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statics SM CH 04

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Geometry: From the geometry of the figure,
u 5 sin21 a
0.75
b 5 30°
1.5
Equations of Equilibrium: Applying the equations of equilibrium to the free-body diagram in Fig. (a),
1cSFy 5 0;
FBA sin 30° 2 200(9.81) 5 0
FBA 3924 N 5 3.92 kN
Ans.
1 SFx 5 0;
S
3924 cos 30° 2 FBC 5 0
FBC 5 3398.28 N 5 3.40 kN
Ans.
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Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram in Fig. (a),
1cSFy 5 0;
3500 sin u 2 200(9.81) 5 0
u 5 34.10°
1 SFx 5 0;
S
3500 cos 34.10° 2 FBC 5 0
FBC 5 2898.37 N 5 2.90 kN
Ans.
y 5 1.5 sin 34.10° 5 0.841 m 5 841 mm Ans.
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2 SFx 5 0;
2T cos 30° 1 8 1 5 sin 45° 5 0
S
T 5 13.32 5 13.3 kN
Ans.
1cSFy 5 0;
F 2 13.32 sin 30° 2 5 cos 45° 5 0
F 5 10.2 kN
Ans.
2 SFx 5 0;
8 2T cos u 1 5 sin 45° 5 0
S
12 2 T sin u 2 5 cos 45° 5 0
1cSFy 5 0;
Solving,
T 5 14.3 kN
Ans.
u 5 36.3°
Ans.
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The300-lb
1500-N
electrical
transformer
with center
of
•5–25.
electrical
transformer
with center
of gravity
•4–5. The
gravity
G is supported
byata pin
at Aa and
a smooth
at
G is at
supported
by a pin
A and
smooth
pad atpad
B.
at B. Determine
the horizontal
and vertical
components
of
Determine
the horizontal
and vertical
components
of reaction
reaction
and the
B on the
at
the pinat
A the
and pin
the A
reaction
of reaction
the pad Bofonthe
thepad
transformer.
transformer.
0.45
1.5 ftm
A
0.9
3 ftm
G
B
Equations of Equilibrium: From the free-body diagram of the transformer, Fig. a, NB and
Ay can be obtained by writing the moment equation of equilibrium about point A and the
force equation of equilibrium along the y axis.
+ ΣMA = 0;
NB (0.9) – 1500(0.45) = 0
NB = 750 N
+↑ΣFy = 0;
Ans
Ay – 1500 = 0
Ay = 1500 N
Ans
Using the result NB = 750 N and writing the force equation of equilibrium along the x axis,
+
→ ΣFx
= 0;
750 – Ax = 0
Ax = 750 N
Ans
0.45 m
1500 N
0.9 m
334
185
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Upper
Saddle
River,
NJ.NJ.
AllAll
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5–26.
4–6. A skeletal diagram of a hand holding a load is shown
in the upper figure. If the load and the forearm have masses
of 2 kg and 1.2 kg, respectively, and their centers of mass are
located at G1 and G2, determine the force developed in the
biceps CD and the horizontal and vertical components of
reaction at the elbow joint B. The forearm supporting
system can be modeled as the structural system shown in
the lower figure.
D
G1
C
B
A
G2
D
Equations of Equilibrium: From the free-body diagram of the structural
system, Fig. a, FCD can be obtained by writing the moment equation of equilibrium
about point B.
�
+ ΣMB = 0;
G1
C
A
100 mm
G2
135 mm
75
B
65 mm
2(9.81)(100 + 135 + 65) + 1.2(9.81)(135 + 65)
–FCD sin 75° (65) = 0
FCD = 131.25 N = 131 N
Ans
Using the above result and writing the force equations of equilibrium along the x
and y axes,
+
S ΣFx = 0;
131.25 cos 75° – Bx = 0
+ ↑ΣFy = 0;
131.25 sin 75° – 2(9.81) – 1.2(9.81) – By = 0
Bx = 33.97 N = 34.0 N
By = 95.38 N = 95.4 N
Ans
Ans
335
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SaddleRiver,
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NJ.All
All
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5–27. As an airplane’s brakes are applied, the nose wheel
4–7.
exerts two forces on the end of the landing gear as shown.
Determine the horizontal and vertical components of
reaction at the pin C and the force in strut AB.
C
30
20
B
400 mm
A
600 mm
2 kN
6 kN
Equations of Equilibrium : The force in strut AB can be obtained
directly by summing moments about point C.
�
+ ΣMC = 0;
2(1) – 6(1 tan 20°) + FAB sin 50° (0.4)
– FAB cos 50° (0.4 tan 20°) = 0
FAB = 0.8637 kN = 0.864 kN
Ans
Using the result FAB = 0.8637 kN and sum forces along x and y axes,
we have,
+ ↑ΣFy = 0; 6 + 0.8637 cos 50° – Cy = 0
S ΣFx = 0;
+
0.8637 sin 50° + 2 – Cx = 0
Cx = 2.66 kN
Cy = 6.56 kN
Ans
Ans
336
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Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
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asas
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portion
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exist.
NoNo
portion
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*5–28. The 1.4-Mg drainpipe is held in the tines of the fork
*4–8.
lift. Determine the normal forces at A and B as functions of
the blade angle u and plot the results of force (vertical axis)
versus u (horizontal axis) for 0 … u … 90°.
0.4 m
A
B
u
+ Q©Fx = 0; NA – 1.4(10)3 (9.81) sin u = 0
NA = 13.7 sin u kN
Ans
+ a©Fy = 0; NB – 1.4(10)3 (9.81) cos u = 0
NB = 13.7 cos u kN
Ans
337
188
188 337
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exist.No
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•5–29.
