.
+ (25
+ 225
lb)(9 in.)
lb -in.
M = 470
35"&
(/;)
With only one
pegs A and D, or B and
string,
lb- in.
^)<
C should be used. We have
sty*
6
90°-
= 36.9<
tan
= 53.1°
8
Direction of forces:
With pegs
/I
53.1° -4
andD:
53.1°
With pegs/? and C:
(c)
The
distance between, the centers of the
2
<J&+6 =10
F
A
c/
d between the forces is
= 10in. + 2|-in.
=11
in.
M = Fd
We must have
is
in.
Therefore, the perpendicular distance
s
two pegs
<
4701b-in. = F(llin.)
F = 42.7
lb
<
20 10 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed,
publisher, or used beyond the limited
reproduced or distributed in anv form or by any means, without the prior written permission of the
you are a student using this Manual,
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. If
you are using
it
©
without permission.
228
PROBLEM
25
3.74
lb
Four pegs of the same diameter are attached to a board as shown.
strings are passed around the pegs and pulled with the
forces indicated. Determine the diameter of the pegs knowing
Two
that the resultant couple applied to the
board
is
485
lb in.
counterclockwise.
SOLUTION
M ^ dADpAD+dBC FBC
485
lb in.
•
= [(6 + rf)in.](35
lb)
+ [(8 + rf)in.j(25
lb)
rf
= 1.250
in.
<
PROPRIETARY MAIERIAL.
2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of Urn Manual mav be
displaved
reproduced or distributed in any jorm or by any means, without the prior written
for their individual course preparation. Ifwit are a student using this Manual
you are using if without permission.
©
229
PROBLEM
3.75
SllH'i
of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
The
shafts
filb.fl
SOLUTION
Based on
where
M M,+M
==
2
M,== -(81b-ft)j
M
M
2
|M|
3l
I2M
==
-(61b-ft)k
==
-(8tb-ft)j-(61b-ft)k
==
2
+(6)
>/(8)
2
=10
or
lb -ft
M = 10.00 lb- <
ft
M
=
"|m:|
-(8 lb
•
ft)j
- (6
lb fl)k
•
10ib-ft
-0.8j-0.6k
or
M
cos0v
jMjl = (101b-ft)(~0.8j-0.6k)
# =90°
=O
v
cos<9v =-0.8
6>
.
=143.130°
(
cos0z --O,6
Z
or
=126.870°
9X - 90.0°
6 =
143. 1 °
Oz =\ 26.9°
<
2010 The McGraw-Hill Companies, Inc. All lights reserved. No part of this Manual may be displayed,
beyond the limited
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. If
you are using
it
without permission.
230
PROBLEM
170
mm
150
mm
If
.1.50
160
P = 0,
3.76
replace the
two remaining couples with a single
its magnitude and the direction
equivalent couple, specifying
mm
of its
axis.
mm
!8N
18
N
3-3
N
SOLUTION
M = M,
We have
where
M,
l
GIC
rt?/c
+ M,
x *i
-(0.3 m)i
(18N)k
M,=-(0.3m)ix(18N)k
= (5.4N-m)J
M
Also,
2
'D/F
=rD/r xF2
-(.15m)i + (.08m)j
(.15m)i + (.08m)j + (.17m)k
(34
2
(.15)
2
+(.08) +(.1.7)
2
N)
m
= 141.421 N-m(.15i + .08j + .l 7k)
M
= 141,421
N-m
i
J
-.15
.08
-.15
.08
k
.17
- 141 .42 1(.01 361 + 0.0255 j)N
•
m
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©
you are using
it
without permission.
231
PROBLEM
3.76 (Continued)
M = [(5.4 N -m)j] + [14 1.42.1 (.01 36i +
and
= (1 .92333 N m)i + (9.0062 N
•
2
|m:|-^(m,) +(m;v )
= V(l-92333) 2
•
.
0255 j)
(1
+(9.0062)
~|Mj~
.92333
m]
2
2
or
•
=
•
m)j
= 9.2093 N m
M
N
N
•
m)i + (9.0062
9.2093
M = 9,2
1
N m <
•
N m) j
•
N-m
= 0.20885 + 0.97795
cos 6>v =0.20885
0,= 77.945°
cos
y
or
Z
<
=0.97795
or
0, =12.054°
cos
9^11.9°
=12.05°*
0,
=0.0
=90°
or
6 =90.0°
'
<
PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
limited
this Manual,
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using
you are using
it
without permission.
232
PROBLEM
If
P~0,
3.77
replace the two remaining couples with a single
of its
its
magnitude and the direction
axis.
SOLUTION
M = M, + M
2
;
/'J
M, = rc x F, - (30
M
2
dOE
m xF
=r
2
2
>/(0)
= 16
x H) 6
in.)*
m=
r
;
F2
lb,
(1
5 in.)i
= 40
lb) jj
- (5
2
2
+ (5) +(10) = 5^
lb
= -(480
lb in.)k
•
in.)j
in.
E,
= 8>/5[(llb)j-(2ib)k]
M
2
-8n/5
k
i
j
15
~5
1
M
- 8>/5[(10
lb in.)i
-(480
in.)k
lb
•
(178.885 lb
M
2
M
M
cos0
v
COS
COS
•
+
in.)i
+ (30
lb in.)j
•
8V5[(1
lb
+ (536.66
^/(178.885) +(536.66)
603.99
K
-2
2
•
+ (1 5
in.)i
•
+ (30
lb in.) j
•
lb in.)k]
lb
- (21
1
•
in.)j
+
(1
5 lb in.)k]
•
.67 lb in.)k
•
+(-21 1.67) 2
lb in
A/
lb -in.
<
,=110.5°
A
= 604
0.29617! + 0.88852 j - 0.35045k
0.29617
0.88852
0..
-0.35045
72.8°
0=27.3°
0,
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©
233
1
PROBLEM
If
P = 20
lb,
3.78
replace the three couples with a single
of its
direction
its
magnitude and the
axis.
SOLUTION
From
the solution to Problem. 3.77
16 lb force:
40
lb force:
F - 20
lb
M, = -(480
M
M
lb in.)k
•
2
= 8>/5 [(10
3
= rc x P
+ (30
lb in.)i
•
lb in.)j
•
+ (1 5
lb in.)k]
•
= (30in.)ix(20lb)k
= (600
lb in.)j
-
M-M, +M +M
2
3
= ~(480)k + 8V5 (1 Oi + 30j + 1 5k) + 600 j
-(178.885
lb -in)i
M = ./{178.885)2 +
= 1169.96
M
M
X„
0.
1
+ (1136.66
2
(1
3.66)
lb -in.)j
+(211 .67)
-(211.67 lb-in.)k
2
M = 1170
lb in
lb- in.
<
6'=100.4°
<
52898i + 0.97 1 54 j - 0. 1 8092 k
cos
ex = 0.152898
cos
^ = 0.97 154
cos
0.= -0.1 8092.1
1
9.
=81.2°
0=13.70°
Manual,
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this
©
234
PROBLEM
P = 20 N,
If
3.79
replace the three couples with a single
its
magnitude and the
direction of its axis.
\
,-
160 nun
18
N
SOLUTION
M-M, + M +M
We have
2
3
where
i
M^r^xF,
k
J
N-m = (5.4N-m)j
0.3
18
i
M
3
=r /fXF2
k
J
,15
.08
.15
.08
141.421
N-m
.17
141.421(.0136i + .0255j)N-m
(See Solution to Problem 3.76.)
M 3= r
(X4
xF3=
0.3
N-m
0.17
20
= -(3.4N-m)i + (6N-m)k
M=[(1.92333-3.4)i + (5.4 + 3.6062)j + (6)k]N.m
= -(1 .47667 N m)i + (9.0062 N m)j + (6 N
•
llVff—
•
nj2L
xj2
M; + M
+ m;
V
1**2
,
,
V(l .47667) + (9.0062) + (6)
10.9221
N-m
distribution to
2
or
M = 10.92N-m A
CO 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part
of this Manual may be displayed,
teachers and educators permitted by McGraw-Hillfor their individual course preparation. you are a student using this
If
Manual,
you are using
in)
it
in
without permission.
235
PROBLEM
X
M
3.79 (Continued)
-1.47667
+ 9.0062 + 6
10.9221
jM|
= -0. 35200i + 0.82459J + 0.54934k
1
cos(?T
= -0.135200
6X =97.770
or
cos
= 0.82459
#,,=34.453
or
=56.678
or
V
cos#. = 0.54934
©
-97.8°
<
= 34.5°
^
= 56.7°
^
V
y
<9
2
20 JO The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
distribution to
Z
<9
in
236
]
200
PROBLEM
N-iii
1.20
A
and B connect the gear box to the wheel assemblies
and shaft C connects it to the engine. Shafts A and
B lie in the vertical yz plane, while shaft C is directed along
the x axis. Replace the couples applied to the shafts with a
Shafts
of a
1
3.80
N«m
tractor,
single equivalent couple, specifying
z
A
} 1600
its
magnitude and the
direction of its axis.
N-m
SOLUTION
Represent the given couples by the following couple vectors:
M^
= -1 600 sin 20° j + 600cos20°k
1
-(547.232
M:
Mc
single equivalent couple
•
1
1200 sin 20°j + 1200 cos 20°k
(41 0.424
The
N m) j + (1 503.5 N m)k
N
•
m)j + (11 27.63
N m)k
•
-(1120N-m)i
is
+M 5 +M C
M^M,,
= -(1 120
M=
N
2
yj(\
1
20)
= 2862.8
•
m)i - (1 36.808
N
m)j + (263 1.1 N- m)k
•
+ (1 36.808) 2 + (263
2
1
.
I)
N-m
-1120
cos
6,
2862.8
COS
tfy
^
-136.808
2862.8
2631.1
cos 0„
2862.8
M = 2860N-m
B =92.7*
'
23.2°
<
to teachers and educators permitted by McGraw-Hillfor their individual course preparation.
distribution
0=113.0°
you are using
it
in
without permission.
237
PROBLEM
3.81
20°
A*'
The
tension in the cabJe attached to the end
adjustable
boom ABC
is
560
lb.
C of an
Replace the force
C with an equivalent force-
exerted by the cable at
couple system (a) at A, (b) at B.
W
Aim
SOLUTION
{a)
mK^*^t^^
TscnSo
ZF:
Based on
F,,=r = 560lb
¥A = 560 lb
or
X^^S
ZM A
:
M^^sinSO
)^,)
= (560lb)sin50°(18ft)
Â
= 7721.7 lb -ft
M
or
(b)
I
&
Based on
XF:
7^ =
7*
= 560
fi
:
MB =(Tsm50°)(dB
= (560
«eA\
3*6
)
lb) sin 50° (10 ft)
= 4289.8
\
=7720.1b-ftjH
F = 560 lb ^C 20° <
ZMB
B
/(
lb
or
^M
^ 20° ^
lb
ft
c
M g =42901b-ftjH
or
\£&<!2t
hc
***
F*
©
PROPRIETARY MATS RIAL,
you are using
if
without permission.
238
PROBLEM
3.82
A
P
60-lb force
1
Replace
P
is
applied at Point A of a structural member.
with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a
and a second force at D.
vertical force at
B
SOLUTION
(a)
Based on
Pc ^P = \60\b
2.F:
XMC
:
Mc —P d
x
Px
where
cy
=(160
= 80
or P,
=160
lb
^T.60°^
+ Py dCx
lb) cos 60°
lb
/;=(1601b)sin60°
= 138.564
rf
tt
lb
=4ft
JCl = 2.75
ft
,
Mc = (80 lb)(2.75
= 220
lb -ft
= 334.26
(b)
EFV
Based on
:
ft)
+ (1 38.564
lb)(4 ft)
+ 554.26 lb- ft
lb -ft
or
M c =334
lb
•
ft
)<
PDx =P cos 60°
= (160lb)cos60 c
= 80 lb
TM,
(Pcos60°)(clDA )
[(160 lb)cos60°](1.5
ft)
= PB (dDB )
= PB (fi ft)
PB = 20.0 lb
or
P„=20.01bH
€> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,
any form or by any means, without the prior written permission of the. publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,
in
239
PROBLEM
ZFV
:
3.82 (Continued)
Psin60 o = PB + PI)}
,
(160 lb) sin 60° = 20.0 lb +
PDy =118.564
PDy
lb
2
^=>/('k) +('y
2
2
2
= 7(8 °) +0 18 564)
= 143.029 lb
-
# = tan
- tan
_1
(p
\
p
J—
118.564^
]
I
—
80
= 55.991°
)
or
©
PD
= 143.0 lb
,
^56.0°^
you are using
it
without permission.
240
PROBLEM
-*-r-HS A
50 nun
F,-
The 80-N
i
(a)
horizontal force.
P
acts
on a
bell crank, as
shown.
with an equivalent force-couple system at B.
two
C and D that are equivalent
vertical forces at
couple found
to the
100
P
Replace
(b) Find the
«i
3.83
in Part a.
mm
SOLUTION
A
(a)
F =F = 80N
IF:
Based on
XM:
JvJ:
or
/?
M
n
--•••-
^
=Fdr
80
N
(.05
4.0000
m)
N m
•
M«-4,00N-m }^
or
(b)
FB = 80.0 N
M
If the two vertical forces are to be equivalent to
B they must be
a couple. Further, the sense of the moment of this couple must be
,
counterclockwise.
^
5
Then, with
Fc
4.0000
acting as shown,
MD =Fc d
XM:
1
«*
FD
and
N-m = Fc 04m)
(.
Fc =100.000 N
2Fy
= FD
:
or
F„ =100.0
nH
2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed,
distribution
Fc
-Fc
F =l 00.000 N
ft
=100.0nH
or
you are using
it
in
without permission.
241
PROBLEM
......'
iSsif^S& &^ -fiftî isKSS
ts
3.84
A dirigible is tethered by a cable attached to its cabin
A
C
B
6.7
m
4
at B. If
040 N, replace the force exerted by
the cable at B with an equivalent system formed by two
parallel forces applied at A and C,
the tension in the cable
is
1
m \
60°/\
......,lA.^
,
SOLUTION
Require the equivalent forces acting
angle of
Then
A and
at
C be parallel and at an
a with the vertical.
for equivalence,
040 N) sin 30° = FA
XFX
:
(1
ZFy
:
-(1
si n
a + FB sin a
(1)
040 N) cos 30° = -FA cos a - FB cos a
Dividing liquation (1) by Equation
(2),
{FA +FB )s\na
~
N) cos 30° -{FA + FB )cosa
(1040 N) sin 30°
-(.1040
(2)
__
a ~ 30°
Simplifying yields
Based on
XMC
[(1040 N)cos30°](4
:
m) = {FA cos30°)(l 0.7 m)
FA = 388.79 N
or
F,,
= 389 N "^ 60°
M
Based on
XM.A
:
- [(1 040 N) cos 30°](6.7 m) = (Fc cos 30°)(l 0.7 m)
Fc
distribution to
©
^60°^
2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed,
any form or by any means, without (he prior written permission of the publisher, or used beyond the limited
you are using
M
FC =651N
or
=651.21
it
in
without permission.
242
PROBLEM
3.85
force P has a magnitude of 250 N and is applied at the end C
of a 500-mm rod AC attached to a bracket at A and B. Assuming
a ~ 30° and j3 — 60°, replace P with (a) an equivalent force-couple
system at B, (b) an equivalent system formed by two parallel forces
The
300
mm
:,-l
A\
200
mm
A
applied at
and B.
SOLUTION
(a)
F=P
£F:
Equivalence requires
1M B
F = 250N^60°
or
M = -(0.3 m)(250 N) = -75 N
:
The equivalent force-couple system at B
m
is
M = 75.0 N
F = 250N "^60°
(b)
•
m)<
Require
9
%
MO
/f/
5^1*
™
NJ
c
Equivalence then requi :es
£FV
F^
2)F
y
Now
--
:
—~FR
f
cos
>
or
= -F
25
F.
if
4
cos
cos0 = O
sin
- Fe sin
= -Fn => -250 =
reject
=
= 90°
or
and
Also
+ FA
= F,, cos
:
F,
£.A/B
:
+ FB = 250
- (0.3 m)(250 N) = (0.2m)F
/4
or
F.,=-375N
and
F =625N
i?