•4–9. The mass of 700 kg is suspended from a trolley
which moves along the crane rail from d = 1.7 m to
d = 3.5 m. Determine the force along the pin-connected
knee strut BC (short link) and the magnitude of force at pin
A as a function of position d. Plot these results of FBC and FA
(vertical axis) versus d (horizontal axis).
d
A
C
2m
B
+ ΣMA = 0;
 4
FBC   (1.5) – 700(9.81)(d) = 0
 5
FBC = 5722.5d
S ©Fz = 0;
+
1.5 m
Ans
 3
–Az + (5722.5d)   = 0
 5
Az = 3433.5d
+ ↑©Fy = 0;
 4
–Ay + (5722.5d)   – 700 (9.81) = 0
 5
Ay = 4578d – 6867
FA = (3433.5d)2 + (4578d – 6867)2
Ans
338
189
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8:43:21
© 2011
Pearson
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Inc.,
Upper
Saddle
River,
rights
reserved.
This
material
is protected
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they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
AllAll
rights
reserved.
This
material
is protected
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allall
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laws
as as
they
currently
exist.
portion
material
may
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form
means,
without
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writing
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exist.
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portion
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4–10. If
5–30.
If the
the force
force of
of FF == 100
500 lb
N is applied to the handle of
the bar bender, determine the horizontal and vertical
components of reaction at pin A and the reaction of the
roller B on the smooth bar.
C
1000
mm
40 in.
F
60
Equations of Equilibrium: From the free-body diagram of the handle of the bar
bender, Fig. a, Ay and NB can be obtained by writing the force equation of equilibrium
along the y axis and the moment equation of equilibrium about point A, respectively.
+↑ΣFy = 0;
A
125 mm
5 in.
Ay – 500 sin 30° = 0
Ans
Ay = 250 N
+ ΣMA = 0;
B
NB cos 60° (125) – 500(1000) = 0
NB = 8000 N = 8.0 kN
Ans
Using the result NB = 8000 N and writing the force equation of equilibrium along the
x axis,
+
→ ΣFx
= 0;
Ax – 8000 + 500 cos 30° = 0
Ax = 7567.0 N = 7.567 kN
Ans
500 N
12
5m
m
1000 mm
339
190
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NJ.All
Allrights
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reserved.
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currently
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exist.No
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5–31.
4–11. If the force of the smooth roller at B on the bar
bender is required to be 1.5
7.5 kip,
kN, determine the horizontal
and vertical components of reaction at pin A and the
required magnitude of force F applied to the handle.
C
1000
40 mm
in.
F
60
Equations of Equilibrium: From the free-body diagram of the handle of the bar
bender, Fig. a, force F can be obtained by writing the moment equation of equilibrium
about point A.
+ ΣMA = 0;
B
7.5 cos 60° (125) – F (1000) = 0
F = 0.46875 kN
A
125 mm
5 in.
Ans
Using the above result and writing the force equation of equilibrium along the x and
y axis,
+
→ ΣFx
= 0;
Ax + 0.46875 cos 30° – 7.5 = 0
Ans
Ax = 7.094 kN
+↑ΣFy = 0;
Ay – 0.46875 sin 30° = 0
Ans
Ay = 0.234 kN
12
5m
m
1000 mm
340
191
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Upper
Saddle
River,
All
rights
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material
protected
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© 2010
Pearson
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Upper
Saddle
River,
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All
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reserved.
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asas
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portion
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*4–12.
*5–32. The jib crane is supported by a pin at C and rod AB.
If the load has a mass of 2 Mg with its center of mass located
at G, determine the horizontal and vertical components of
reaction at the pin C and the force developed in rod AB on
the crane when x = 5 m.
A
3.2 m
C
Equation of Equilibrium: Realizing that rod AB is a two-force member, it will
exert a force FAB directed along its axis on the beam, as shown on the free-body
diagram in Fig. a. From the free-body diagram, FAB can be obtained by writing
the moment equation of equilibrium about point C.
+ ΣMC = 0;
4m
0.2 m
B
D
G
x
 3
 4
FAB   (4) + FAB   (0.2) – 2000(9.81)(5) = 0
 5
 5
FAB = 38 320.31 N = 38.3 kN
Ans
Using the above result and writing the force equations of equilibrium along the x and y axes.
+
S ©Fx = 0;
+ ↑©Fy = 0;
 4
Cx – 38 320.31   = 0
 5
Cx = 30 656.25 N = 30.7 kN
Ans
 3
38 320.31   – 2000(9.81) – Cy = 0
 5
Cy = 3372.19 N = 3.37 kN
Ans
341
192
192 341
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•5–33.
•4–13. The jib crane is supported by a pin at C and rod AB.
The rod can withstand a maximum tension of 40 kN. If the
load has a mass of 2 Mg, with its center of mass located at G,
determine its maximum allowable distance x and the
corresponding horizontal and vertical components of
reaction at C.
A
3.2 m
C
Equation of Equilibrium: Realizing that rod AB is a two-force member, it will
exert a force FAB directed along its axis on the beam, as shown on the free-body
diagram in Fig. a. From the free-body diagram, the distance x can be obtained by
writing the moment equation of equilibrium about point C.
+ ΣMC = 0;
4m
0.2 m
B
D
G
x
 3
 4
40 000   (4) + 40 000   (0.2) – 2000(9.81)(x) = 0
 5
 5
x = 5.22 m
Ans
Writing the force equations of equilibrium along the x and y axes,
+
S ©Fx = 0;
+ ↑©Fy = 0;
 4
Cx – 40 000   = 0
 5
Cx = 32 000 N = 32 kN
Ans
 3
40 000   – 2000 (9.81) – Cy = 0
 5
Cy = 4380 N = 4.38 kN
Ans
342
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allcopyright
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5–54.
4–14. The
The uniform
uniform rod
rod AB
AB has
has aa weight
weight of
of 15
75 lb
N and the
spring is unstretched when u = 0°. If u = 30°, determine
the stiffness k of the spring.
1.86m
ft
A
u
30.9
ft m
k
Geometry : From triangle CDB, the cosine law gives
B
l=
0.7608 2 + 0.5196 2 – 2(0.7608) (0.5196) cos 120° = 1.1154 m
Using the sine law,
0.45 m
sin α
sin 120°
=
0.7608
1.1154
= 36.21°
75 N
0.45 m
Equations of Equilibrium : The force in the spring can be
obtained directly by summing moments about point A.