F^=375N ^60°
FB = 625 N
^
60°
^
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distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. Ifyou are a student using this Manual,
243
PROBLEM
Jk\
300
3.86
Solve Problem 3.85, assuming
a-fi~
25°.
mm
1'
'A\
200
mm
SOLUTION
P= 2SO
id)
£F:
XM B
The
:
¥B = P
M
fi
equivalent force-couple system
F«
(b)
N\
or
F = 250 N X'
25.0°
fi
= ~-(0.3 m)[(250 N)sin 50°] = -57.453 N
at.
•
m
B is
= 250 N
^
25.0
M„^57.5N-m )<
C
Require
Z5"o
25"0
n!
M
0-2. w>
MB = dAE Q
(0.3
m)[(250 N) sin 50°]
= [(0.2m)sin50°]£
= 375N
Adding
the forces at B:
F,,
= 375
N
^
25.0°
©
F»=625N X:
25.0°
<
244
—
PROBLEM
A
3.87
force and a couple are applied as
shown
to the
end of a
cantilever beam, (a) Replace this system with a single force
F
d from C to
a
applied at Point C, and determine the distance
drawn through Points D and E. (h) Solve Part a
directions of the two 360-N forces are reversed.
line
450
mm
if the
rA
pi
i
.v
150
mm
SOLUTION
A3fcON
(a)
We have
IF:
F = (360N)j-(360N)j-(600N)k
A
or
SMD
and
3(>ON
COON
E
d - 0.09 m
or
T
(b)
*-J
w
d - 90.0 mm below ED
A
r
c
—
A
(360N)(0.]5m) = (600N)(rf)
p
1
E
:
F = -(600N)k
We have from Part a
F = -(600N)k 4
150 win
1MD
and
:
-<360N)(0.15m) = -(600N)(rf)
d = 0.09 m
or
H
-
---' r '" "
d=
90.0
mm above ED A
fe
d
D
3&0N
©
you are using
it
in
without permission.
245
|*-]2()inm-*)*i +
i
'
'
[
— ~— -"'-^î^;-;
250
;
N
mm
j
y
The shearing
C
1 900 N
90 ram
250
3.88
--::; :-: ;
M
I
90
PROBLEM
forces exerted on the cross section of a steel channel can
be represented by a 900-N vertical force and two 250-N horizontal
forces as shown. Replace this force and couple with a single force F
applied at Point C, and determine the distance x from C to line BD.
(Point
C is defined as the shear center of the section.)
N
SOLUTION
Replace the 250-N forces with a couple and move the 900-N force to Point
moment of the couple
C such that its moment about 11 is
equal to the
25"o
k^
M
Ph
O.tftwi
?J
TÔO M
iOoM
•ZSoM
D
Mu = (0.18)(250 N)
= 45N-m
M
Then
lf
= x(900 N)
45N-m = x(900N)
x- 0.05 m
or
F = 900NJ
©
x = 50.0
mm A
246
PROBLEM
2.8
3,89
in.
While tapping a hole, a machinist applies the horizontal forces
to the handle of the tap wrench. Show that these forces
are equivalent to a single force, and specify, if possible, the
point of application of the single force on the handle.
X
shown
3.2 in'
¥
1)
2.05
I!)
SOLUTION
B are parallel, the force at B can be replaced with the sum of two forces with one of
Since the forces at A and
the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple.
F -
Have B 2.9
parallel forces,
lb- 2.65
lb
= 0.25 lb, where the
2.65 lb force be part of the couple. Combining the two
Couple
M
and
couple
= (2.65
lb)[(3.2in.
= 14.4103
lb -in.
14.41 03
lb -in.
+ 2.8 in.)cos25°]
1
H
A
6
*
f
A
is^h
-Q
X
A single equivalent force will be located in the negative z-direction
ZMn
Based on
-1
4.4 1 03 lb
•
in.
= [(.25 lb) cos 25°](a)
a = 63.600
F'= (.25
F' = (0.227 lb)i
+ (0. 057
distance of 63.6
1
in. to
lb)k and
the right
is
in.
lb)(cos 25°i
+ sin 25°k)
applied on an extension of handle
BD at a
of B
2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of (his Manual may be displayed,
247
48
40
2o:
PROBLEM
lb
3.90
Three control rods attached to a lever ABC exert on it the
forces shown, (a) Replace the three forces with an equivalent
force-couple system at B. (b) Determine the single force that
is equivalent to the force-couple system obtained in Part a,
and specify its point of application on the lever.
in.
20
20 ib
SOLUTION
IK
(a)
First note that the
two 20-lb forces form A couple. Then
F = 48
Ib
a\6
ALd
*E>S°
= 180° -(60°
where
+ 55°) = 65°
M = 2,MB
and
= (30in.)(481b)cos55°- (70 in.)(20 lb)cos20 c
= -489.62
The equivalent force-couple system
B
at
The
single equivalent force F' is equa
between A and B. For equivalence.
EM,:
where a
is
the distance from
lb
to F. Further, since the sense
applied to the lever 17.78
left
of pin
of
M
is
clockwise
,
=
=
490
lb-
in.jH
F' must be applied
M = -aF' cos 55°
lb
•
in.
= -tf(48.01b)cos55°
o = 17.78
To the
M
^65°
.
oi-
is
in
B to the point of application of F' Then
-489.62
and
I
•
is
F = 48.0
0>)
lb
in.
F' = 48.0
lb
^L
65.0°
<
in.
<
B
you are using
it
without permission.
248
PROBLEM
300
N
3,91
A hexagonal plate is acted
upon by the force P and the couple shown.
Determine the magnitude and the direction of the smallest force P for
which this system can be replaced with a single force at is.
300 N
SOLUTION
From the statement of the problem, it follows that XME =
for P mi „, must require that P be perpendicular to r
B/E Then
for the given force-couple system. Further,
.
XME
:
(0.2 sin 30° +
0.2)mx300N
3<x>tv\
3ooM
+ (0.2m)sin30°x300N
~(0-4m)P- =0
or
R,,
-300N
P.,
distribution to
N
30.0°
<
© 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may he. displayed,
you are using
300
it
in
without permission.
249
PROBLEM
3,92
A
rectangular plate is acted upon by the force and couple
shown. This system is to be replaced with a single equivalent
force, (a) For a - 40°, specify the magnitude and the line of
action of the equivalent force, (b) Specify the value of or if the line
of action of the equivalent force
to the right of ZX
is to intersect line
CD 300 mm
SOLUTION
(a)
The given
force-couple system
(fi\
M) at B is
CJ
O
F = 48Nj
and
M = ZMB = (0.4 m)(l 5 N)cos 40° + (0.24 m)(1 5 N)sin 40
or
M =6.91.03 N -m
The
single equivalent force F'
is
equal to F. Further for equivalence
~ZM B
or
6.9.1.03
:
M = dF'
N -m = dx 48 N
d~ 0.1 4396 m
or
and the
(b)
line
of action of/"
7
intersects line
Following the solution to Part a but with
J.MB
:
AB
(0.4 m)(l 5
m
1
48N A
mm to the right of A.
144
d ~ 0.
F' =
N) cos
and
a unknown, have
a + (0.24 m)(l 5
N) sin a
= (0.!m)(48N)
5 cos
or
a + 3 sin a - 4
Rearranging and squaring
Using cos
2
a~
1
-sin
2
a
25 cos
34 sin
Then
2
or ~
a ~ (4 - 3
sin.
a)
2
and expanding
25(1
or
2
24
2
-sin a) -16 -24
sin or -
sin
9
sin «r
+9
2
sin
or
=
a
2(34)
sin
a = 0.97686
a = 77.7°
or
©
or
sin
a = -0.27098
a = -15.72°
250
PROBLEM
3.93
An
eccentric, compressive 1 220-N force P is applied to the end
of a cantilever beam. Replace P with an equivalent force-couple
system at G.
100
""60
mm
mm
SOLUTION
We have
-(1220N)i=F
ZF:
F = -(1220N)i
<
M=(73.2N-m)j-(122N-m)k
<
0, trvn
Also,
we have
IM G
i>c
J
k
-.3
-.06
i
1220
:
xP = M
N-m = M
-1
M=
(1
220
N
m)[(~0.06)(-l) j - (~0. l)(-l)k]
or
©
you are using
it
without permission.
251
PROBLEM
3.94
To keep a door
mm
.67
594
1.00
closed, a wooden stick is wedged between the
and the doorknob. The stick exerts at B a 175-N force
directed along line AB. Replace that force with an equivalent
force-couple system at C.
floor
mm
mm
SOLUTION
We have
Kt
IF:
P
lfi
^
(,!-*.«
=Fc
iat
where
*
AB
I.
= *"AB'AB
(33
mm
)i
+ (990mm)j- (594mm) k
1155.00
mm
or
We have
JM. C
:
rm::
xPAli
=
3>
()75N)
Fc =(5.00N)i + (150N)j-(90.0N)k
<
MC
»
J
M c = 5 0.683
k
N-m
-0.860
30
1
= (5){(-0.860)(-1 8)i
-18
- (0.683)(-l 8)j
+ [(0.683)(30)- -(0.860)(l)]k}
or
M c = (77.4 N-m)i + (61.5 N •m)j + (106.8 N-m)k A
©
252
PROBLEM
An
3.95
guyed by three cables as shown. Knowing
AB is 288 lb, replace the force
exerted at A by cable AB with an equivalent force-couple
system at the center O of the base of the antenna.
antenna
is
16
ft
SOLUTION
We have
dAB =yl(r64f +(-12Sf
Then
T^=~^(-64i-128j + 16k)
= (32
M =M
Now
+{16)
2
= 144
ft
1b)(-4i-8j + k)
=r }/0 xT,„,
= 128jx32(-4i-8j + k)
- (4096
at
lb ft)i
•
+ (1 6, 384
lb ft)k
•
O is
F = -(1 28.0
M=
(4. 1
lb)i
- (256
kip
•
fl)i
lb) j
+ (32,0 lb)k
+ (1 6.38
kip ft)k
•
-4
<
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©
you are using
it
without permission.
253
PROBLEM
3.96
guyed by three cables as shown. Knowing
AD is 270 lb, replace the force
exerted at A by cable AD with an equivalent force-couple
system at the center O of the base of the antenna.
An
antenna
is
J.6ft
SOLUTION
2
We have
2
*ad = VC-* 4 ) +C-128) +(-128)
= 192
2
ft.
T^=~5Jb(~64i-128j + 128k)
Then
= (90 1b)(-i-2j-2k)
M-M
Now
= r^xT^
= 128jx90(-i~2j-2k)
= -(23,040
The equivalent
for ;e-couple
system
at
lb ft)i
•
+ (1
1,520 lb ft)k
•
O is
F = -(90.0
lb)i
- (1 80.0
M = -(23 .0 kip
distribution to
ft)i
+ (1
lb)k
<
.52 kip ft)k
<
- (1 80.0
1
•
'& 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
you are using
•
lb) j
it
in
without permission.
254
PROBLEM
3.97
Replace the J50-N force with an equivalent force-couple
at A.
system
120
mm
Y\
20 mm
60
mm
SOLUTION
LF:
F = (150N)(~cos35 o j-sin35 o k)
= -(122.873 N)j - (86.036 N)k
XMA
where
:
M=rm xF
rDIA
= (0.
1
8 m)i - (0. 1 2 m)j
J
k
-0.12
0.1
-122.873
-86.036
i
M=0.18
Then
+ (0. 1 m)k
N-m
- [(-0. 2)(-86.036) - (0. l)(-i 22,873)]i
1
+ [-(0.18)(-86.036)]j
+ [(0.18)(-122.873)jk
= (22.6 N m)i + (1 5.49 N m)j - (22. N m)k
•
The equivalent force-couple system at A
•
1
•
is
F = -(1 22.9 N)j- (86.0 N)k
M = (22.6 N
•
m)i + (1 5.49
N
•
m)j - (22. 1
N m)k
•
<
<4
in
255
i
PROBLEM
3.98
A 77-N force F
m
couple Mi are
and a 31-N
E of the bent plate shown. If
Fj and Mi are to be replaced with an equivalent
force-couple system (F2,
2) at corner B and if
•
f
applied to corner
83.3
M
mm
(M2)7 -
0,
_
and
M
determine (a) the distance d, (b)
F2
2.
SOLUTION
(a)
XM&
We have
:
M
=0
2z
M^xFJ + MÔ
where
rHfB
(1)
= (0.3 lm)i- (0,0233) j
_
(0.06 m)i + (0.06 m)j
0.
1 1
- (0.07 m)k
(nn Kn
m
= (42 N)i + (42 N) j - (49 N)k
M
l2
=k-M.,
M, = Ku M
m)1* - (0.07 m)k
J
„
- _~d\ + (0.03
(3
^
\
—
—
j
2
V^ +0.0058
Then from Eq nation
.
]
.
VE
|s|
.
.
m)
m
dX
1
Solving for d, Equat on
(I
)
0.31
-0.0233
42
42
(-0.07m)(31N-m)J n
-0
+-49
Vc/
+0.0058
reduces to
(13.0200 + 0.978 , }
2.17N,m
4d
2
d = 0.1350 in
From which
2
©
_^
+0.0058
or
d ~ 135.0mm
A
256
1
PROBLEM
3.98 (Continued)
F2 = F.= (42i + 42j- 49k)N
(/>)
or
-
M
2
- VWB x.F, + M,
k
i
J*
0.31
(1.
+
-0.0233
42
14 70i
(0.1350)1
-(25.858
+ 1 5. 1900J + 1 3.9986k)
N
•
m)i + (21.190
Nm
distribution to
•
•
M* =-
(25.9N-m)i + (21.2N-m)j
<
any form or by any means, without the prior mitten permission of the publisher, or used beyond the limited
you are using
N m
N m)j
or
(3IN-m)
0.155000
-49
42
+ 0.03 j -0.07k
+ (-27.000i + 6. 0000 j - 14.0000k)
M,
F2 = (42 N)i
it
in
without permission.
257
PROBLEM
3.99
A 46-lb
F and a 2120-lb in. couple M are
A of the block shown. Replace
force
•
applied to corner
the given force-couple system with an equivalent
force-couple system at
3
comer //.
in.
SOLUTION
We have
z
dAJ
18)
46
F
Then
+(-14)-+(-3r =23
in.
1b
-(18i-14j-3k)
23
(36
tb)i
- (28
lb) j
2
Also
d 4C
Then.
M
- (6
2
1b)k
7(-45) +(0) +(-28)
:
2120
lb
in.
(-451
2
= 53
in.
- 28k)
53
-(1800 1b-in.)i-(l120 1b-in.)k
M'^M + r^xF
Now
rt
where
Then
=(45in.)i + (14in..)]
r
/J/y/
M' = (-1800i~I120k) +
- (-1 8001 -
1 1
20k) +
k
i
J
45
14
36
-28
-6
{[(1 4)(-6)]i
+ [-(45)(-6)]j + [(45)(-28) - (1 4)(36)]k}
= (- 800 - 84)i + (270) j + (-1120-1 764)k
1
= -(1 884
= -(1.57
lb in.)i
•
at
lb
•
H
ft)i
+
+ (270
lb in.)j
•
(22.5 lb
•
- (2884
ft)j - (240
lb
in.)k
•
lb ft)k
•
F' = (36.0
is
M' = -(] 57
©
lb ft)i
•
lb)i
- (28.0
+ (22.5
lb)j
lb ft) j
•
lb)k
<
lb ft)k
4
- (6.00
- (240
•
258
PROBLEM
3.100
The handpiece
for a miniature industrial grinder weighs 0.6 lb, and
center of gravity
its
handpiece
is
located on the
is
offset in the
xz plane
in
y
axis.
such a
way
The head of
that line
the
BC forms
an angle of 25° with the x direction. Show that the weight of the
handpiece and the two couples M.j and
2 can be replaced with a
M
M
- 0.68 lb in.
assuming that
and 2 ~ 0.65 lb in., determine (a) the magnitude and the direction
of the equivalent force, (/>) the point where its line of action
intersects the xz plane.
single equivalent force. Further,
M
•
}
•
SOLUTION
First
assume
that the given force
W and couples M, and M
Now
w = ™^
and
M = M, + M
that since
act at the origin.
M &
2
= ~(M2 cos 25°)i + (M, Note
2
W and M are perpendicular,
M
2 sin
W
follows that they can be replaced
it
ft
25°)k
M
2
with a single equivalent force.
(a)
We have
(b)
Assume that
F-W
the line
V^-iV. =-(0.6
or
lb)j
of action of F passes through Point P(x,
M =r
/vo
m=
where
r
xi
F = -(0.600 lb)j
or
0, z).
Then
<
for equivalence
xF
+ zk
-(M2 cos 25°)i + (M, - M2 sin 25°)k
i
k
J
X
(Wz)\
z
-~
(Wx)k
-w
Equating the
(/;)
I
and k coefficients,
z
=
~^ COs25
W = 0.61b M
For
°
x=
and
W
( M 'M
{
= 0.68
x
lb
•
in.
M
2
*
"
-0.65
sin
25
'
W
lb -in.
0.68 -0.65 sin 25°
0.67550
-0.6
in.
or
•0.65 cos 25°
-0.98 1 83 in.
0.6
2
A
= -0.982 in.
^
The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
teachers and educators permitted by McGraw-Hillfor their individual course preparation. you
If
distribution to
or
x = 0.675 in,
in
259
1
PROBLEM
3.101
4-m-Iong beam is subjected to a variety of loadings, (a) Replace each loading with an equivalent forcecouple system at end A of the beam, (b) Which of the loadings are equivalent?
A
-100
*
4
Mi
ni
mo n
X
000
N
900 N-ni
\
:
i';00N.100
(W
=SCK>
100
N
N
200
N
.
200 N-
•'i.iN
800
\
;i00N-ri)
"A"
2(Hi
300
N
\
(/)
(e)
ifl)
•on
N-m
100 N-tn
1.
OON-iii
I
800 N
300
2300 N-nt
.-'
N
N
,100
N
?-'\m Nmii
/
V
300 N- us
MJON-iu
(ft)
(s)
&
SOLUTION
A
(o)
We have
(a)
M A
s/y -400N-20()N = /?o
R =600NH
or
£JW/.
and
1
We
M
£7y -600N =
have
L/V/^
and
and
distribution to
=
fl
lOOON-m^)^
tf,,
or
R 6 =600NH
or
M, -900 N -m)<
or
R c = 600 N <
or
M, = 900 N m ") <
-900N-m = A/
ft
Z/V/^
4500
J
N-m - (900 N)(4 m) = M c
1
•
€: 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
this Manual,
teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using
you are using
M
I/y 300N~900N^#,.
We have
(c)
tt
in.) = M„
800 N-m - (200 N)(4 m)
a
or
(/?)
/w
it
in
without permission.
260
PROBLEM
{d)
We have
ZM,
and
:
t
We have
(e)
My
(800
- 400
ZMA
and
:
(Continued)
N + 800 N = Rd
-400
£F„:
3.101
N)(4m)- 2300 N-m =
or
R,,=400NH
or
M,,=900N-m)^
or
R e =600NJ^
M
rf
N - 200 N = Re
N m + 400 N m - (200 N)(4 m) = Me
200
•
M
or
We have
(,/')
SF
>
-800N + 200N = /?/
:
Jt
=200N-m )^
e
R =600nJ^
or
/
ZMA
and
:
- 300 N
•
m + 300 N m + (200 N)(4 m) = M.f
•
>
We have
(g)
£F
ZM A
and
-200
:
:
200
or
M / =800N-m )^
or
R^=1000n|^
N- 800 N = tf,
N m + 4000 N m - (800 N)(4 m) = M
•
•
>
We have
(A)
M £ =1000N-m )^
or
RA = 600 N <
Z/y -300N-300N = ifA
LMA
and
or
:
2400
N m - 300 N m - (300 N)(4 m) = Mh
•
•
or
(b)
J
M,,=900N-m^H
Therefore, loadings (c) and (h) are equivalent.
-4
MjW
distribution to
m
«3 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> /wr/ o//Afe
«M y /« <fo^, /,
any jorm or by any means, without the prior written permission of the publisher, or used beyond the
limited
teachers and educators permitted by McGraw-Hillfor their individual course preparation,
ifyou are a student using this Manual
you are using it
m
without permission.
261
200
N
100
4
\
:
100
N
PROBLEM
m
N-m
>
3.102
\
a
-'
Problem 3.101 which
4-m-long beam
is
loaded as shown. Determine the loading of
is
equivalent to this loading.
:'S00N'in
SOLUTION
XFV
We have
ZMA
and
:
:
-200N-400N = fi
-400 N
•
or
R
m + 2800 N-m - (400 N)(4 m) = M
M=800N-nO
or
R = 600NH
M
HF
Mm.
Equivalent to case (/) of Problem 3.101 -4
Problem 3.101 Equivalent force-couples
at
A
M
Case
R
(a)
600 N|
(*)
600
(c)
600 N
id)
400 N
(*)
600
NJ
200 N-m}
if)
600
N|
800
(g)
1000
N
1.000
(A)
600
N
N
1000
N-m)
900
N m)
900
N m
900
N
|
•
•
•
^)
m. ^)
|
|
|
N m
•
")
Nm")
900 N
•
m ^)
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the limited
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou
PROPRIETARY MATERIAL,
you are using
it
without permission.
262
PROBLEM
3.103
Determine the single equivalent force and the distance from Point
loading of (a) Problem 3 1 1 Z>, (/?) Problem 3.10k/, (c) Problem 3. 1
A
.
PROBLEM
A
3.101
4-m-long beam
equivalent force-couple system at end
400
N
4 in
is
to
its
of action for the beam and
line
e.
subjected to a variety of loadings, (a) Replace each loading with an
(b) Which of the loadings are equivalent?
A of the beam,
w.\ \
X
-Hit)
1
900
>' \-lll
".'
,'
1
N
KOI)
N
i>>
!
(«)
-J0U
"Oil
(W
X
101)
N
200 \r
c
n
A
:«iu
:
i
'<
\
in
..,
11,'
\
800
N
X-
'/:
200 V:..
W)
./lit)
N
300 N- mi
.
v.xt
.-'MN-in
i,.
NO!)
i')(H)N-m
A
\
vj
:/)
»iN
i,.\
MJON-ii)
c
200
""
N-ii
''
1000 N-
«
(g>
:
N-ni
(A)
SOLUTION
(a)
I
GVCH
<\OOH«r\
For equivalent single force
We have
ZF„:
-600
at distance
d from A
N^^
^~~ c
h--4
a£
R ~ 600 N ^
or
w
XMC
and
W~t
^
400 n
2.2>oorvm
:
(600N)(rf)-900N-m =
or
(b)
We have
-«
£/y -400N + 800N = #
or
ZMC
and
8<30rt
d^ 1.500 m
:
R = 400Nf^
(400N)(rf) + (800N)(4~</)
\*Y\
-2300N-m =
[v~d -^t'
or
(c)
We have
lFy
:
ZOOiM
i%
200fc<W
lm
til!
rf»c
s
= 2.25
m «
-400N-200N = /f
or
and
c/
R = 600 N ^
J
ZMC 200N-m + (400N)(t/)
-(200N)(4-c/) + 400Nm =^0
:
je
or
&
tf
= 0.333 m
<
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you are using
it
without permission.
263
j
PROBLEM
3.104
Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the
shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment
M - (15 lb
•
ft)]
+ (15
lb
•
ft)k located at the origin.
lb.lt
SOLUTION
system
First note that the force-couple
force of the other four systems
is
(10
at
F cannot
lb)i].
be equivalent because of the direction of the force [The
to the origin 0\ the forces remain
Next move each of the systems
unchanged.
A:
M^ =IM
=(5
lb-ft)j
-(25
D:
M D = 2M
= -(5
+ (15
1b-fl)j
+ (15
lb ft)]
•
+ [(4.5
ft)j
lb-ft)k
+ (25
+ (l
+ (2 ft)kx(10
Ib)i
1b-ft)k
lb
ft)]
•
ft)k
+ (2
ft)k"jxl.O lb)i
= (151b-ft)i + (151b-ft)k
G
:
M 6 = XM
/:
- (1 5
.
M =SM
;
/
lb ft)i
•
+
(1
5 lb ft)
•
=(151b-ft)]-(51b-ft)k
+ [(4.5ft)i + (lft)jjx(101b)j
= (15
is
1b-ft)]-(15 1b-ft)k
the system at corner D,
distribution to
you are using
^
it
in
without permission.
264
,
6
6
PROBLEM
m
ft"
ft'
The weights of two
are 84 lb
fe&
3,105
B
and 64
children sitting at ends
lb, respectively.
A and B of a seesaw
Where should
a third child sit
so that the resultant of the weights of the three children will
pass through
if she weighs (a) 60 lb, (/?) 52 lb.
C
_
,
SOLUTION
&\\b
w„
feMlb
t
^£±
K.
(a)
For the resultant weight, to act at C,
Then
(b)
(84 lb)(6
£Afc
- 60
ft)
Then
(84 lb)(6
ft)
\h{d)
- 52
\h{d)
- 64
- 64
distribution to
.
c
lb(6 ft)
= 60
„
lb
=
d = 2.00
ft
to the right
of C
d = 2.31
ft
to the right
of C
Wc - 52 lb
lb(6
ft)
=
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teachers and educators permitted by McGraw-Hillfor their individual course preparation.
you are using
W
=0
ZMC =
For the resultant weight to act at C,
w
%
it
in
without permission.
265
kl
PROBLEM
ii
mounted on a pipe as shown.
at A and B each weigh 4.1 lb, while the one
Three stage
34
in.
The
lights
3.106
lights are
weighs 3.5 lb. (a) If d = 25 in., determine the
to the line of action of the resultant
distance from
of the weights of the three lights, (b) Determine the
value of d so that the resultant of the weights passes
through the midpoint of the pipe.
at
C
D
P
SOLUTION
LU
4.Wb
•
A
<b
\h
L
3.£
K
C-
3
1
!
D
E
D
For equivalence
TFy
:
X/^:
-4.1-4.1-3.5 = -/?
-(10
-[(4.4
rf
We have
The
375.4 + 3.5(25)
= 1 1.71
= 25
=
!
We have
(rf.iinin.)
or
I = 39.6
375.4 + 3.5</
in.
of D.
in.
= 11.7(42)
or
</
= 33.lin.
A
2010 The McGraw-Hill Companies, Inc. Ail rights reserved. No part of this Manual may be displayed,
distribution to
in.)(11.7 lb)
in.
1.7 1
1 = 42
|
in.)(4.1 lb)
resultant passes through a Point 39.6 in. to the right
(b)
you are using
R = 11.7 lb
+ rf)in.](3.5 lb)«-(£
375.4 + 3.5^
or
-(44
in.)(4.1 lb)
or
it
in
without permission.
266
a
PROBLEM
3.107
A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position.
If b = .5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that
1
the distance from support
equivalent force and
its
A
of action of the equivalent force
to the line
is
maximum,
(/?)
the magnitude of the
point of application on the beam.
U00N
400
~
9
in
400
N
I
1
600
N
N
SOLUTION
4ooM
i
j
liooN
r
toco N
4oo| N
L
~i
R
B.
For equivalence
2.FV
-1 300 + 400 b
:
- 400 - 600 = -R
/?= 2300 -400-
or
ZM A
^400^
:
2
N
0)
* (400) ~(a + b)(6()Q) = -LR
b
1
1000a + 6006 -200
or
2300-400-
10a + 9
6=
Then with
1.5
m
—
2
3
L=
(2)
8
r*
23a
3
Where
(a)
a,
L
are in
m
Find value of a to maximize
L
s
dL
=
3
V
23--al-M0a + 9— -o
,
da
'<n
23
;
©
—
\2
8
a
267
PROBLEM
„_
230
or
184
3.107 (Continued)
80
a
64
3
3
,80
— a + 24n
a +—a~+
9
.
32
>
a~
-0
,,
9
3
2
16a -276a + 1143 =
or
a~
Then
2
276 ± J(-276) - 4(1 6)(.l 143)
—
2(16)
a
or
Since
(b)
^42?
Using Eq.
.#
(1)
=
1
0.3435
-9
tn,
= 2300
m
and
a must be
a
~ 6,9065
less than
9
m
m
6 9065
"
400
m
-4
K-458N
<«
a
or
= 6.9
1
1.5
1
and using Eq.
(2)
,t
=
0(6.9065)
+ 9 - - (6.9065) 2
3
= 3.16
m
23— (6.9065)
R is applied 3.16 m to the right of A. M
distribution to
you are using
it
in
without permission.
268
PROBLEM
Gear
3.108
C is rigidly attached to arm AB. If the forces and couple
shown can be reduced
to a single equivalent force at A,
determine the equivalent force and the magnitude of the
couple M.
SOLUTION
We have
For equivalence
YFX
:
-1 8 sin 30° + 25 cos 40° =
R =10.1511
or
£F
:
or
lb
-1 8 cos 30° - 40 - 25 sin 40° =
ff
=-71.658
v
R,.
lb
R = yl(\QA5\\) 2
Then
Rx
+(7l.65$y
= 72.416
tan# =
71.658
10.1511
R = 72.4
= 81.9°
or
IMA
Also
:
M - (22
in.)(l
8
lb) sin
35° - (32
~(48in.)(25 1b)sin65
M = 2474.8
lb -in.
"%"
81.9°^
25°
= ()
or
A/
= 2061b-ft
<
© 2010 The McGraw-Hill Companies, Inc. Al! rights reserved. No part of this Manual may be displayed,
to teachers and educators permitted by McGraw-Hillfor their individual coarse preparation. Ifyou are a student using this Manual,
distribution
i.n.)(40 lb) cos
lb
you are using
it
in
without permission.
269
j
PROBLEM
**>»»
•12
.
3,109
in.
(Y
U"
A
couple of magnitude
M = 54
lb
•
in.
and the three forces shown are
applied to an angle bracket, (a) Find the resultant of this system of
!
i
|
j
I
j
8in.
forces.
I
--*
(/?)
Locate the points where the line of action of the resultant
AB and
intersects line
line
BC.
I'Mh
SOLUTION
(a)
We have
LF
R = (-1 Oj) + (30 cos 60°)i
:
+ 30sin60°j + (-45i)
= -(30
lb)i
+ (15.9808
lb)
or
(b)
First
R = 34.0
reduce the given forces and couple to an equivalent force-couple system (R,
We have
2Af
fl
:
M
fi
=(54
lb -in)
+ (12
in.)(10 lb)
-(8
R at D
LM
:
-186
lb
•
in
f}
= a(.l 5.9808
/f
) at
28.0° <«
B.
in.)(45 lb)
|\
Then with
M
^
lb
iv_l—
—
S>-
lb)
1*
t>
c
a = U .64
or
and with
R at E
LM
tf
:
-1 86 lb in
= C(30
C - 6.2
or
The line of action of R
below B.
intersects line
AB
1 1
^
lb)
«
Jc
in.
.64 in. to the left of #
and
intersects line
BC 6.20 in.
<
distribution to
~
in.
in
270
10 lb
30
-12
PROBLEM
II
3.110
in.
60'
,
A couple M and the three forces shown are applied to an angle bracket.
Find the moment of the couple if the line of action of the resultant of the
force system is to pass through (a) Point A, (/?) Point B, (c) Point C.
Sin.
M
'-^
,!.3li.
SOLUTION
In each case, must have
{a)
Mf -
+)M*=2MA
=
M+
(1
2 in.)[(30 lb)sin 60°] - (8 in.)(45
M--= 4-48.231
(b)
+)M*=!MB
=
:
A/ + (12
M--= +240
(c)
+)m£=z,mc
=
M
=
lb
M + (12
lb)
=
lb -in.
in.)(10 lb)
-(8
in.)(45 lb)
lb)
- (8
<
M = 240
<
=
in.
in.)(l
M = 48.21b -in.)
in.)[{30 lb) cos 60°]
lb
in.)
=
M= <
©
271
3 10
500
N
PROBLEM
N
XA
u
3.111
on a 700 x 375-mm plate as shown, (a) Find
of these forces, (b) Locate the two points where
of action of the resultant intersects the edge of the
Four forces
act
the resultant
/I
the line
375
mm
plate.