+ ΣMA = 0;
75 cos 30° (0.45) – Fsp cos 36.21° (0.9) = 0
0.9
cos 30°
= 0.7608 m
Fsp = 40.25 N
1.8 –
Spring Force Formula : The spring stretches x = 1.1154 – 0.9
= 0.2154 m
Fsp
40.25
k=
=
= 186.9 N/m
0.2154
x
0.9
m
cos 30°
0.9 m
Ans
0.9 tan 30°
= 0.5196 m
5–55. The horizontal beam is supported by springs at its
ends. Each spring has a stiffness of k = 5 kN>m and is
originally unstretched so that the beam is in the horizontal
position. Determine the angle of tilt of the beam if a load of
800 N is applied at point C as shown.
800 N
C
A
B
1m
3m
Equations of Equilibrium : The spring force at A and B can be
obtained directly by summing moments about points B and A, respectively.
+ ΣMB = 0;
800(2) – FA (3) = 0
FA = 533.33 N
+ ΣMA = 0;
FB (3) – 800(1) = 0
FB = 266.67 N
Spring Formula : Applying ∆ =
F
, we have
k
533.33
= 0.1067 m
5(103)
266.67
= 0.05333 m
∆B =
5(103)
∆A =
Geometry : The angle of tilt a is
 0.05333
a = tan–1 
= 1.02°
 3 
Ans
360
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AM
4–15. The airstroke actuator at D is used to apply a force of
5–23.
F = 200 N on the member at B. Determine the horizontal
and vertical components of reaction at the pin A and the
force of the smooth shaft at C on the member.
C
15
600 mm
B
A
60
200 mm
600 mm
D
F
Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a,
NC can be obtained by writing the moment equation of equilibrium about point A.
�
:21 AM
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+ ΣMA = 0;
200 sin 60° (800) – NC (600 + 200 sin 15°) = 0
Ans
NC = 212.60 N = 213 N
Using this result and writing the force equations of equilibrium along the x and y axes,
+
S ΣFx = 0;
+ ↑ΣFy = 0;
–Ax + 212.60 cos 15° – 200 cos 60° = 0
Ans
Ax = 105 N
–Ay – 212.60 sin 15° + 200 sin 60° = 0
Ans
Ay = 118 N
332
195
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8:43:05
© 2011
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Saddle
River,
All
rights
reserved.
This
material
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copyright
laws
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currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
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material
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asas
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portion
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*5–24.
*4–16. The airstroke actuator at D is used to apply a force
of F on the member at B. The normal reaction of the
smooth shaft at C on the member is 300 N. Determine the
magnitude of F and the horizontal and vertical components
of reaction at pin A.
C
15
600 mm
B
A
60
200 mm
600 mm
D
F
Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a,
force F can be obtained by writing the moment equation of equilibrium about point A.
�
+ ΣMA = 0;
F sin 60° (800) – 300(600 + 200 sin 15°) = 0
Ans
F = 282.22 N = 282 N
Using this result and writing the force equations of equilibrium along the x and y axes,
+
S ΣFx = 0;
+ ↑ΣFy = 0;
–Ax + 300 cos 15° – 282.22 cos 60° = 0
Ans
Ax = 149 N
–Ay – 300 sin 15° + 282.22 sin 60° = 0
Ans
Ay = 167 N
333
196
196 333
05a SM_CH04.indd
Ch05a 319-360.indd
4/8/118:43:08
11:48:20
6/12/09
AM AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
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underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
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formor
orby
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anymeans,
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withoutpermission
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writingfrom
fromthe
thepublisher.
publisher.
4–17. Outriggers A and B are used to stabilize the crane
*5–36.
from overturning when lifting large loads. If the load to be
lifted is 3 Mg, determine the maximum boom angle u so that
the crane does not overturn. The crane has a mass of 5 Mg
and center of mass at GC, whereas the boom has a mass of
0.6 Mg and center of mass at GB.
4.5 m
GB
5m
GC
u
2.8 m
0.7 m
+ ©MA = 0;
A
B
2.3 m
– 5(9.81)(2.3) + 3(9.81)(9.5 sin u – 0.7)
+ 0.6(9.81)(5 sin u – 0.7) = 0
u = 26.4°
Ans
345
197
SM_CH04.indd
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Ch05a 319-360.indd
345
4/8/11
6/12/09 11:48:20
8:43:39 AM
AM
2011
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Allrights
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reserved.This
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©©
2010
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River,
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•5–61.
•4–18. If spring BC is unstretched with u = 0° and the bell
crank achieves its equilibrium position when u = 15°,
determine the force F applied perpendicular to segment
AD and the horizontal and vertical components of reaction
at pin A. Spring BC remains in the horizontal postion at all
times due to the roller at C.
C
k
2 kN/m
B
F
150
u
300 mm
A
D
Spring Force Formula: From the geometry shown in Fig. a, the stretch of spring
BC when the bell crank rotates u = 15° about point A is x = 0.3 cos 30° – 0.3 cos 45°
= 0.04768 m. Thus, the force developed in spring BC is given by
400 mm
Fsp = kx = 2000(0.04768) = 95.35 N
Equations of Equilibrium: From the free-body diagram of the bell crank, Fig. b,
F can be obtained by writing the moment equation of equilibrium about point A.
+ ΣMA = 0;
95.35 sin 45° (300) – F(400) = 0
Ans
F = 50.57 N = 50.6 N
Using the above result and writing the force equations of equilibrium along the x and y
axes,
+
S ©Fx = 0;
Ax – 50.57 sin 15° – 95.35 = 0
Ans
Ax = 108.44 N = 108 N
+ ↑©Fy = 0;
Ay – 50.57 cos 15° = 0
Ans
Ay = 48.84 N = 48.8 N
366
198
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366
4/8/11
6/12/09 11:48:20
8:45:35 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
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copyrightlaws
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currently
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exist.No
Noportion
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withoutpermission
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writingfrom
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thepublisher.
publisher.