:>/.,
TfiON
-500
mm
200
mm
GOO N
SOLUTION
= (-400 N + 160 N - 760 N)i
+(600N + 300N + 300N)j
= ~(1000N)i + (1200N)j
OOOM
~V*Ot-\
R = ^(1000 N) 2 +(1200 N) 2
= 1562.09 N
1
200
N^
tan<9
1000N
-1.20000
R = 1562N ^50.2°^
-50.194°
M£=£rxF
(t>)
= (0.5m)ix(300N + 300N)j
= (300
N m)k
•
-lOOO N
(300N-m)k=xix(1200N)j
x = 0.25000
x = 250
(300
m
mm
N m) = y\ x (-1 000 N)i
v
= 0.30000
y - 300
m
mm
Intersection
250
mm to right of C and 300 mm above C -4
PROPRIETARY MATERIAL, €> 20)0 The McGraw-Hill Companies?, lac. AH rights reserved. No part of this Manual may be displayed,
Ill
;Vfo
500
N
N
PROBLEM
3.112
-i
/
Solve Problem 3.1
1
1,
assuming
760-N force
that the
is
directed
to the right.
/
s
/
j
/
I
375
;/':
mm
C»
N
7<j0
— 500 mm
200
600
PROBLEM 3.111 Four forces act on a 700 x 375-mm plate as
shown, (a) Find the resultant of these forces, (b) Locate the two
points where the line of action of the resultant intersects the edge
of the plate.
mm
N
SOLUTION
-
v
H
V
c
= (520N)i + (1200N)j
fc>
n
/\
5»h\
+(600N + 300N + 300N)j
-
/
>U
= (-400N + J60N + 760N)i
-
J&
/sw\
TiÔw
C?0O M
R = 7(520 N) 2 +(1200 N) 2 = 307.82 N
1
tan#
/l 200N
..
2.3077
N
520
R = 1308N ^66.6
= 66.5714'
M«
(b)
C
IrxF
= (0.5m)ix(300N + 300N)j
*~
S20 N
- (300 N m)k
•
(300N-m)k = xix(1200N)j
x = 0.25000
x = 0.250
or
(300
m
mm
N m)k - [/i + (0.375
•
m)j]x[(520 N)i + (1200 N)j]
= (1200x'~l95)k
/ = 0,4 1250 m
x' ~ 412.5 mm
or
Intersection. 41.2
©
mm to the right of A and 250 mm to the right of C
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distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,
you are using ii without permission.
273
j
4
8
If
8ft
PROBLEM
300
1.00
240 lb\
j
A truss supports the loading shown.
II)
A
on the
force acting
70
&-~
\\
It
— —8
*\*
180
line
7
\\
8
3.113
ft-
of action with a
Determine the equivalent
and the point of intersection of its
drawn through Points y4 and G.
truss
line
6ft
8
ft
ft
ih
SOLUTION
R = IF
We have
R = (240 lb)(cos70°i - sin 70°j) -
(1
60
v«*o
lb)j
lb
+ (300 lb)(- cos 40°i - sin 40° j) - (1 80 lb)
R = -(147.728 lb)i - (758.36 Ib)j
= 7(147.728) 2 +(758.36) 2
= 772.62
$=
lb
rA
tan
^y
-758.36
tan"
-147.728
= 78.977
We have
YMA = dR
where
ZM
c
^
79.0°
<
= 9.54 ft to the right of A
<
or
R = 773 lb
v
-[240 lb cos 70°](6
-(160 lb)(12
ft)
ft)
- [240 lb sin 70°](4 ft)
+ [300 lbcos40°](6
ft)
-[300 lb sin 40°](20 ft) - (1 80 lb)(8 ft)
-7232.5
lb -ft
-7232.5 lb
d
-758.36
9.5370
-ft
lb
or
ft
©
rf
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you are using
it
without permission.
274
,
PROBLEM
21011)
r
150
lb
A and B
mounted on bracket CDEF. The tension
two belts is as shown. Replace the four
forces with a single equivalent force, and determine where its
line of action intersects the bottom edge of the bracket.
Pulleys
I
]
on each
ill
-
(>
3.114
in
are
side of the
SOLUTION
Equivalent, force-couple at A
We have
due
to belts
-120
IF:
lb
R = 280 lb
on pulley A
-160
lb
=
AL«*HW<P^? ,b
J"?
R =280 lb
/(
A
IM A
We have
-40 lb(2 in.) =
;
MA
M„=801b-.in.J)
Equivalent force-couple at B due to belts on pulley
We have
£F:
B
Rr=f?
+ 50 lb) ^L 25° = R„
(2 1.0 lb
1.
R„=3601b^l!l25
1M B
We have
:
-601b(1.5in.) =
<
MB
M g =90Ib-nO
Equivalent force-couple at
We have
F
LF:
R F = (- 280 lb)] + (360 lb)(cos 25°i + sin 25°j)
= (326.27
lb)i- (127.857 lb)j
R = RF
= 7(326.27) 2 +(127.857) 2
= 350.43 lb
r
d
\
= tan -i
1 in octN
f -127.857
tan"
326.27
-21.399'
or R,,
=R = 3501b^ 2.1.4° «
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you are using
it
without permission.
275
PROBLEM
TM F
We have
:
3.114 (Continued)
MF = ~(280 lb)(6
in.)
- 80 lb
in.
-[(360lb)cos25°](1.0in.)
+[(360
lb) sin 25°](l
2
in.)
- 90
lb
•
in.
M F =-(350.56 lb -in.)k
To determine where a single resultant force will
MF =
clR
intersect line
FE,
y
M,
-127 -857
2.74 18
lb
or
in.
© 20)0 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manna! may be displayed,
to teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,
distribution
d = 2.1A in.
you are using
it
in
without permission.
276
420 N
PROBLEM
240 nun
•loo
n h
fv
50 mm
'-
\
i
•
>
SI!
520
3.115
:
A machine component is subjected to the forces and couples
shown. The component is to be held in place by a single rivet
that can resist a force but not a couple. For P — 0, determine the
location of the rivet hole if it is to be located (a) on line FG,
(b) on line GIL
v
mm
?u.\-m.
:
ISO
mm
50 mm
50 nun
640 nun
SOLUTION
We have
tZO
N
+F
ftaKl
ten-™
b
.R,
4oW«n^
First replace the applied forces
ZFX
Thus
and couples with an equivalent force-couple system
200cos 5° - 20cos 70° + P = Rx
:
1
1
or
tf
ZFr
- 200 sin
:
1
v
5° - 1 20sin 70° - 80
or
R.
ZMC
at G.
:
= (152.1. 42 + .P)N
=R
-244.53
N
- (0.47 m)(200 N) cos 1 5° + (0.05 m)(200 N) sin
5°
.1.
+ (0.47 m)(!20 N) cos 70° -(0.1 9 m)(120 N) sin 70°
-
(0.
3
1
m)(P N) -
+ 40N-m =
(0.59
P-0 inEq.(l):
Now with R at /
N m
-
MG
MG =-(55.544 +
or
m)(80 N) + 42
0.1
3P)N-m
(I)
Setting
ZM G
:
- 55.544
N m = -a(244.53 N)
•
a- 0.227 m
or
and with
SMC
R at J
:
- 55 .544 N
or
(a)
The
rivet
is
0.365
m above G.
(b)
The
rivet hole is
0.227
m to the right of G.
hole
m = -b(\ 52. 42 N)
b = 0.365 m
1
4
distribution to
you are using
it
in
without permission.
Ill
'120
240
N
mm
^
200
50.
80
PROBLEM
3.116
Solve Problem
3.
1 1.5,
assuming that P = 60 N.
PROBLEM
N
3.115 A machine component is subjected to the
and couples shown. The component is to be held in
place by a single rivet that can resist a force but not a couple.
For P = 0, determine the location of the rivet hole if it is to
be located (a) on line FG, (b) on line GIL
forces
520 mir
I
^.O-Nvm.
^kd
mm
180
mm
50
mm
-
640
.50
mm
mm
SOLUTION
See the solution to Problem 3.115 leading
to the
development of Equation (1)
MG = -(55.544 +
and
/?,
=(152.142 +
0. 1 3P)
/*)
N m
•
N
P = 60 N
For
We have
Rx
=(152.142 + 60)
= 2.12.14N
A/c =-[55.544 + 0.13(60)]
= -63.344 N-m
Then with
ZMG
R at /
-63.344
:
a = 0.259
or
and with
XMG
R at J
:
-63.344
(b)
m above G.
The rivet hole is 0.259 m to the right of G.
The
rivet hole is
m
N-m = -£(212.14 N)
b = 0.299
or
(a)
N-m = -o(244.53N)
m
A
0.299
A
©
278
PROBLEM
3.117
A 32-lb motor is mounted on the floor. Find the resultant of the
weight and the forces exerted on the
the line
of action of the resultant
and determine where
belt,
intersects the floor.
f
2
vi
in.
i2
r
in.
eJpM'--
-
)
Si
SOLUTION
We have
l4-OSth30°
£F:
*>
(60
lb)i
- (32 lb) j + (1 40 lb)(cos30°i + sin 30° j) = R
R = (181. 244 lb)i + (38.0 lb)j
I4-Ocos30°
or
ZM
We have
R = ]85.2Ib^1J.84°^
XM =xRy
:
-[(140 lb)cos30°][(4 + 2cos30°)in.] -[(140 lb)sin30°][(2
~(60
lb)(2 in.)
= x(38.0
in.) sin
30°]
lb)
1
(-694.97 -70.0 -120)
in.
38.0
x
and
= -23.289 in.
Or, resultant intersects the base (x axis) 23.3
in. to
the left of
the vertical, centerline (y axis) of the motor.
© 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed
distribution to
you are using
it
in
without permission.
279
:
PROBLEM
»-»('-$)
3.118
follower AB rolls along the surface of member C,
As
F perpendicular
a constant force
to the surface, (a)
it
exerts
Replace
with an equivalent force-couple system at the Point
F
D
obtained by drawing the perpendicular from the point of
contact to the
x
axis, (b)
For a
the value of x for which the
couple system
at
=
1
m and b — 2 m, determine
moment of the
equivalent force-
D is maximum.
SOLUTION
(a)
The
slope of any tangent to the surface of member
dy
dx
Since the force
F
__
d
C is
~2b
dx
*
v
;
Wb(.-s)
perpendicular to the surface,
is
2
tana
2b\x
For equivalence
F=R
IF:
ZMD
:
(F cos a)(yA ) =
MD
where
ZbX
2bx
cos cr
2 2
^l(a
y A ~b
)
+(2bxy
'<4'
.3\
2Fb<
M,
Va 4 +4* 2 jc 2
Therefore, the equivalent force-couple system at £>
is
R = F7
'« 2
^
tan"
v
2to y
„3>
2.F//
M
2 ..2
V^+4//x
©
2010 The McGraw-Hill Companies, Inc. All rights reserved. A'o /km-/ o///ik Mmim/ /hot te displayed,
PROPRIBTARY MATERIAL.
you are using
it
without permission.
280
1
PROBLEM
(/?)
To maximize M,
the value of x
must
3.118 (Continued)
—— =
satisfy
dx
a = 1 m,
where, for
8F(jk-*
M
/i
/?
=2m
3
)
+ 16jc 2 (1-3jc 2 )-(x-;c 3 )
1
(32x)(l
+ 16x 2 )" 1/2
= 8Fdx
(1
(l
+ .16;r)
+
1
2
6,y )(1
- 3x 2 ) - 6x(x - x 3 ) =
4
32x +3x
or
-3±V9-4(32)(-l)
0.13601
]
m?
and
2
-l=0
-0.22976 nV
2(32)
Using the positive value of .r
x
= 0.36880 m
©
or
x = 369mm
-^
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you are using
it
without permission.
281
PROBLEM
\i
,
mm
200
::%,
3.119
Four forces are applied to the machine component
shown. Replace these forces by an equivalent
force-couple system at
ABDE as
A
'
20
mm
fijli 300 N
'
25!
X \^f7-.^MM0'^
!
f
j
:
100
?;"|
1
60
i
i
i
m
A
mm
SOLUTION
R = -(50 N)j - (300 N)i -
(1
20 N)i - (250 N)k
R = -(420 N)i - (50 N)j - (250 N)k
rw
=(0.2m)i
rD
=(0.2m)i + (0.16m)k
rB
= (0.2 m)i - (0. 1 m) j + (0.
1
-(3oon)i
m)k
6
-(ztorji
M*=r ,x[-(300N)i-(50N)j]
;
+ rD x (-250 N)k + r x ( - 20 N)i
1
i
0.2
m
-50
»
0.2
m
-(1.0
N
k
J
m
N
0.16m
-250
N
k
J
-0.1m
0.16
m
N
-120
=
i
0.2
N
-300
+
k
J
•
m)k + (50 N m)j - (1 9.2 N m)j - (1 2 N m)k
•
Force-couple system at A
is
R = -(420 N)i - (50 N)j - (250 N)k M*
©
= (30.8 N m)j -- (220 N m)k
•
•
<
you are using
it
without permission.
2S2
PROBLEM
225
Two
i» hi
50-mm-diameter pulleys are mounted
AD. The belts at B and C lie in
vertical, planes parallel to the yz plane. Replace
the belt forces shown with an equivalent forceon
^r
i(>"
(
'i[}
'<.Qj
:/|5
145
3.120
i!>*
s
]
line shaft
couple system at A.
^r ii0N
\
N
mm
ISO
SOLUTION
Equivalent force-couple at each pulley
Pulley B
R = (i 45 N)(-cos 20° j + sin 20°k) - 2
/f
= -(35
1
.26 N)j
1
Nj
5
+ (49.593 N)k
M B = -(21 5 N-.I45 NX0.075 m)i
2lS>4
= -(5.25N-m)i
Pulley
C
R c = (1 55 N + 240 N)(-sin
1
0°j
- cos 0°k)
1
tS5e*
= -(68.591 N)j- (389.00 N)k
M c = (240 N - 55 N)(0.075 m)i
(toi
&c
1
= (6.3750 N
Then
R=R
M. A =
/;
+
•
2HOfi
m)i
Rc = -
(4.1
9.85 N)j
R = (420N)j-(339N)k <
or
- (339.4 l)k
M B + M c + xm x R s + xcu x R c
i
- -(5.25 N m)i + (6.3750 N m)i +
•
k
J
N-m
0.225
-351.26
*
J
49.593
k
N-m
0.45
-68.59:
-389.00
= (1.1 2500 N m)i + (1 63.892 N m)j - (1.09.899 N m)k
•
•
or
MA =
©
•
(.1
.
1
25
N
m)i + (1 63.9
N
•
m)j - (1 09.9
N
•
m)k
<
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283
PROBLEM
3.121
While using a pencil sharpener, a student applies the forces and
couple shown, (a) Determine the forces exerted at B and C knowing
..1!
that these forces
and the couple are equivalent
system at A consisting of the force
R
and the couple
X \ +(1.0 lb
A
the corresponding values of R and
M =M
a force-couple
to
R = (2.61b)i +./? ,j~(0.71b)k
>
ft)j
•
M
x
- (0.72
lb
•
(b) Find
ft)k.
.
SOLUTION
(a)
From
the statement of the problem, equivalence requires
or
IF
B+C=R
ZFX
# + C, = 2.6
Y,F
y
-C\,=Ry
XFZ
-C
=-0.7
(rB/A
xB + M. B ) + raA xC=M«
and
IM A
or
LM
z
:
lb
(1)
(2)
Cz
or
1b
1b
'1.75
V
:
(llb-ft)
+
(C V ) = M,
ft
12
ZM y
'3.75
1.75
ft
:
(/?,)
+
12
tt|(C v
12
(3)
H-|—
'
I
(l
(0.7 lb)
Using Eq. (1)
1
lb
•
ft
+ 1.75(2.65,) = 9.55
3.755,
or
Bx = 2.5
and
Cx
—
'35
ZM„
^
(C,.)
ft
12
}
lb
=0.1. lb
--0.721b- ft
)
Cy = 2.4686
or
B = (2.5
lb)i
lb
C = (0.
1
000
lb)i
- (2.47
lb) j
- (0.700
lb)k
<
R y =*-2Al\b <
Eq.(2)=>
'
Using Eq.
=
12
3.75£ T + I.75C V =9.55
or
(b)
=0.7
(3)
1
.75^
+
(2.4686)
=
M
x
12
or
©
A/, =1.360 lb
-ft
^
284
j
PROBLEM
3,122
C
A
mechanic uses a crowfoot wrench to loosen a bolt at
The
mechanic holds the socket wrench handle at Points A and B
and applies forces at these points. Knowing that these forces
are equivalent to a force-couple system at C consisting of the
force C - (8 lb)i + (4 lb)k and the couple
c = (360 lb in.)i,
determine the forces applied at A and at B when A — 2 lb.