5–38.
4–19. Spring CD remains in the horizontal position at all
times due to the roller at D. If the spring is unstretched
when u = 0° and the bracket achieves its equilibrium
position when u = 30°, determine the stiffness k of the
spring and the horizontal and vertical components of
reaction at pin A.
D k
0.6 m
C
0.45 m
B
u
F
300 N
A
Spring Force Formula: At the equilibrium position, the spring elongates x =
0.6 sin 30° m. Using the spring force formula, the force in spring CD is found to be
Fsp = kx = 0.3k.
Equations of Equilibrium: From the free-body diagram of the bracket, Fig. a, the
stiffness k of spring CD and Ay can be obtained by writing the moment equation of
equilibrium about point A and the force equation of equilibrium along the x axis,
respectively.
+ ΣMA = 0;
0.3k cos 30° (0.6) – 300 cos 30° (0.45) – 300 sin 30° (0.6) = 0
Ans
k = 1327.35 N/m = 1.33 kN/m
+ ↑ΣFy = 0;
Ay – 300 = 0
Ans
Ay = 300 N
Using the result k = 1327.35 N / m and writing the force equation of equilibrium along
the x axis,
+
S ©Fx = 0;
Ax – 0.3(1327.35) = 0
Ans
Ax = 398.21 N = 398 N
347
199
SM_CH04.indd
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347
4/8/11
6/12/09 11:48:21
8:43:43 AM
AM
2011
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Education,
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Saddle
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NJ.All
Allrights
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reserved.This
Thismaterial
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©©
2010
Pearson
Education,
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Upper
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River,
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exist.
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5–39. Spring CD remains in the horizontal position at all
*4–20.
times due to the roller at D. If the spring is unstretched
when u = 0° and the stiffness is k = 1.5 kN>m, determine
the smallest angle u for equilibrium and the horizontal and
vertical components of reaction at pin A.
D k
0.6 m
C
0.45 m
B
u
Spring Force Formula: At the equilibrium position, the spring elongates x =
0.6 sin u. Using the spring force formula, the force in spring CD is found to be
Fsp = kx = 1500 (0.6 sin u) = 900 sin u.
F
300 N
A
Equations of Equilibrium: From the free-body diagram of the bracket, Fig. a, the
equilibrium position u and Ay can be obtained by writing the moment equation of
equilibrium about point A and the force equation of equilibrium along the y axis,
respectively.
+ ΣMA = 0;
900 sin u cos u (0.60) – 300 sin u (0.6) – 300 cos u (0.45) = 0
540 sin u cos u – 180 sin u – 135 cos u = 0
Solving by trial and error yields
Ans
u = 23.083° = 23.1°
+ ↑ΣFy = 0;
Ay – 300 = 0
Ans
Ay = 300 N
Using the result u = 23.083° and writing the force equation of equilibrium along
the x axis,
+
S ©Fx = 0;
Ax – 900 sin 23.083° = 0
Ans
Ax = 352.86 N = 353 N
348
200
SM_CH04.indd
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348
4/8/11
6/12/09 11:48:21
8:43:45 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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orby
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withoutpermission
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fromthe
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4–21. The
The platform
platformassembly
assemblyhas
has
a weight
of 1.25
kN
*5–40.
a weight
of 1.25
kN and
and center
of gravity
intendedto
to support
support a
center
of gravity
at at1.GIf
it itis isintended
1. If
placed at point
point G22, determine the
maximum load of 2 kN placed
smallest counterweight W that should be placed at B in
order to prevent the platform from tipping over.
G2
0.6 m
When tipping occurs, RC = 0
+ ΣMD = 0;
1.8 m
–2 (0.6) + 1.25 (0.3) + WB (2.1) = 0
Ans
WB = 0.393 kN
G1
2 kN
1.25 kN
2.4 m
B
1.8 m
0.6 m
C
2.4 m
0.3 m
D
0.3 m
1.8 m
0.3 m
1.8 m
0.3 m
349
201
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349
4/8/11
6/12/09 11:48:22
8:43:48 AM
AM
2011
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Education,
Inc.,
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Saddle
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NJ.All
Allrights
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reserved.This
Thismaterial
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lawsasasthey
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currently
©©
2010
Pearson
Education,
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Saddle
River,
NJ.
exist.
No
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this
material
may
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•4–22. Determine the horizontal and vertical components
•5–41.
of reaction at the pin A and the reaction of the smooth
collar B on the rod.
C
300 lb
1500
N
2250
N
450 lb
B
30
A
1 ftm
0.3
0.6
2 ftm
D
1.2
4 ftm
10.3
ft m
Equations of Equilibrium: From the free-body diagram, Ay and NB can be
obtained by writing the force equation of equilibrium along the y axis and the
moment equation of equilibrium about point A.
+↑ΣFy = 0;
Ay – 1500 – 2250 = 0
Ans
Ay = 3750 N
+ ΣMA = 0;
NB (1.2 sin 30°) – 1500(0.3) – 2250(0.9) = 0
Ans
NB = 4125 N
Using the result NB = 4125 N and writing the force equation of equilibrium
along the x axis,
+
→ ΣFx
= 0;
Ax – 4125 = 0
Ans
Ax = 4125 N
1500 N
2250 N
1.2 m
0.3 m
0.6 m
0.3 m
350
202
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350
4/8/11
6/12/09 11:48:22
8:43:50 AM
AM
©
©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underall
allcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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orby
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publisher.
4–23. Determine the support reactions of roller A and the
5–42.
smooth collar B on the rod. The collar is fixed to the rod
AB, but is allowed to slide along rod CD.
A
900 N
1m
1m
C
45
2m
B
D
600 N m
45
Equations of Equilibrium: From the free-body diagram of the rod, Fig. a, NB
can be obtained by writing the force equation of equilibrium along the y axis.