M
•
SOLUTION
We have
A + B==C
F- A +B =8 lb
£F:
or
X
0ib
Bx =-(Ax +8)b)
0)
Cm.
IF-
A. +
Ay =-By
or
LFZ
# =0
2lb + ^=4lb
:
B =2
SM C
i
r8/c
:
J
*,
lb
(3)
xB + r,/c xA = M c
k
i
+
2
8
Bx
(2)
8
4
2
j
k
8
4-
lb -in.
= (360
lb -in.)!
2
(25 y - SAy )i + (25, -16 + %AX - 1 6)
or
+(85J ,+8^)k = (360lb-in.)i
From
i-coefficient
j- coefficient
k-coefficient
2Z?j;
-8^, =360
-2.fi,+8/f v
(5)
8.^+8^=0
(6)
iMM
2010 The McGraw-Hill Companies, Inc. AH rights reserved. JVo/wr/ o/z/ife
may 6e displayed,
distribution to
(4)
=32UVin.
you are using
lb -in.
it
in
without permission.
285
PROBLEM
From Equations
(2) and (4):
2B y
3.122 (Continued)
-8(--5,)
Z?
From Equations
(1)
and
(5):
v
= 360
=361b
A r =36 lb
2("4~8) + 8,4 t =32
4 = 1.6 lb
From Equation
Bx
(1):
=-(1.6 + 8) = -9.6
lb
A = (1 .600 lb)i - (36.0 lb)] + (2.00 lb)k <
B = -(9.60 Ib)i + (36.0 lb) j + (2.00 Ib)k
©
<4
you arc using it without permission.
286
PROBLEM
64
3.123
in.
As an
lfl}
96
?
BC
adjustable brace
is
shown
force-couple system
used to bring a wall into plumb, the
exerted on the wall. Replace this
is
force-couple system with an equivalent force-couple system at
in.
if
R = 2 1 .2
and
lb
M=
1
3.25 lb
•
A
ft.
SOLUTION
5
We have
R - R^ - RXBC
2F:
(42
,
where
in.)i
- (96 in.) j - (1 6 in.)k
106
ra =
or
We have
R^,
£MV,:
rc//J
where
v<>
(42i
S*
- 96 Jj - 6k)/
\\
"v,
j
\\ j
1
j06
c
= (8.40 lb)i - (1 9.20 lb)j - (3.20 lb)k
xR + M=M
xCIA
A
in.
<
-4
= (42 in.)i + (48 in..)k =
—
(421
+ 48k)ft
= (3.5ft)i + (4.0ft)k
R = (8.40 lb)i - (19.50 lb)j - (3.20 lb)k
=
-42 i + 96 j + 1 6k
106
= -(5.25 lb
i
Then
•
4.0
3.5
M. A
-19.20
= (71.55 lb
•
ft)j
+ (2 lb
•
ft)k
k
j
8.40
+ (1 2 lb
ft)i
lb-ft
+ (-5.25i + 12j-f
2k)lb-ft
=M
i4
-3.20
•
ft)i
+ (56.80 lb
or
©
M ^ - (7
ft)j
1
- (65.20
.6 lb
•
ft)i
lb
ft)k
+ (56.8 lb
•
ft)j
- (65.2
lb
-
ft)k
<
2010 The McGraw-Hill Companies, Inc. All rights reserved, to part of this Manual may be displayed,
287
j
PROBLEM
3,124
mechanic replaces a car's exhaust system by firmly clamping the catalytic converter FG to its mounting
brackets
and ./ and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB,
he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent
force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise
relative to muffler DE, as viewed by the mechanic.
A
H
«
0.33 .<f
0.14 in
('
:-:i
SOLUTION
(a)
RÂ + B
SF:
(100 N)(cos 30°j - sin 30° k) - (1 1 5 N)j
-(28.4N)j-(50N)k
SM. D
and
:
MD
where
l
'aid
k â +ybid x ^b
AID
-(0.48 m)i
- (0.225 m) j + (1
B/D
-(0.38 m)i
+ (0.82 m)k
.
1.2
m)k
Then
j
k
-0.225
1.1.2
cos 30°
-sin 30°
i
M n =100
1
0.48
00[(0.225 sin 30° - 1
.
1
i
+ 115
2 cos 30°)i
+ (-0.48 cos 30°)k] + 1 1 5[(0.82)i
J
-0.38
k
0.82
„1
+ (-0.48 sin
30°)
+ (0.38)k]
8.56i-24.0j + 2.13k
any form or by any means, without the prior written permission of the publisher, or used beyond the. limited
distribution to
you are using
it
in
without permission.
288
PROBLEM
The equivalent force-couple system at D
3.124 (Continued)
is
R = -(28.4 N)j - (50.0 N)k
MD = (8.56 N
(b)
Since
(MD ) Z
is
positive, pipe
•
m)i - (24.0
m)j + (2. 1 3
N m)k
•
<
A
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teachers and educators permitted by McGraw-Hillfar their individual course preparation. Ifyou are a student using this Manual,
distribution to
•
CD will tend to rotate counterclockwise relative to muffler DE.
you are using
N
<
it
in
without permission.
2 SO
PROBLEM
3.125
For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple
at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF
tends to rotate clockwise or counterclockwise, as viewed by the mechanic.
system
0..M
0.10
0.30
m
0.33
m
m
m
0.56 in
o
\
SOLUTION
(a)
R=A+B
SF:
= (100 N)(cos 30°j - sin 30° k) - (1 1 5 N)j
= -(28.4N)j-(50N)k
M
and
where
M^r^xA + r^xB
'A/F
y
BIF
-(0.48
m)i - (0.345 m)j + (2.10 m)k
-(0.38
m)I -
(0.
1
2 m)j + (1 .80
m)k
Then
J
k
-0.345
2.3.0
cos 30°
-sin 30°
i
M;, =100
MF =
1
0.48
00[(0.345 sin 30° - 2. 1
+ (-0.48 cos 30°)k] +
1
i
+ 115 -0.38
5[(1 .80)1
k
0.12
1.80
-1
cos 30°)i
1
J
+ (-0.48 sin
30°)j
+ (0.38)k]
= 42.4i-24.0j + 2.13k
©
2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manned may he displayed,
you are using
it
without permission.
290
PROBLEM
at
3.125 (Continued)
F is
R = -(28.4N)j~(50N)k
M
(b)
Since
(MF ) Z
positive, pipe
is
distribution to
= (42.4 N m)i - (24.0
•
N
•
m)j + (2. 1 3
N
•
m)k
EF will tend to rotate counterclockwise relative to the mechanic.
<
-4
CO 20
The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed
teachers and educators permitted by McGraw-Hillfor their individual course preparation. you are a student using
If
this Manual,
you are using
,,
A
it
1
in
without permission.
291
PROBLEM
3.126
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the
chuck and bit parallel to they axis. The assembly was then rotated
25° about the y axis and 20° about the centerline of the horizontal
arm AB, bringing it into the position shown. The drilling process
was started by switching on the motor and rotating the handle to
bring the bit into contact with the workpiece. Replace the force
and couple exerted by the drill press with an equivalent force-
couple system at the center
O of the base of the vertical column.
SOLUTION
R=F=
We have
(l 1
lb)[( sin
= (3.4097
R = (3.4
or
M
We have
1
20° cos 25°)]i - (cos 20°) j - (sin 20° sin 25°)k]
lb)i
- (1 0.3366
lb)j
- (1 .58998 lb)k
- (1 0.34 lb) j - (1 .590 Ib)k
lb)i
=rB/0 x¥xM. c
where
rm)
= [(l4in.)sin25°]i + (15
in.)j
+ [(14in.)cos25°]k
- (5.9167 in.)i + (15 in.)j + (12.6883 in.)k
M c = (90 lb in.)[(sin 20° cos 25°)i - (cos 20°)j - (sin 20° sin 25°)k]
•
= (27.898
lb in.)i
•
- (84.572
lb
•
in.)j
- (13.0090 ib
•
in.)k
p=
M.
i
J
k
5.9167
15
12.6883
3.4097
-10.3366
1.58998
+(27.898 - 84.572 - 13.0090)
= (1 35.202 lb
•
in.)*
- (3
1
ll
tb
lb -in.
lb
-
in.
.90 1 lb in.) j
•
or
- (1 25.3
M
1
3 lb in.)k
= (1 35.2 lb
in.)i
- (3 1 .9 lb
in.)j
- (125.3
lb in.)k
-
PROPRIETARY MATERIAL. €) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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this Manual,
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you are using
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without permission.
292
PROBLEM
3.127
Three children are standing on a 5x5-tn raft. If the weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively, determine the magnitude and the point
of application of the resultant of the three weights.
SOLUTION
F
We have
XF:
f
F,+Fa +FC =R
-(375 N)j - (260 N)j - (400 N)j = R
-(1035N)j-R
or
We have
^.v
:
FaM + F*(zb) + fc(zc) = R(zD
(375 N)(3
zD
We have
Z,MZ
:
= 1035N <
)
m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1 035 N)(z )
D
= 3.0483 m
or
Zf)
= 3.05 m
<
PA (x A ) + FB (xB ) + Fc {xc ) = R(xD )
375 N(l m) + (260 N)(
xD = 2.5749
1
.5
m
m) + (400 N)(4.75 m) = (1035 N)(.^ )
r
x n = 2.57
m <
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/?
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it
in
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293
PROBLEM
3.128
0.5 in
Three children are standing on a 5x5-m raft. The weights
of the children at Points A, B, and C are 375 N, 260 N, and
400 N, respectively. If a fourth child of weight 425 N
climbs onto the raft, determine where she should stand if
the other children remain in the positions shown and the
of action of the resultant of the four weights
through, the center of the raft.
line
0.25 ill"
P
'
0.25
is
to pass
m
SOLUTION
We have
L.F:
FA +FB +$C
=R
-(375 N) j - (260 N)j - (400 N) j - (425 N) j -
R
R=-(1460N)j
We
1MX FA (zA
have
:
)
+ FB (zB ) + Fc (zc ) + FD {z D ) = R{zH )
(375 N)(3
SMZ
We have
:
m) + (260 N)(0.5 m) + (400 N)(4.75 m)
+(425 U)(zD )
= (1460 N)(2.5 m)
z D =1.16471
m
or
=1.165
m M
x n = 2.32
m A
Zt,
FA (x A ) + FB (xa ) + Fc (xc ) + FD (x D ) = R{xH )
(375 N)(l
m) + (260 M)(l .5 m) + (400 N)(4.75 m)
+(425 N)(x/} ) = (1460 N)(2.5 m)
x D = 2,3235
m
or
20] The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior mitten permission of the publisher, or used beyond the limited
using this Manual,
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are. a student
©
you are using
it
without permission.
294
PROBLEM
"!
ir
2.5
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine the magnitude and the
point of application of the resultant of the four wind forces
when a -\ ft and 6 = 1.2 ft.
ft
on.'ij
feii
3.129
Ufi
)}•
1:
l'i><m
5
8
f:-.i
If"
ft
<)ii
5.5
ft
SOLUTION
We have
1
.»
Soft,
Assume
that the resultant
R is applied at Point P whose coordinates are (x,y, 0).
Equivalence then requires
E/<;
:
- 05 - 90 - 60 - 50 = -R
1
1
or
XM
X
:
(5
ft)(1
+ (5.5
05
lb)
- (1
ft)(50 lb)
= -j/(405
y = -2.94
or
ZMy
:
(5.5 ft)(105 lb)
+
+ (12
(22.5 ft)(50 lb)
+ (3
it)(l
60
lb
4
lb)
lb)
ft
ft)(90 lb)
= -j(405
+ (14.5
ft)(160 lb)
lb)
x~ 12.60 ft
or
R acts
ft)(90 lb)
R - 405
1
2.60
ft
to the right
of member AB and 2.94
ft
below member BC.
C3 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,
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it
in
without permission.
295
PROBLEM
Four signs are mounted on a frame spanning a highway,
and the magnitudes of the horizontal wind forces acting on
the signs are as shown. Determine a and b so that the point
of application of the resultant of the four forces is at G.
H).5ft
'>>><:.. p.
5
3.130
i
ft
9
5.5
S
ft
it
SOLUTION
Since
R acts at G, equivalence then
i
equires that
G :EM V
at
1M
-(a + 3)
:
+ (2.5
of the applied system of forces also be zero Then
.
ft
x (90
ft)(50 lb)
lb)
+ (2
ft)(l
05
lb)
=
or
XMy
:
-(9ft)(105ft)- -(14.5-6)
ft
x (90
a~ 0.722
ft
<
b-= 20.6
ft
A
lb)
+ (8ft)(501b) =
Ol
PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
you are using
it
without permission.
2%
PROBLEM
224 X
3.131*
A
group of students loads a 2x3.3-m flatbed trailer with two
x 0.66 x 0.66-m boxes and one 0.66 x 0.66 x .2-m box.
Each of the boxes at the rear of the trailer is positioned so that
0.66
.1,5
1
it is aligned with both the back and a side of the trailer.
Determine the smallest load the students should place in a
second 0.66x0.66x1. 2-m box and where on the trailer they
should secure it, without any part of the box overhanging the
sides of the trailer, if each box is uniformly loaded and the line
of action of the resultant of the weights of the four boxes is to
m
pass through the point of intersection of the centerlines of the
trailer
and the
axle. {Hint:
placed either on
its
side or
Keep
on
its
in
mind
that the
box may be
end.)
SOLUTION
For the smallest weight on the
so that the resultant force of the four weights acts over the axle at the
trailer
intersection with the center line
of the
trailer, the added 0.66x0.66x1. 2-m box should be placed adjacent to
one of the edges of the trailer with the 0.66 x 0.66-m side on the bottom. The edges to be considered are
based on the location of the resultant for the three given weights.
We have
EF
- (224 N)j - (392 N)j~(l 76 N) j = R
:
R = -(792N)j
£M
We have
Z
:
-(224N)(0.33 m)-(392 N)(1.67 m) - (176 N)( 1.67 m) = (-792 N)(j)
x R =1.29101
XMX
We have
:
(224 N)(0.33
m
m) + (392 N)(0.6 m) + (1 76 N)(2.0 m) = (792 N)(z)
z„= 0.83475 m
From
the statement of the problem,
it
is
known
that the resultant
of
R
from the original loading and the
W passes through G> the point of intersection of the two center
Thus, IM a —
Further, since the
load W
to be as small as possible, the fourth box should be placed as
lightest load
lines.
lightest
as possible without the
is
box overhanging the
(0.33
trailer.
m<x<
1
0.
far
from
G
These two requirements imply
m)(l .5
m < z < 2.97 m)
©
you are. using
it
without permission.
297
PROBLEM
XL =
With
G:
at
ZMZ
:
3.131* (Continued)
0.33
m
o- 0.33) mxWL
-(1.29101-1)
mx(792N) =
wL ~ 344.00 N
or
Now must check
if this is
physically possi ble,
G:
at
IM
:
V
(Zz - 1 .5)mx344 N) - (1 .5 - 0.83475)m x (792 N) =
^ -3,032 m
or
which
is
not acceptable.
^ = 2.97 m:
With
G:
at
I,M X
(2.97
- 1 .5)m x WL - (1 .5 - 0.83475)m x (792 N) =
)
K = 358.42 N
or
Now check if thi s is physically possible
G:
at
XM
-
Z
^)mx(358.42 N)-(l .29101 -l)mx(792 N) =
^ -0.357 m
or
The minimum
And
(1-
it is
weight, of the fourth
placed
box
WL = 358 N <
is
on end (A 0.66x0.66-m
side
down) along
side j42? with the center of the
box 0.357
m
<
from side AD.
PROPRIETARY MATERIAL, ©
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20
reproduced or distributed in any form or by any means, without the. prior written permission of the publisher, or used beyond the limited
1
298
PROBLEM
224
3.132*
3. 31 if the students want to place as much
weight as possible in the fourth box and at least one side of
Solve Problem
N
the
392
1
box must coincide with a side of the
trailer.