+ ↑ΣFy = 0;
NB sin 45° – 900 = 0
NB = 1272.79 N = 1.27 kN
Ans
Using the above result and writing the force equation of equilibrium and the
moment equation of equilibrium about point B,
+
S ©Fx = 0;
1272.79 cos 45° – Ax = 0
Ax = 900 N
+ ΣMB = 0;
Ans
–900(1) + 900(2) sin 45° – 600 +MB = 0
MB = 227 N · m
Ans
351
203
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4/8/11
6/12/09 11:48:22
8:43:54 AM
AM
2011
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Allrights
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©©
2010
Pearson
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River,
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exist.
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5–43.
*4–24. The
Theuniform
uniformrod
rodAB
ABhas
hasaaweight
weightof
of 15
75 lb.
N. Determine
the force in the cable when the rod is in the position shown.
B
75 N
1.5 sin 40°
5 ft
30
C
A
10
0.75 cos 40°
+ ΣMA = 0;
+
→ ΣFx
= 0;
NB (1.5 sin 40°) – 75 (0.75 cos 40°) = 0
T cos 10° – 44.69 = 0
NB = 44.69 N
T = 45.38 N
T
Ans
·
4–25. Determine the horizontal and vertical components
*5–44.
of force at the pin A and the reaction at the rocker B of the
curved beam.
500 N
200 N
10 15
2m
A
+ ΣMA = 0;
NB (4) – 200 cos 15° (2) – 500 cos 10° (2) = 0
NB = 342.79 = 343 N
+ ↑ΣFy = 0;
Ans
Ay – 500 cos 10° – 200 cos 15° + 342.79 = 0
Ay = 342.8 = 343 N
+
S ©Fx = 0;
B
Ans
– Ax + 500 sin 10° – 200 sin 15° = 0
Ax = 35.1 N
Ans
352
204
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4/8/11
6/12/09 11:48:23
8:43:57 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
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copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
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publisher.
•5–45.
•4–26. The floor crane and the driver have a total weight
of 2500
lb with
12.5 kN
with aa center
center of
of gravity
gravity at
at G. If the crane is
required to lift the 2.5-kN
500-lb drum,
drum, determine the normal
reaction on both the wheels at A and both the wheels at B
when the boom is in the position shown.
F
3.6
12 m
ft
0.93mft
D
C
30
1.8
m
6 ft
G
E
A
2.2 ft
0.66 m1.4
0.42
ft m
B
2.52
m
8.4 ft
Equations of Equilibrium: From the free-body diagram of the floor crane, Fig. a,
+ ΣMB = 0;
12.5 (0.42 + 2.52) – 2.5 (4.5 cos 30° – 2.52) – NA (0.66 + 0.42 + 2.52) = 0
Ans
NA = 9.252 kN
+↑ΣFy = 0;
9.252 – 12.5 – 2.5 + NB = 0
Ans
NB = 5.748 kN
4.5 m
12.5 kN
2.5 kN
2.52 m
0.42 m
0.66 m
353
205
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353
4/8/11
6/12/09 11:48:23
8:44:00 AM
AM
2011
Pearson
Education,
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Upper
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River,
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
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lawsasasthey
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©©
2010
Pearson
Education,
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Upper
Saddle
River,
NJ.NJ.All
exist.
No
portion
this
material
may
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means,
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5–46.
4–27. The floor crane and the driver have a total weight of
2500kN
lb with a center of gravity at G. Determine the largest
12.5
weight of the drum that can be lifted without causing the
crane to overturn when its boom is in the position shown.
F
12 ft
3.6
m
0.93 m
ft
D
C
30
6 ftm
1.8
G
E
A
2.2 ft
0.66 m1.40.42
ft m
B
2.52
8.4 ftm
Equations of Equilibrium: Since the floor crane tends to overturn about point B,
the wheel at A will leave the ground and NA = 0. From the free-body diagram
of the floor crane, Fig. a, W can be obtained by writing the moment equation of
equilibrium about point B.
+ ΣMB = 0;
12.5 (0.42 + 2.52) – W(4.5 cos 30° – 2.52) = 0
Ans
W = 26.69 kN
4.5 m
12.5 kN
2.52 m
0.42 m
0.66 m
354
206
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354
4/8/11
6/12/09 11:48:23
8:44:02 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
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orby
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withoutpermission
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writingfrom
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publisher.
*4–28. The motor has
5–47.
has aa weight
weightofof4.25
850 kN.
lb. Determine the
force that each of the chains exerts on the supporting hooks
at A, B, and C. Neglect the size of the hooks and the
thickness of the beam.
0.3 m
4.25
kN
850 lb
0.15
0.5m
ft
4.25 kN
0.15 m
0.45 m
A
C
10
+ ΣMB = 0;
+
→ ΣFx
= 0;
+↑ΣFy = 0;
0.45
1.5 ftm
0.3
1 ftm
FA cos 10° (0.3) + 4.25 (0.15) – FC cos 10° (0.6) = 0
(1)
FC sin 10° – FB sin 30° – FA sin 10° = 0
(2)
4.25 – FA cos 10° – FB cos 30° – FC cos 10° = 0
(3)
B
10
30
Solving Eqs. (1), (2) and (3) yields:
FA = 2.16 kN
FB = 0
Ans
FC = 2.16 kN
4–29. Determine the force P needed to pull the 50-kg roller
*5–48.
over the smooth step. Take u = 60°.
P
u
0.1 m
0.6 m
B
A
20
 0.5
f = cos–1   = 33.56°
 0.6
+ ΣMB = 0;
50 (9.81) sin 20° (0.5)+ 50 (9.81) cos 20° (0.3317) – P cos 60° (0.5)
– P sin 60° (0.3317) = 0
P = 441 N
Ans
355
207
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4/8/11
6/12/09 11:48:24
8:44:06 AM
AM
2011
Pearson
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Allrights
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Thismaterial
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©©
2010
Pearson
Education,
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River,
NJ.
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•5–49. Determine the magnitude and direction u of the
•4–30.
minimum force P needed to pull the 50-kg roller over the
smooth step.
P
u
0.1 m
0.6 m
B
A
20
For Pmin,
NA S 0,
+ ΣMB = 0;
 0.5
f = cos–1   = 33.56°
 0.6
50 (9.81) sin 20° (0.5) + 50 (9.81) cos 20° (0.3317) – P cos u (0.5)
– P sin u (0.3317) = 0
236.75 – P cos u (0.5) – P sin u (0.3317) = 0
P=
236.75
(0.5 cos u + 0.3317 sin u)
For Pmin;
– 236.75 (–0.5 sin u + 0.3317 cos u)
dP
=
=0
(0.5 cos u + 0.3317 sin u)2
du
tan u =
0.3317
0.5
u = 33.6°
Ans
Pmin = 395 N
Ans
356
208
SM_CH04.indd
208
05a Ch05a 319-360.indd
356
4/8/11
6/12/09 11:48:24
8:44:09 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formororby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–31. The winch cable on a tow truck is subjected to a
5–50.
force of T = 6 kN when the cable is directed at u = 60°.
Determine the magnitudes of the total brake frictional
force F for the rear set of wheels B and the total normal
forces at both front wheels A and both rear wheels B for
equilibrium. The truck has a total mass of 4 Mg and mass
center at G.
+
S ©Fx = 0;
A
2m
F
B
2.5 m
T
1.5 m
Ans
– NA (4.5) + 4(103)(9.81) (2.5) – 6000 sin 60° (3) – 6000 cos 60° (1.5) = 0
NA = 17 336 N = 17.3 kN
+ ↑©Fy = 0;
1.25 m
3m
u
6000 sin 60° – F = 0
F = 5196 N = 5.20 kN
+ ΣMB = 0;
G
Ans
17 336 – 4(103)(9.81) – 6000 cos 60° + NB = 0
NB = 24 904 N = 24.9 kN
Ans
357
209
SM_CH04.indd
209
05a Ch05a 319-360.indd
357
4/8/11
6/12/09 11:48:25
8:44:12 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,
anyform
formororbybyany
anymeans,
means,
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
exist.
No
ininany
without
5–51. Determine the minimum cable force T and critical
*4–32.
angle u which will cause the tow truck to start tipping, i.e., for
the normal reaction at A to be zero. Assume that the truck is
braked and will not slip at B. The truck has a total mass of
4 Mg and mass center at G.x
G
1.25 m
A
2m
+ ΣMB = 0;
3m
u
2.5 m
T
F
B
1.5 m
4(103)(9.81)(2.5) – T sin u (3) – T cos u (1.5) = 0
T=
65 400
(cos u + 2 sin u)
dT
–65 400 (sin u + 2 cos u)
=
=0
du
(cos u + 2 sin u)2
– sin u + 2 cos u = 0
u = tan–1 2 = 63.43° = 63.4°
T=
Ans
65 400
= 29.2 kN
(cos 63.43° + 2 sin 63.43°)
Ans
4–33.
*5–52. Three uniform books, each having a weight W and
length a, are stacked as shown. Determine the maximum
distance d that the top book can extend out from the
bottom one so the stack does not topple over.
a
d
Equilibrium : For top two books, the upper book will topple
when the center of gravity of this book is to the right of point A.
Therefore, the maximum distance from the right edge of this book
to point A is a/2.
Equation of Equilibrium : For top entire three books, the top
two books will topple about point B.
+ ΣMB = 0;

a
W(a – d) – W d –  = 0

2
3a
d=
4
Ans
358
210
SM_CH04.indd
210
05a Ch05a 319-360.indd
358
4/8/11
6/12/09 11:48:25
8:44:15 AM
AM
©©2011
rights
reserved.
This
material
is is
protected
under
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–34.
5–63. The cart supports the uniform crate having a mass of
85 kg. Determine the vertical reactions on the three casters
at A, B, and C. The caster at B is not shown. Neglect the
mass of the cart.
B
0.1 m
A
0.2 m
0.5 m
0.4 m
0.2 m
0.6 m
C
0.35 m
0.35 m
Equations of Equilibrium : The normal reaction NC can be
obtained directly by summing moments about x axis.
©Mx = 0;
©My = 0;
©Mz = 0;
NC (1.3) – 833.85 (0.45) = 0
NC = 288.64 N = 289 N
Ans
833.85 (0.3) – 288.64 (0.35) – NA (0.7) = 0
NA = 213.04 N = 213 N
Ans
NA + 288.64 + 213.04 – 833.85 = 0
NB = 332 N
Ans
368
211
SM_CH04.indd
211
05b Ch05b 361-400.indd
368
4/8/11
11:48:25
AM AM
6/12/09
8:45:39
© 2011
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Education,
Inc.,
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Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
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form
any
means,
without
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writing
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publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
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or or
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without
permission
in in
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4–35.
*5–64. The pole for a power line is subjected to the two
cable forces of 60 lb, each force lying in a plane parallel to
the x- y plane. If the tension in the guy wire AB is 80 lb,
determine the x, y, z components of reaction at the fixed
base of the pole, O.
z
300
60 lbN
0.3 m
A
45
1 ft
45
4 ftm
1.2
60 N
lb
300
400
80 N
lb
3 mft
10
B
3 ft
0.9 m
O
y
x
Equations of Equilibrium :
ΣFx = 0;
Ox + 300 sin 45° – 300 sin 45° = 0
Ox = 0
ΣFy = 0;
Oy + 300 cos 45° + 300 cos 45° = 0
Oy = –424.3 N
ΣFz = 0;
Ans
Oz – 400 = 0
Oz = 400 N
ΣMx = 0;
Ans
Ans
(MO)x + 400 (0.9)
0.3 m
– 2[300 cos 45° (4.2)] = 0
(MO)x = 1422 N · m
ΣMy = 0;
300 N
Ans
(MO)y + 300 sin 45° (4.2)
300 N 1.2 m
– 300 sin 45° (4.2) = 0
(MO)y = 0
ΣMz = 0;
Ans
400 N
0.9 m
(MO)z + 300 sin 45° (0.3)
3m
– 300 sin 45° (0.3) = 0
(MO)z = 0
Ans
369
212
212 369
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:45:41
11:48:26
6/12/09
AM AM
©©2011
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2010Pearson
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Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
•5–65. If P = 6 kN, x = 0.75 m and y = 1 m, determine
*4–36.
the tension developed in cables AB, CD, and EF. Neglect
the weight of the plate.
z
B
F
P
x
D
A
y
E
x
C
2m
2m
y
Equations of Equilibrium: From the free-body diagram, Fig. a, TCD and TEF
can be obtained by writing the moment equation of equilibrium about the x and
y axes, respectively.