N
PROBLEM
A group of students loads a 2x3.3-m
two 0.66 x().66x().66-m boxes and one
0.66 x 0.66 x 1 .2-m box. Each of the boxes at the rear of the
trailer is positioned so that it is aligned with both the back
and a side of the trailer. Determine the smallest load the
students should place in a second 0.66x0.66x1 .2-m box
and where on the trailer they should secure it, without any
part of the box overhanging the sides of the trailer, if each
box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through.
the point of intersection of the centerlines of the trailer and
the axle. (Hint: Keep in mind that the box may be placed
either on its side or on its end.)
3.131*
flatbed trailer with
1..S
m
SOLUTION
First replace the three
known
loads with a single equivalent force
R applied at coordinate (XR
,
0,
Z
/(
)
£/y
Y
224-392-176 = -/?
R = 792 N
or
IM
V
:
(0.33
m)(224 N) + (0.6 m)(392 N)
+ (2m)(176N) = Zfl(792N)
z R -0.83475
or
ZMZ
:
m
A*»-tr
-(0.33 m)(224 N)-(l .67 m)(392 N)
- (1 .67 m)(176 N) = xs (792 N)
x D =1.29101
or
From
the statement
of the problem,
it
is
known
that the resultant
through G, the point of intersection of the two center
£M C
m
lines.
of
R
and the heaviest loads
W#
passes
Thus,
=
W/y is to be as large as possible, the fourth box should be placed as close to G as possible while
keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are
Further, since
0.6
m
or
2.7
©
m
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou area student using this Manual,
299
PROBLEM
3.132* (Continued)
Now consider these two possibilities
With x„ -0.6 m:
G:
at
X.M,:
(\~0,6)mxWH - (1.29 101 -l)mx (792 N) =
W„ = 576.20 N
or
Checking
if this is physically possible
G:
at
SM
V
:
{z H
- 1.5)mx (576.20 N)-(1.5-0.83475)mx(792 N) =
zfI =2.414
or
which
is
m
acceptable.
With z n = 2.7
m
G:
at
TM
or
Since this
is
less than the first case, the
X
:
Wn -
(2.7
-
WH
= 439 N
1
.5)
(1 .5
maximum weight of the
- 0.83475)m x (792 N) =
fourth
box
is
WH
and
it
is
placed with a 0.66x1 .2-m side down, a 0.66-m edge along side AD, and the center 2.41
are.
using
m
20)0 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
any farm or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to
<
<
from side DC.
you
= 576 N
it
in
without permission.
300
PROBLEM
3.133
Three forces of the same magnitude P act on a cube of side a as
shown. Replace the three forces by an. equivalent wrench and
determine (a) the magnitude and direction of the resultant force
R, (/>) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
Force-couple system at O:
!
i
^
r = n + pj + Pk = P(i + j + k)
Mj =
c/j
x Pi + ak x Pj + c/i x Pk
y
Pak-Pai-Paj
m:
Since
is
R and M^
the diagonal
have the same
OA.
-Pa(i
<?;- -
+ j + k)
form a wrench with
direction, they
M, =M^.
Thus, the axis of the wrench
We note that
cos
X
/?
M
}
Pitch
- cos
- cos
= Pfi
X
=
Z
#,,
—
—
t=t
= ez =
= ~y=r
54.7
C
= Mo = -PoV3
=p
r^pS
M,
-PaS
tf
PV3
ex
-—a
êY ~e ^5A.T
z
-a
(6)
(c)
*H
Axis of the wrench
is
4
diagonal 0yf
©
301
PROBLEM
3.134*
A piece of sheet metal is bent into the shape shown and is acted
upon by three forces, if the forces have the same magnitude P,
replace them with an equivalent wrench and determine (a) the
magnitude and the direction of the resultant force R,
of the wrench, (c) the axis of the wrench.
\Qj.\
(h) the pitch
SOLUTION
First
reduce the given forces to an equivalent force-couple system (R,
Mj)
at the origin.
We have
£F:
-P\ + I >j + Pk =
/K|FA
R = Pk
or
1M
:
-i*P)\ +
-(*/>)!
M.g
or
(a)
/?
Then
for the
+ -«/>
(
=qp(-\- j +
*
= "2
k
1
=
^
Kx
J^
r
V
wr ench
R=P
and
/v
cos
or
(b)
axis
6>
v
V
~
<
~K
=
cos
^9O°
6y =
0,,
=90°
cos
g
9Z =
1
<
=O°
Now
= k -opf -l-j+|kj
= *,/>
2
Then
^- M i-!*
rt
P
m
.
r>~
,-,
-4
2
/>
CO 2010 The McGraw-Hill Companies, Inc. Ail rights reserved. A'o /*>/•/ o//Aw Manual may be displayed,
302
PROBLEM
(c)
The components of
assumed
to intersect
the
wrench are (R,
ihexy plane
at Point
M
z
3.134* (Continued)
M
),
where M, =
M
x
X axi&>
and the axis of the wrench
is
=rfi xR*
M =M
Where
(
Q whose coordinates are (x,y, 0). Thus require
2
xM,
Then
aP\
-i-j + ~k -~aPk=;(xl + y$) + Fk
Equating coefficients
i
:
j:
The
axis
of the wrench
is parallel
- aP — yP
or
y — -a
~aP~-xP
or
x~a
to the z axis
and intersects the xy plane
©
at
x = a, y — —a.
-^
20 JO The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
303
m
PROBLEM
The
3.135*
forces and couples
of sheet metal
is
shown
are applied to
two screws as a piece
fastened to a block of wood. Reduce the forces and
the couples to an equivalent
wrench and determine (a) the resultant
where the axis of
force R, (b) the pitch of the wrench, (c) the point
wrench
the
100
mm
SOLUTION
First,
reduce the given force system to a force-couple system.
-(20N)i-(15N)j = R
We have
XF:
We have
IM a
:
# = 25N
x F) + XM C = Mg
S(r
M J = -20 N(0.
1
m)j - (4
N m)i -(IN- m)j
•
= -(4N-m)i~(3N-m)j
R = -(20.0 N)i-(1 5.0 N)j <
(b)
We have
M,
=V M 5
(-0.81
/U
- 0.6j)
•
[~(4
N
•
- (3 N
m)]i
m)j]
5N-m
p=
Pitch
M,
R
N
5
25
N
- 0.200
m
or
(c)
From above note
p = 0.200
m <
that
M,
=M
/e
Therefore, the axis of the wrench goes through the origin.
The
line
of action of the wrench
lies in
thcxy
plane with a slope of
©
2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed
reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limited
304
j
PROBLEM
filh'in.
3,136*
The
forces and couples shown are applied to two screws as a piece
of sheet metal is fastened to a block of wood. Reduce the forces and
I .11 lit
wrench and determine (a) the resultant
the pitch of the wrench, (c) the point where the axis of
the couples to an equivalent
R, (b)
wrench intersects the xz plane.
force
the
SOLUTION
First,
reduce the given force system to a force-couple at the origin.
We have
IF:
-(101b)J-(111b)j
=R
R=-(211b)j
IM
We have
:
Z(r x F)
+ ZM C = M J
i
k
J
M£ =
20
i
lb- in.
k
j
+
-.15 lb in. - (12 lb- in)
-11
-10
= (351b-in.)i~(12lb-in.)j
R = -(21lb)j
(b)
M,=VM»
We have
kR =
or
R = -(21.0 lb) j
or
p~ 0.571 in.
R
R
= (-j)-[(351b-in.)i-(121b-in.)j]
= 1 2 lb
•
in.
and
M
,
p=«L = Hî = 0.57I43i„.
and pitch
R
in .) j
211b
distribution to
= -(1 2 lb
in
305
PROBLEM
(c)
Mg*=M,+M 2
M 2 =Mj-M,
We have
M
Require
(35
lb
2
in.)i
35i
From
35
i:
z
From
axis of the
=(35lb-in.)i
=re/0 xR
= (jri + zk)x[-(2Ilb)j]
= -(21x)k + (21z)i
= 21z
= 1.66667 in.
= -21x
k:
z
The
3.136* (Continued)
wrench
is
=
parallel to the y axis
and
intersects the xz plane at
x — 0, z
~
=
1
.667
in.
^
306
\
PROBLEM
IJ
0.1
30
m
3.137*
N'rtt
bolts at A and B are tightened by applying the forces
and couples shown. Replace the two wrenches with a
single equivalent wrench and determine (a) the resultant
R, (b) the pitch of the single equivalent wrench, (c) the
point where the axis of the wrench intersects the xz plane.
Two
m M
0.6
\
jum
SOLUTION
M?
3
\RJ
y
2
*
First,
We have
EF:
and
£M
-(84N)j~(80N)k = R
:
i
I(r
x F) + SM C =
k
J
0.6
.1
= ll6N
MJ
k
i
J
0.4
0.3
+
rt
+ (-30j-32k)N-m =
;
80
84
Mi = -(1 5.6 N
•
m)i
+ (2
N
m)j - (82.4
N m)k
R = -(84.0 N)j- (80.0 N)k ^
(a)
(b)
M
.M,=^-Mg
We have
*,,=
-84j-80k
[-(15.6
N
•
m)i
+ (2 N m)j-(82.4 N m)k]
•
•
116
55.379
X
X
M,
Then pitch
P
R
N-m
116 N
55.379
©
0.47741m
or
p = 0.477 m <
teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual
distribution to
•
M =M AR --(40.102 N-m)j-(38.192N-m)k
and
you are using
N m
it
in
without permission.
307
PROBLEM
(c)
3,137* (Continued)
mJ=Mj+m
M = JVf£~M., = [(-J.5.6i + 2j-82.4k)-(40.102j-38.I92k)]N-m
We have
2
2
= -(15.6 N
m)i + (42.102
•
M
Require
(-
1
5.6i
+ 42.
1
N m)j- (44.208 N -m)k
•
= rQIO X R
2
02j - 44.208k)
= (xi. + 2k) x (84 j - 80k)
= (84z)i + (80x)j-(84x)k
From
-15.6
i:
or
From
= 842
z
= -0.185714
2
= -0.1 857
or
jc
axis <if the
m
-44.208 = -84x
k:
x = 0.52629
The
m
wrench
m.
= 0.526 m
intersects the xz plane at
x = 0.526
m y=
2
= -0.1857 m
<
you
are.
using
it
without permission.
308
PROBLEM
Two
3.138*
A and B
bolts at
by applying the forces and
are tightened
couples shown. Replace the two wrenches with a single equivalent
wrench and determine
wrench
of the
where the axis of the
(a) the resultant R, (b) the pitch
single equivalent wrench, (c) the point
SOLUTION
E^b^&i
First,
(a)
reduce the given force system to a force-couple
We
have
-(26.4 lb)k -(17
ZF:
lb)|
at the origin at B.
— i+—
j
=
|
R
R = -(8.00 lb)i and
7?
We have
XM B
:
rAIB
5.00 lb) j
- (26.4 lb)k
<
= 31.41b
x FA +
i
i\r
(1
MA + MB = M%
k
J
-10
220k -2381
—
i
i
+ -^jj = 2641 - 220k - 1 4(8i + J 5 j)
]7
j7
™26.4
M J = (152 lb
(/;)
We have
•
M^X.-M.^
in.)i
- (21
XR
lb
•
in.)j
- (220 lb
•
in.)k
R
R
_-8.()Oi-15.00j- 26.4k
[(1
52
ib in.)i
•
-(210 lb
•
in.)j
- (220
ib
•
in.)k]
31.4
= 246.56
lb- in.
PROPRIETARY MATERIAL. ©2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means; without the prior written permission of the publisher, or used beyond the limited
you are using
it
without permission.
309
PROBLEM
M, =
and
Then
(c)
M XR = -(62.8
M,
R
—
246.56
M
2
lb -in.
•
- (1 1 7.783 lb
in.)i
•
in.)j
- (207.30 lb
ocoo in.
= „7.8522
•
in.)k
.
or
31.41b
MJ=M,+M
We have
8 lb
1
X
p- —l-
pitch
3.138* (Continued)
in.
^
A
z~ 14.32 in,
A.
» = noc
7.85
,
'
2
= Ml ~ M, = (1 52i - 21 Oj -220k)™ (-62.8.181-1 17.783 j- 207.30k)
= (214.82
lb
in.)i
- (92.21. 7
M
Require
2
lb
in.)j
•
- (12.7000
lb
•
in.)k
p xR
=r
i
*
J
214.821- 92.2 17J-1 2.7000k = x
™8
z
-15
-26.4
= (15z)i-(8z)j + (26.4*)j-(15*)k
From
i:
From
k:
The
214.82
= 152
-12.7000 = -15*
axis of the
wrench
intersects the
xz plane
at
z
= 14.3213 in.
x
= 0.84667 in.
x~ 0.847 in. y -
in
310
PROBLEM
Two
3.139*
A and B are used to move the trunk of a
Replace the forces exerted by the ropes with an
equivalent wrench and determine (a) the resultant force R,
(/>) the pitch of the wrench, (c) the point where the axis of the
ropes attached at
fallen tree.
wrench
intersects the yz plane.
SOLUTION
(a)
First replace the
given forces with an equivalent force-couple system
f
R,
ftlf, )
at the origin.
We have
2
2
2
4,c=V(6) +(2) + (9) ==llm
Then
1650
TAC
U
:
N=
(6i + 2j + 9k)
(900 N)i + (300 N)j
+ (1 350 N)k
and
500
N
(14i
T,
BO
+ 2j + 5k)
15
(1
400 N)i + (200 N)j + (500 N)k
Equivalence then requires
R = T/JC +T/W
LF:
= (900i + 300j + 1350k)
+(I400i
= (2300
£M
:
+ 200j + 500k)
N)i
+ (500 N)j + (1 850 N)k
M.g-r,xT, c +r xTfiD
fi
- (12 m)kx[(900 N)i + (300 N)j + (1350 N)k]
+ (9 m)ix[(1400
N)i
+ (200 N)j + (500 N)k]
= -(3600)i + (1 0800 - 4500) j + (1 800)k
= -(3600 N m)i + (6300 N m)j + (1.800 N m)k
•
•
The components of the wrench
are (R, M,),
•
where
R = (2300 N)i + (500 N)j + (1 850 N)k
distribution to
you are using
it
in
without permission.
311
—
1
PROBLEM
(b)
3.139* (Continued)
We have
tf
= 1007(23) 2 + (5) 2 + (I8.5) 2 = 2993.7
N
Let
R
K
Then
!
(23i
R
+ 5j + 18,5k)
29.937
î=^„i.-MS
(23i
+ 5j +
1
8.5k) (-36001
•
+ 6300 j + 1 800k)
29.937
1
0.29937
-601.26
M,
[(23X-36) + (5X63) + (18.5X1 8)]
N-m
-601.26
Finally
R
N-m
N
2993.7
or
(c)
M,=M X
We have
}
P = -0.201 m
<«
axis
= (-601.26 N-m) x
:
(23i
+ 5j + 18.5k)
29.937
= -(461.93 N m)i - (1 00.42 1 N m)j - (37
or
M.
Now
M -Mj-M,
•
•
1
.56
N
•
m)k
2
= (-36001 + 6300J + 1800k)
-(-46 ,93i - 00.42 j - 37
1
1
1
.56k)
= -(3138.1 N m)i + (6400.4 N m)j + (2
•
1
71
.6
N m)k
•
For equivalence
&
reproduced or distributed in any form or by any means; without the prior written permission of the publisher, or used beyond the limited
312
PROBLEM
M
Thus require
3
3.139* (Continued)
=r,>xR
r
- (y\ + zk)
Substituting
i
-3 1 38. li
+ 6400.4 j + 2 1 7 1.6k
2300
j
k
y
z
500
1850
The
axis
of the wrench
6400.4
= 2300 z
or
z
k:
2171.6
= -2300^
or
y = -0.944
intersects the yz plane at
distribution to
y ~ -0.944
m
z
m
- 2.78
in
<
© 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,
teachers and educators permitted by McGraw- Hi 11for their individual course preparation. Ifyou are a student using this Manual,
you are using
= 2.78
m
j:
it
in
without permission.
313
k
PROBLEM
A
3.140*
guyed by three cables. If the tensions in the
cables have the same magnitude P, replace the forces exerted
on the pole with an equivalent wrench and determine (a) the
resultant force R, (/>) the pitch of the wrench, (c) the point
where the axis of the wrench intersects the xz plane.
flagpole
is
SOLUTION
,iM,
(a)
First
reduce the given force system to a force-couple
We have
PX BA + PkDC + P
ZF:
R=P
4.