ΣMx = 0;
TCD (2) – 6(1) = 0
Ans
TCD = 3 kN
ΣMy = 0;
TEF (2) – 6(0.75) = 0
Ans
TEF = 2.25 kN
Using the above results and writing the force equation of equilibrium along the z axis,
ΣMz = 0;
TAB + 3 + 2.25 – 6 = 0
Ans
TAB = 0.75 kN
370
213
SM_CH04.indd
213
05b Ch05b 361-400.indd
370
4/8/11
11:48:26
AM AM
6/12/09
8:45:43
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
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in in
writing
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exist.
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writing
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5–66.
4–37. Determine the location x and y of the point of
application of force P so that the tension developed in
cables AB, CD, and EF is the same. Neglect the weight of
the plate.
z
B
F
P
x
D
A
y
E
x
C
2m
2m
y
Equations of Equilibrium:
Equilibrium: From
Fromthe
thefree-body
free-bodydiagram
diagramofofthe
the
plate,
Fig.
plate,
Fig.
a, a,
and writing the moment equations of equilibrium about the x9
y9 axes,
axes,
x¿ and y¿
ΣMxœ = 0;
T(2 – y) – 2T(y) = 0
Ans
y = 0.667 m
ΣMyœ = 0;
2T(x) – T(2 – x) = 0
Ans
x = 0.667 m
371
214
05b SM_CH04.indd
Ch05b 361-400.indd
214 371
6/12/09
AM AM
4/8/118:45:45
11:48:26
©©2011
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asas
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currently
2010Pearson
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Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
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laws
they
currently
exist.
No
ininany
without
exist.
Noportion
portionofofthis
thismaterial
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reproduced,
anyform
formororbybyany
anymeans,
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thepublisher.
publisher.
5–67.
4–38. Due to an unequal distribution of fuel in the wing
tanks, the centers of gravity for the airplane fuselage A
and wings B and C are located as shown. If these
kN, lb,
WBW
=B 40
components have
have weights
weights WW
45 000
= kN,
8000and
lb,
AA ==225
WC =W30
kN,
determine
the normal
reactions
of the wheels
and
the normal
reactions
of the
6000
lb, determine
C =
D, E, and
theF ground.
wheels
D, F
E,on
and
on the ground.
z
D
B
A
C
E
2.48mft
ΣFx = 0;
40 (1.8) – RD (4.2) – 30 (2.4) + Rg (4.2) = 0
ΣFy = 0;
40 (1.2) + 225 (2.1) + 30 (1.2) – RF (8.1) = 0
ΣFz = 0;
RD + Rg + RF – 400 – 30 – 225 = 0
F
1.86 m
ft
8 ft
2.4
m
x
ft
1.86 m
4 ft m
1.2
6 mft 0.9
20
3 ftm
y
40 kN
Solving,
RD = 113.15 kN
Ans
Rg = 113.15 kN
Ans
RF = 68.0 kN
Ans
30 kN
225 kN
372
215
SM_CH04.indd
215
05b Ch05b 361-400.indd
372
4/8/11
11:48:27
AM AM
6/12/09
8:45:46
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
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reserved.
This
material
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under
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asas
they
currently
exist.
portion
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material
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any
means,
without
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writing
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publisher.
exist.
NoNo
portion
of of
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material
may
bebe
reproduced,
in in
any
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or or
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4–39.
*5–68. Determine the magnitude of force F that must be
exerted on the handle at C to hold the 75-kg crate in the
position shown. Also, determine the components of reaction
at the thrust bearing A and smooth journal bearing B.
z
0.1 m
A
B
x
0.6 m
0.5 m
y
0.2 m
0.1 m
Equations of Equilibrium: From the free-body diagram, Fig. a, F, Bz, Az, and
Ay can be obtained by writing the moment equation of equilibrium about the y, x,
and x¿ axes and the force equation of equilibrium along the y axis.
C
F
©My = 0; –F(0.2) + 75(9.81)(0.1) = 0
Ans
F = 367.875 N = 368 N
©Mx = 0; Bz (0.5 + 0.6) – 75(9.81)(0.6) = 0
©Mx¿ = 0;
©Fy = 0;
Ay = 0
Bz = 401.32 N = 401 N
–Az(0.6 + 0.5) + 75(9.81)(0.5) = 0
Ans
Az = 334.43 N = 334 N
Ans
Using the result F = 367.875 N and writing the moment equations of equilibrium
about the z and z¿ axes,
©Mz = 0; –Bx(0.5 + 0.6) – 367.875(0.2 + 0.1 + 0.5 + 0.6) = 0
©Mz¿ = 0;
Bx = –468.20 N = –468 N
Ax(0.6 + 0.5) – 367.875(0.2 + 0.1) = 0
Ans
Ax = 100.33 N = 100 N
Ans
The negative signs indicate that Bx act in the opposite sense to that shown on the free-body
diagram.
373
216
216 373
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:45:51
11:48:27
6/12/09
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©©2011
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2010Pearson
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UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
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This
material
protected
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laws
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exist.No
Noportion
portionofofthis
thismaterial
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formororbybyany
anymeans,
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withoutpermission
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thepublisher.
publisher.