3
+
DE
3,
i
at the origin.
=R
—
4.^
5
5
i
l
J
+
— —4. + —
(~9.
l
12
i
U5
1
k
25
5
R=
—
(2i-20j-k)
A
25
«=— V(2)
We have
EM:
( -AP
3P
x
+( 2 °)
I(r
+(!)
xP) = M
3P
4P
+ (20a)jx|— j-_j| + (20fl)jx
-9P.
25
m:
24/J«
5
+
YIP.
m;
25
teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou cue a student using this Manual,
distribution to
.
1
(-i-k)
you are using
AP
1
it
in
without permission.
314
PROBLEM
(b)
We have
M
=X8 >M*
where
X„
R ~
X
Then
25
M,
and pitch
P
M, =
(c)
3.140* (Continued)
(2i-20j-k)
(2I-20J-k)
nfip
25
—M2i~20}~k)
—
9V5
5
My
-SPa
(
R
\5yf5
.21
25
-%Pa
MX,
9V5
(~i~k) =
-8a
^
or
Sp
M
2
1
(2i-20j-k) =
675
M
Require
( %Pa
I
2
~r
8 Pa
(-2l
+ 20j + k.)
=—
(-4301- 20j- 406k)
xR
e/(?
f
675
—
675
\
(-4031 - 20j - 406k)
-*
675
= M*-M, = -?^H-k)-— (-21 + 20J + k)
5
p = ~ 0.098 8a
81
15^5 v.-'v-'/
,9>/5
Then
r
15V5
3P^l
= (xi + 2k) x -(2l-20J-k)
[20zi + (x + 2z)j-20xk]
^-Wi
25j
From
8(~403)
i:
Pa
675
From
8 (_4()6
k:
r
w
V
25y
= 20z
Pa
)^l
^ _20jc 3P
675
The
axis
of the wrench
\
z
= -1,990! 2a
x = 2.0049a
25
intersects the jez plane at
x = 2.00a,
<
distribution to
z~ -1.990a
in
315
PROBLEM
60
Determine whether the force-and-couple system shown can be
to a single equivalent force R. If it can, determine R
and the point where the line of action of R intersects the yz
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
I4N-IH
.1
M
/"
3.141*
mm
40
[
reduced
mm
8M
v" V
10 N-m ^^Pq
160mm
£
/^¥
40
the point
.1
where
its
axis intersects the yz plane.
^
mm
SOLUTION
First,
We have
XF:
F^+FG =R
"
R = (50N)k + 70N
(40 mm)i
+ (60 mm) j - (120 mm)k
140
mm
= (20N)i + (30N)j-(10N)k
and
= 37.417 N
/?
1M
We have
:
£(r
M.g
xF) + 2M
Mi
(
= [((). 12 m)j x (50 N)k] + {(0. 16 m)i x[(20 N)i + (30 N) j - (60 N)k]}
+ (I0N-m)
+ (14N-m)
(160mm)i-( 120 mm)j
200 mm
(40
mm)i - (1 20 mm) j + (60 mm)k
1
m;
To be able to reduce
perpendicular. Thus,
(1
8
N
•
m)i
the original forces
R
•
M=
- (8.4 N m)j +
•
and couples
40 mm
(.1
0.8
N m)k
•
to a single equivalent force,
R and M must be
0.
Substituting
(20i
or
+ 30j - 0k)
(20)(1 8)
1
-
(1 8i
- 8.4 j +
0.8k)
=
+ (30)(-8.4) + (-10)(10.8) =
0^0
or
R and M
1
are perpendicular so that the given system can be reduced to the single equivalent force
R = (20.0 N)i + (30.0 N)j-(1 0.00 N)k <
distribution
you are using
it
in
without permission.
316
PROBLEM
Then
3.141* (Continued)
for equivalence
MS=r_xR r=yl + zk
Thus require
Substituting
j
i
18i-8.4j + 10.8k
20
k
y
z
30
-10
j:
k:
The
line
-8.4 = 20z
10.8
= -20^
of action of R ntersects the yz plane
i
or
at
or
x
=
PROPRIETARY MATERIAL, ©
= -0.42
m
y = -0.54 m
2
y = -0.540 m
z
= -0.420 m
you are using
it
in
without permission.
317
—
1
1
PROBLEM
3.142*
Determine whether the force-and-coupie system shown can be
reduced to a single equivalent force R. If it can, determine R
intersects the yz
and the point where the line of action of
plane. If it cannot be so reduced, replace the given system with
an equivalent wrench and determine its resultant, its pitch, and
R
the point
where
axis intersects the yz plane.
its
SOLUTION
First
We have
determine the resultant of the forceis at D.
d DA
dm
=
VH2)
2
^y](-~6)
:
2
2
+(9) +(8)
2
2
+(Q) +(-%)
2
=17in.
=\0m.
Then
34lb
*DA
-(
=
=
12i
+ 9j + 8k)
17
=
-(241b)i
+ (18Ib)j + (16
1b)k
and
*w
10
= -{18
1b)i-(241b)k
Then
If
:
R = FZM +FaD
= (-241
+ 1 8 j + 6k + (-1 8i - 24k)
= -(42
1b)i
+ (18lb)j-(8!b)k
For the applied couple
dAK
= V(- 6 )
2
2
2
+ (" 6) + o 8 ) = 6
^
in
-
Then
_
1601b-in.,
M
..
,.
(-61-61 + 10I
18k),
671
-
To be able to
rixiuce
the original ibrces
must be perpendicular.
1
^
[
(lib-
and couple
in,)"
-
lb i«-)j
•
+ (3
lb
to a single equivalent force
in.)k]
R and M
Thus
R-M=0
©
2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual inay be displayed,
distribution to teachers and educators permitted by McGraw-Hillfar their individual course preparation. Ifyou are a student using this Manual,
you are using
it
without permission.
318
PROBLEM
3.142* (Continued)
Substituting
(-42i + 18j-8k)-Ê(-i-j + 3k)
160
or
=
[(-42)H) + (18)(~l) + (-8)(3)] =
0^0
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force
R = -(42.0 Ib)i + (18.00 lb)j -(8.00 lb)k
Then
for equivalence
o
T
"
5
M-r
Thus require
p/D
where
vPID
xR
= -(12
in.)i
+ [{y - 3)in.Jj + (z
in.)k
Substituting
160
H-J + 3k)=
i
j
k
-12
(y-3)
z
-42
1
-8
8
= [(>>~3X-8)-(z)(.l.8)]i
+ [(z)(-42)-(-12)(-8)]j
+ [(-12)(1.8)-(.y-3)(-42)]k
160
-42z-96
z
or
y = 11.59 in.
480
k:
The
line
of action of R
-216
intersects the yz plane at
+ 42(^-3)
x
—
= -1.137 in.
or
y-1
1
.
59
in.
z ~ -1
.
1
37
in.
©
you
are.
using
it
without permission.
319
PROBLEM
u
3.143*
Replace the wrench shown with an equivalent system consisting of two
A and B.
1.,
forces perpendicular to the y axis and applied respectively at
A* 1-
\
^J
b
a
"S
"
s
\
*
X
SOLUTION
y
Express the forces at A and
B^^
B as
A = Ax + Az k
i
B = Bx i + Bz k
Then, for equivalence
to the
given force system
ZFX
Ax + Bx =0
0)
ZFZ
AZ + BZ
=R
(2)
Az (a) + Bz (a + b) =
(3)
-Ax (a)-Bx (a + b) = M
(4)
ZMX
ZM
Z
From Equation
(
I
^*
£
:
BX ^~AX
),
Substitute into Equation (4)
-Ax (a) + Ax (a + b) =
M
A
x
From Equation
and Equation
(2),
(3),
A
D
*
b
M
M
b
K~--R~AZ
Az a + (R~Az )(a + b) = Q
^K)
and
Bz
=
--
R-R
K
Bz
b)
=
b
A=
Then
i
[bj
+ Rn+-)k 4
~(2Mf > «
distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using (his Manual,
320
M
PROBLEM
3.144*
Show
that, in general, a wrench can be replaced with two forces chosen in such a
through a given point while the other force lies in a given plane.
way
that
one force passes
SOLUTION
First, choose a coordinate system so that the xy plane coincides with the given plane. Also, position the
coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a.
Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the
scalar
A
components of R and
force system to be
given Point P and
M are known relative to the shown coordinate system.
shown
as equivalent
is illustrated in
B be the force that lies in the given plane.
b. Let A be the force passing through the
b be the x-axis intercept of B.
Figure
Let.
The known components of the wrench can be expressed as
and
M = MJ + MJ+
and
B - Bx + Bz k
R = Rx + Ry j + Rz k
i
while the
unknown
forces
k
A and B can be expressed as
A = Ax + Ay $ + Az k
\
P is given,
Since the position vector of Point
position vector i> are also
z
it
i
follows that the scalar components (\%y, z) of the
known.
Then, for equivalence of the two systems
+BX
ZFy
V
Ar
(2)
/?,=
A2 + Bz
(3)
£M
T
ZMy
:
:
;
2M_:
=
-•
M = yA
My ~ zA
M = xA
x
z
z
-zAy
x
- xAz
v
-yAx
six independent equations for the six
exists a unique solution for
From Equation
AX
**
X/y
Based on the above
==
£/V
(1)
(4)
bBz
(5)
(6)
unknowns (A x , Ayy
Az Bx B B2
,
,
,
,
b), there
A and B.
(2)
a.
©
-$
-4
reproduced or distributed in any form or by any means, without the prior writ/en permission of the publisher, or used beyond the limited
you are using
it
without permission.
321
PROBLEM
3.144* (Continued)
4=
Equation (6)
r
o (xR ~M
y
z)
A
,yj
Bx
Equation (1)
=
--
*,-
(O (xR -M 4
y
2)
,yj
Az = flW+zR,)
Equation (4)
Bz ~-K-
Equation (3)
(
i
<
>
~
(Mx + zRy ) <
,yj
b
Equation (5)
©
(xM\ + yMy
+zMJ
(M x -yRz +zRy )
you are using
it
without permission.
322
PROBLEM
3.145*
Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
SOLUTION
M
u*-
6iv«*> point
Co)
observe that
(b)
y
always possible to construct a line perpendicular to a given line so that the constructed
line also passes through a given point. Thus, it is possible to align one of the coordinate
axes of a rectangular
coordinate system with the axis of the wrench while one of the other axes passes through the given point.
First,
See Figures a and
it
is
b.
We have
R = />j
The unknown
forces
M = JWj
and
A and B can be expressed as
A = AJ + AJ + AM
The
distance
a
is
known.
It
is
assumed
tt?x
:
XFy
:
XM
V
XMZ
The
for equivalence
(I)
R = Ay +By
(2)
0Â +B
(3)
z
z
-zB..
M
-aAz
(4)
~xBz +zB
(5)
x
= aAy + xB
:
(6)
A and B are made perpendicular,
A-B =
There are eight unknowns:
AX B X + A V B V + AR
or
Ax Ay
,
,
A.,
Bx B Bz
,
But only seven independent equations. Therefore, there
Next consider Equation
If
0, z).
+Bx
= Ax
:
2Af„:
B = BJ + BJ + Bs k
and
B intersects the xz plane at (x,
that force
-LBV.
Since
and are known.
B v - 0,
(I.)
and
exists
(3) this equation
,
x,
an
z
infinite
number ofsolutions.
0~-zB
(4):
Equation (7) becomes
Using Equations
>
(7)
A.B.
becomes
+ A,B ~0
At
+ At=0
The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed,
you are using
it
1
in
without permission.
323
PROBLEM
Since the components of
A must be real, a nontrivial solution is not possible. Thus,
so that from Equation (4), z
To obtain one possible
(Note: Setting
Ay Az
,
Bz
it
is
required that
By * 0,
= 0.
solution, arbitrarily let
or
,
3.145* (Continued)
Ax - 0.
equal to zero results in unacceptable solutions.)
The defining equations then become
or
= 5.
R = Ay +By
(2)
Q~AZ + B
(3)
Z
Then Equation
(2) can
be written
M = ~aA~xB,
(5/
0-aAy +xBy
(6)
A v B ¥ + Az Bz =0
(7)'
Ay
=R~By
Equation (3) can be written
aAy
Equation (6) can be written
Substituting into Equation (5)',
R-By
M=~aAz
By
(~AZ )
J
(8)
or
aR
>
Substituting into Equation (7)',
(R-By )By +
By s
or
Then from Equations
(2), (8),
}
aR
a
a
2
R
2
R
±M
2
and (3)
a
A„=Ra
2
R
2
R
+M
a
A..
aR
aR
B.
a
2
R
2
a
2
2
a
2»3
2
M
2
rm:
R 2 +M
A
R
R +
7
:
aR
l
M
£
r>2
cfR + M"
J2.
M
+M 2
2
© 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No pan of this Manual may be displayed,
to teachers and educators permitted by McGraw-Hillfor their individual course preparation. If
distribution
}
[aR
in
324
PROBLEM
In
3.145* (Continued)
summary
A
RM
„2
-(Mj-aRk)
n A2
a'R + AT
aR l
B
a
Which shows
that
it is
possible to replace a
2
R2
+M
-(aRj
+ Mk)
:
wrench with two perpendicular
forces,
one of which
is
applied at
a given point.
Lastly, if
R>
From Equation
and
(6),
M>
x<0
0,
it
follows from the equations found for
A and B that A >
(assuming a>0). Then, as a consequence of
plane parallel to theyz plane and to the right of the origin, while force
but to the left to the origin, as shown in the figure below.
B
letting
lies in
and B,
Ax -0,
force
> 0.
A
lies in
a
a plane parallel to the yz plane
you are using
it
without permission.
325
PROBLEM
3.146*
Show that a wrench can be replaced with two forces, one of which has a prescribed
line
of action.
SOLUTION
choose a rectangular coordinate system where one axis coincides with the axis of the wrench and
another axis intersects the prescribed line of action (AA'). Mote that it has been assumed that the line of action
of force B intersects the xz plane at Point P(x, 0, z). Denoting the known direction of line AA' by
First,
it
follows that force
Force
B can
A can be expressed as
be expressed as
B = Bx + By j + Bz k
i
Next, observe that since the axis of the wrench and the prescribed line of action AA' are known,
the distance a can be determined. In the following solution, it is assumed that a is known.
it
follows that
Then, for equivalence
IFX
Since there are six unknowns (A,
0=AA +B
:
x
(!)
(
ZIy.
R = AA y + By
(2)
IF'
= AA, + B„
(3)
SM.,.:
zB.
(4)
JM
V
EM
V
M = -aAX. + zB
:
x
- xB.
(5)
Q = -aAZy + xBy
:
Bx> By Bz
,
,
x, z)
and six independent equations,
(6)
it
will
be possible to
obtain a solution.
beyond the limited
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it
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32(»
X
PROBLEM
Case
/:
Let z
-
3.146* (Continued)
to satisfy Equation (4)
Now Equation (2)
AXy
=R-B
y
Bz = ~AX
Equation (3)
Z
(
aAX,
a
Equation (6)
\
(R-B y )
v^v
Substitution into Equation (5)
M
a
-
-aAX7
K
(R-BY-AX)
M
(
J
A
B>>
B.
X2\aR
I
#=-
M )bi+b
(
y
aR)
X.z V
yv
y
XaR<
B.
1
Xz aR~Xv M
A=—
Then
MR
Xz aR~Xy M
aR
M X,
XX MR
BX =-AXX
Xz aR-Xv M
XZ MR
Xz aR-Xy M
B=~AX,
In
^
summary
P
A=
,
Ayv ~
B
R
XaR-XM
(Xx
aR
M
,
XAA <
"
A.
M + X aR] + X Mk) <
z
z
x~a
and
y
(
\~R
J
z
aR~Xy M
XmR 2
or
x=
KM
™
—
.
A.
X,R
Note
that for this case., the lines
of action of both
A
and
B
intersect the
a-
axis.
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without permission.
327
PROBLEM
Case
2:
B ~
Let
to satisfy
Now Equation (2)
3.146* (Continued)
Equation (4)
A
f
X.^
Equation (1)
B=-R
Equation (3)
(
B=-R X^
\.
yJ
aAX y ~
Equation (6)
which requires a -
r
M=z
R
In
last
expression
is
the equation for the line
R
yJ
K.
This
(
x?
x
X^
or
X^x — Xr
xv
ih
of action of force B.
summary
r
A
R^
K
v*yj
f
B
<.
Assuming
that
Xx Xy Xz > 0,
,
,
R^
M,i-4k)
yJ
the equivalent force system
that the
component of A
in the
xz plane
is
as
shown below.
A.
3
Note
is
parallel to B.
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328
0.6
PROBLEM
m -*U_ 0.6 in
3.147
A crate of mass 80 kg is held in the position shown. Determine
0.5 in
moment produced by the weight
(a) the
V'
\\
(b) the smallest force applied at
C\
B
W of the
crate about E,
that creates a
moment of
equal magnitude and opposite sense about E.
O.V>
m
SOLUTION
o.2Sm
(a)
By
W = mg = 80 kg(9.8
definition
ZME
We have
:
mis
1
2
)
W
^r
= 784.8 N
ME = (784.8 N)(0.25 m)
M £ = 196.2 N-m^
(/?)
For the force
at
B
to
be the smallest, resulting
must be perpendicular
produces a counterclockwise
a moment.
in
to the line connecting
E
to B.
(M£) about .£, the
F
fl
that the force
d = ^(0.85 m) 2 + (0.5 m) 2 = 0.9861 5 m
ZME
We have
:
1
96.2
N m = FB (0.986
•
1
5
A
m)
0,5 in
Fs =198.954 N
and
= tan
_1
m
0.5 m
C
0.85
or
59.534
£
c
1
99.0
N^
59,5°
<
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If
using this Manual,
distribution to
of action of force
moment about E.
Note:
you are using
line
The sense of F« must be such
it
in
without permission.
329
PROBLEM
3.148
known that the connecting rod AB exerts on the crank BC a 1 .5-kN force
directed down and to the left along the centerline of AB. Determine the moment
It is
of the force about C.
iXi
Ba™r
n (m.
28
nm
\%
21
mm
SOLUTION
Using
(a)
MC =y (FAB
(&>
l
(0.028
+X\(FAB)y
)x
m)
—
X1500N
25
24
+
(0.021
m)|— X1500N
1
25
42N-m
Using
or
M c =42.0N-m)*«
or
M c = 42.0 N m <
(b)
MC =y2(FAB)x
X1500N = 42N-m
= (0.1 m)
25
distribution to
*)
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PROPRIETARY MATERIAL. ©2010 The
you are using
•
it
in
without permission.
330
PROBLEM
3.149
A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting
force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
F^^^S^^RSilSi^R®
~-^î
^/i ;->"^-~ -------~-.r^t;~~- r
v
-
SOLUTION
We have
TX2 = (6
Then
Tx = Txz sin 30° = 2.9708 lb
Tv = Tsc sin 8° = -0.83504 lb
T,
Now
™*A
where
lb) cos
=r
~"
8° = 5.9416 lb
cos 30<
rBfA
-5.14561b
X TffC
(6sin45°)j-(6cos45°)k
[
/?//»
__
6
ft
(j-k)
J
k
1
-1
-0.83504
-5.1456
i
M
Then
6
2.9708
6
._
M
or
(-5.1456~0.83504)i
-~~ (2.9708) j-~ -(2.9708)k
s
~(25.4 lb-ft)i-(I2.60 lb-ft)j-(12.60 lb-ft)k
©
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distribution to teachers and educators permitted by McGraw-Hillfor their individual course, preparation- Ifyou are a student using this Manual,
yon are using if without permission.
331
PROBLEM
Ropes
AB
support a
3
and
tent.
3.150
BC are two of the ropes
The two ropes
used
to
are attached to a
AB
540 N,
and
the stake, (b) the projection on the stake of the
force exerted by rope AB at Point B.
stake at B. If the tension in rope
m
is
determine (a) the angle between rope
x
0.16
AB
m
Detail of the stake at
B
SOLUTION
2
BA = V("3) 2 + Of + (-1 -5) - 4.5
First note
BD = ,J(rO.QS) 2 +(0M) 2
+(0.16)
2
m
=0.42
m
TiM =^-(-3I + 3J-1.5k)
Then
= Zk ( _2i + 2j~k)
Xm
lin=:
=
BD =
BD
—
]
(
0.081
+ 0.38J + 0.1 6k)
0.42
(-41
+ 191J + 8k)
21
(a)
We have
%li^BD= TBA COS0
^L(~2l + 2j - k)
or
cos
or
•
—
(-41
+ 9 j + 8k) = TBA cos 6
1
- J-K-2X-4) + (2)(19) + (-1X8)]
63
-0,60317
or
(b)
(9
= 52.9°
^
We have
= TBA cos6
= (540N)(0.60317)
,
or
©
(7 JM ) JD
=326N ^
332
PROBLEM
3.151
A
former uses cables and winch pullers B and E to plumb
one side of a small barn. If it is known that the sum of the
moments about the x axis of the forces exerted by the
cables on the barn at Points A and D is equal to 4728 lb ft,
determine the magnitude of TDE when TAti - 255 lb.
•
SOLUTION
The moment about
the x axis due to the two cable forces can be found using the z components of each
force acting at their intersection with the xy-plane (A and D). The x components of the forces are parallel to
the x axis, and the y components of the forces intersect the x axis. Therefore, neither the x oi-y components
produce a moment about the x
axis.
We have
£M,:
where
\'ab)z ~*- '*AB
M
(T
)M
m
(yA ) + (T
)z
(y D )
=
Mx
= k-{T4B AAB )
k
2551b
-i-12j-H2k
17
801b
(TD/i ) z
~k-%DE
=k
1.5i-14j + 12k N
T,
DE
8.5
0.64865r,DE
yA
12
ft
y»
14
ft
4728
(1
80
Ib)(12 ft) +
lb -ft
(0.648657^ )(1 4 ft) = 4728 lb
TDE = 282.79 lb
and
•
ft
or
©
TDE
nF -2831b
<
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you are using
it
without permission.
3M
PROBLEM
/;,;:,..,,..,..
3.152
Solve Problem 3.151 when the tension in cable AB
is
306
lb.
PROBLEM 3.151 A farmer uses cables and winch pullers B
E to plumb one side of a small barn. If is known that the
and
H
/
'
-{
Hft/hb
-
x axis of the forces exerted by
the cables on the barn at Points A. and D is equal to 4728 lb ft,
determine the magnitude of T;^ when Tab ~ 255 lb.
^
i"-
it
sum of the moments about
the
•
SOLUTION
x axis due to the two cable forces can be found using the z components of each force
The x components of the forces are parallel to the x axis,
and the y components of the forces intersect the x axis. Therefore, neither the x or y components produce a
moment about the x axis.
The moment about
the
acting at the intersection with the xy plane (A and D).
We have
m
iTAB ) z {yA ) + {T
£M,:
Where
(TAB )x
{yD )
)z
= Mx
=k-rAB
M^sV)
-i-12j
306
k
+ 12k
lb
17
= 21.6
lb
-k T
(TDE )
0.5i-14j + 12k^
k
T,DE
8.5
0.648657^DE
>>A
yu
12
ft
14
ft
M .=47281b-ft
A
(2 1 6 lb)(12 ft)
+ (0.648657^ )(14 It) = 4728 lb
TDE = 235.211b
and
-
ft
or 7V„,
/j/;
= 235 lb
^
PROPRIETARY MATERIAL €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
you are using
it
without permission.
334
PROBLEM
'3
Hi
A
3.153
wiring harness
is
made by
routing either two or
three wires around 2-in. -diameter pegs
mounted on a
sheet of plywood. If the force in each wire
is
3
lb,
determine the resultant couple acting on the plywood
when a ~ 8
in.
place,
three wires are in place.
1
(//) all
and (a) only wires
AB and CD are in
HMt^fiffi^v
SOLUTION
In general,
M - "ZdF,
where d
is
the perpendicular distance between the lines of action of the
two forces
acting on a given wire.
(a)
A»
2.4 »*•
We
M = dAB FAB+da> FCD
have
(2
+ 24)in.x3
1b
+ |2 + -x28
in.x3
1b
or
M = 151.2
or
M = 67.2
lb- in. ")
<
(b)
Af
tlM
M î dab fab+ dCD FCD
We have
\
+ dEF FEF
= 151.2 lb-in.-28in.x3
lb -in.
)<
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teachers and educators permitted by McGraw-Hillfor their individual course preparation, ifyou are a student using this Manual,
distribution to
lb
in
335
PROBLEM
A
worker
3.154
tries to
move
a rock
by applying a 360-N
force
Replace that force with an
equivalent force-couple system at D. (b) Two workers
to a steel bar as
shown,
(a)
move the same rock by applying a vertical force
and another force at D. Determine these two forces if
they are to be equivalent to the single force of Part a.
attempt to
at
I
A
SOLUTION
(a)
We have
360 N(-sin40°i - cos40°j) = -(23 1.40 N)i - (275.78 N)j = F
LF:
or
We have
JM D
:
r/;//)
where
F = 360N
^50°^
xR = M
r
e/D
= -[(0.65 m) cos 30°]i + [(0.65 m)sin 30°].)
= -(0.56292 m)i + (0.32500 m)j
i
M
J
-0.56292
0.32500
-231.40
-275.78
k
N-m
= [1 55.240 + 75.206)N-m]k
- (230.45 N m)k
(b)
£M D
We have
where
:
M = r,
'hid
/D
= 40
or
M = 230N-m X) A
xF,
-
05 m)cos30°]i + [(1 .05 m)sin30°]j
= -(0.90933 m)i + (0.52500 m)j
FA =
i
J
-0.90933
0.52500
k
N-m
-I
= [230.45 N-m]k
©
336
PROBLEM
(0.90933F, )k
or
3.154 (Continued)
= 230.45k
FA = 253.42 N
We have
F^=253NH
F^F^+F^
XF:
-(231.40 N)i -(275.78 N)j = -(253.42
From
or
i:
23 1 .40
j:
22.36
N^ + FoC-cosî-sin^j)
N = FD cos (9
(D
= FD sm0
(2)
fi
Equation (2) divided by Equation (1)
= 0.096629
tan
(9
= 5.5193°
or
#-5.52°
FD° =
mM
= 232.48 N
cos 5.5 193°
distribution to
FD
=
232N ^5.52°^
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or
it
in
without permission.
337
PROBLEM
1.10
N
3.155
A
1
is
applied to the
10-N force acting
in a vertical plane parallel to the yz plane
220-mm-long horizontal handle
AB
of a
socket wrench. Replace the force with an equivalent force-
couple system at the origin
O of the coordinate system.
SOLUTION
We have
P S =F
SF:
y
where
B
= 11
N[~ (sin 1 5°) j + (cos
1
5°)k]
= -(28.470 N) j + (1 06.252 N)k
or
YM
We have
where
hio
F = -(28.5 N) j + (1 06.3 N)k
<
%,xPB
\
= [(°- 22 cos 35 °)» + (°-
]
5 )J
- (0.22sin 35°)k]m
= (0. 802 1 3 m)i + (0. 1 5 m) j - (0. 26 87 m)k
1
1.
i
j
k
0.180213
0.15
0.126187
-28.5
106.3
Mo
or
M
©
1
N m = ML
•
[(12.3487)i-(19.1566)j-(5.1361)k]Nm
= (1 2.35 N m)i - (1 9.
•
1
6
N
•
m)j - (5
.
1
3
N m)k <
•
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338
PROBLEM
fiO Ht
3.156
Four ropes are attached to a crate and exert the forces shown.
If the forces are to be replaced with a single equivalent force
applied at a point on line AB, determine (a) the equivalent
force and the distance from A to the point of application of
the force when a - 30°, (b) the value of a so that the single
equivalent force
is
applied at Point B.
SOLUTION
We have
\U3
tOOIfes *
i_
K
..
"1^ &
a
>t>
1
-
f
(a)
For equivalence
XF
or
Rx = 120.480
lb
LFy
a+
V
-100 cos 30° + 400coi 65° + 90 cos
:
1
:
00
sin
1
60 + 400 sin 65° + 90 sin 65° -
R y = (604.09 + 100 sin a)
or
a -30°
With
R = 4 J20.480) 2
0)
lb
+(654.09)
2
tan
1,M A
Also
LMA
or
(b)
is
applied 64.9
in.
(46
:
or
in.)(l
To the
60
lb)
+ (66
in.)(400 lb) cos 65°
+(36
in.)(90 lb) cos 65°
right
654.09
=
lb -in.
and
= 79.6°
in.)(400 lb) sin 65°
+(26
=42,435
„
120.480
= 665 lb
R
Ry
54.09 lb
Then
and
K° = RX
c/
+ (66
in.)(90 lb)sin65°
= </(654.09
=
64.9
in.
lb)
R~= 665
lb
^79.6°^
<
of A.
We have d — 66 in.
LM A
Then
:
42, 435 lb
•
in
= (66 m.)Ry
Ry = 642.95
or
Using Eq.
642,95
(I)
= 604.09 + 100sin a
or
a = 22. 9°
^
(!) 2010 The McGraw-Hill Companies, Inc. All rights reserved. /Vo /«/// o//Aw Mww/ may be displayed,
to teachers and educators permitted by McGraw-Hillfor their individual course, preparation. Ifyou are a student using this Manual,
distribution
lb
you are using
it
in
without permission.
339
PROBLEM
200
mm
H 200
A
mm
3.157
is used to tighten a screw at A.
Determine the forces exerted at B and C, knowing
blade held in a brace
(a)
these forces are
that
equivalent
to.
a force-couple
A consisting of R = -(30 N)i + R v \ + Rz k
and Mj =-(l2N m)i. (b) Find the corresponding
values of R and Rz (c) What is the orientation of the
y
system at
.
slot in the
head of the screw for which the blade
likely to slip
when
the brace
is in
the position
is
least
shown?
SOLUTION
(")
R =B+C
IF;
-(30 N)i + R j +
y
or
Equating the
i
coefficients
-30N = -CV
i:
JM A
Also
Rz k = -j?k + (-Cr i + Cy j + C2 k)
C =30N
or
V
M*=r^xB + r ,xC
:
c7
-(l
2
N
•
m)i
= [(0.2 m)i + (0.
1
5 m)j] x (~B)k
+(0.4 m)ix[-(30 N)i
-12 N-m = -(0.15 m)/?
i:
k:
j:
or
+ C,,j + Cz k]
5 = SON
= (0.4m)Cy
or
Cy =0
= (0.2m)(80N)-(0.4m)C,
or
C,=40N
B = -(80.0N)k C = -(30.0 N)i + (40.0 N)k <
(b)
Now we have for the equival ence of forces
-(30 N)i +
(c)
First note that
when
R = -(30 N)i
the slot in the head
-
- (40
Ry ^0
<
Rz = -40.0 N
<
Ry =0
j:
k
RJ + Rz k = -(80 N)k + [(-30 N)i + (40 N)k]
:
Rz = -80 + 40
N)k. Thus, the screw
of the screw
is vertical.
or
is
best able to resist the lateral force
R.z
4
distribution to teachers and educators permit ted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
340
1
PROBLEM
A
3.158
concrete foundation
12
ft
mat in
the shape of a regular
hexagon of side
supports four column loads as shown. Determine the magnitudes
of the additional loads that must be applied at B and F if the
of all six loads is to pass through the center of the mat.
resultant
SOLUTION
From
the statement
of the problem
it
can be concluded that the six applied loads are equivalent to the resultant
R at 0. It then follows that
£M o =0
1MX
or
=
ZM ^Q
Z
For the applied loads.
f—
*
K
/
\
,
= 0:
(6>/3
&)FS + (6>/3
x
\
—17. «L
SM
1
7t>
\
Then
E
J
—
ft)(10 kips)
- (6>/3
ft)(20 kips)
-(673 ft)/>=0
FB -FF
or
SM, = 0:
(1.2 ft)(15
-(12
=10
kips)
+ (6
ft)(30 kips)
(1)
ft)FB
- (6
-(6
ft)(10 kips)
ft)(20 kips)
Ffl+ 7^=60
or
Then
+ (6 ft) FF =
(l)
+ (2)=>
and
©
(2)
FB = 35.0 kips j
4
Ff = 25.0 kips J
-^
you are using il without permission.
341

Solucionario Mecanica Vectorial para ingenieros Estatica

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