•5–69. The shaft is supported by three smooth journal
*4–40.
bearings at A, B, and C. Determine the components of
reaction at these bearings.
z
900 N
600 N
450 N
0.9 m
0.6 m
x
0.9 m
500 N
0.9 m
C
0.6 m
A
0.9 m
B
0.9 m
y
Equations of Equilibrium: From the free-body diagram, Fig. a, Cy and Cz can be obtained by
writing the force equation of equilibrium along the y axis and the moment equation of
equilibrium about the y axis.
ΣFy = 0;
Cy – 450 = 0
Ans
Cy = 450 N
ΣMy = 0;
Cz(0.9 + 0.9) – 900(0.9) + 600(0.6) = 0
Ans
Cz = 250 N
Using the above results
ΣMx = 0;
Bz(0.9 + 0.9) + 250(0.9 + 0.9 + 0.9) + 450(0.6) – 900(0.9 + 0.9 + 0.9) – 600(0.9) = 0
Ans
Bz = 1125 N = 1.125 kN
ΣMx¿ = 0;
600(0.9) + 450(0.6) – 900(0.9) + 250(0.9) – Az (0.9 + 0.9) = 0
Ans
Az = 125 N
ΣMz = 0;
–Bx(0.9 + 0.9) + 500(0.9) + 450(0.9) – 450(0.9 + 0.9) = 0
ΣFx = 0;
Ax + 25 – 500 = 0
Ans
Bx = 25 N
Ans
Ax = 475 N
374
217
SM_CH04.indd
217
05b Ch05b 361-400.indd
374
4/8/11
11:48:28
AM AM
6/12/09
8:45:53
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
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copyright
laws
asas
they
currently
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portion
this
material
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means,
without
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writing
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publisher.
exist.
NoNo
portion
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material
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5–70.
4–41. Determine the tension in cables BD and CD and
the x, y, z components of reaction at the ball-and-socket
joint at A.
z
D
3m
300 N
B
x
A
1.5 m
0.5 m
C
rBD = {– 1i + 1.5j + 3k) m;
1m
y
rBD = 3.50 m
r 
TBD = TBD  BD  = – 0.2857 TBD i + 0.4286 TBD j + 0.8571 TBD k
 rBD 
In a similar manner,
r 
TCD = TCD  CD  = – 0.2857 TCD i – 0.4286 TCD j + 0.8571 TCD k
 rCD 
Thus, using the components of TBD and TCD, the scalar equations of equilibrium become:
ΣFx = 0;
Ax – 0.2857 TBD – 0.2857 TCD = 0
ΣFy = 0;
Ay + 0.4286 TBD – 0.4286 TCD = 0
ΣFz = 0;
Az + 0.8571 TBD + 0.8571 TCD – 300 = 0
ΣMAx = 0; – (0.8571 TBD) (1.5) + (0.8571 TCD) (1.5) = 0
ΣMAy = 0; 300 (1) – (0.8571 TBD) (1.5) – (0.8571 TCD) (1.5) = 0
Solving
TBD = TCD = 117 N
Ans
Ax = 66.7 N
Ans
Ay = 0
Ans
Az = 100 N
Ans
375
218
218 375
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:45:57
11:48:28
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©©2011
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2010Pearson
PearsonEducation,
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Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
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This
material
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laws
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exist.No
Noportion
portionofofthis
thismaterial
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reproduced,ininany
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formororbybyany
anymeans,
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withoutpermission
permissionininwriting
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thepublisher.
publisher.
5–71.
usedtotosupport
support
1.25-kN
5–71.The
Therod
rodassembly
assembly is used
thethe
250-lb
cylinder.
4–42.
(= 125
kg) cylinder.
Determine the
Determine
the components
of components
reaction at of
thereaction
ball-andat the
ball-and-socket
A, the
smooth
journal
bearing
socket
joint A, the joint
smooth
journal
bearing
E, and
the E,
force
anddeveloped
the force developed
along
rodconnections
CD. The connections
along rod CD.
The
at C and Datare
C and
D are ball-and-socket
joints.
ball-and-socket
joints.
z
D
Equations of Equilibrium: Since rod CD is a two – force member, it exerts a force
FDC directed along its axis as defined by uDC on rod BC, Fig. a. Expressing each of
the forces indicated on the free – body diagram in Cartesian vector form,
FA = Axi + Ay j + Azk
x
FE = Exi + Ezk
C
1 ftm
0.3
A
0.3
1 ftm
0.3
1 ftm
W = [–1.25k] kN
F
0.31 m
ft
E
0.45
1.5 ftm
y
FDC = –FDCk
Applying the force equation of equilibrium
ΣF = 0;
FA + FE + FDC + W = 0
(Axi + Ay j + Azk) + (Exi + Ezk) + (–FDC k) + (–1.25k) = 0
(Ax + Ex)i + (Ay)j + (Az + Ez – FDC)k = 0
Equating i, j, and k components,
Ax + Ex = 0
(1)
Ay = 0
(2)
Az + Ez – FDC – 1.25 = 0
(3)
In order to apply the moment equation of equilibrium about point A, the position
vectors rAC, rAE, and rAF, Fig. a, must be determined first.
rAC = [–0.3i + 0.3j] m
rAE = [0.6j] m
rAF = [0.45i + 0.9j] m
Thus,
0.3 m
0.3 m
E(0, 0.6, 0) m
0.3 m
ΣMA = 0; (rAC × FDC) + (rAE × FE) + (rAF × W) = 0
(–0.3i + 0.3j) × (–FDCk) + (0.6j) × (Exi + Ezk) + (0.45i + 0.9j) × (–1.25k) = 0
0.45 m
(–0.3 FDC + 0.6 Ez – 1.125)i + (0.5625 – 0.3 FDC)j + (–0.6Ex)k = 0
Equating i, j, and k components,
–0.3 FDC + 0.6 Ez – 1.125 = 0
(4)
0.5625 – 0.3 FDC = 0
(5)
–0.6 Ex = 0
(6)
W = 1.25 kN
Solving Eqs. (1) through (6), yields
FDC = 1.875 kN
Ans
Ex = 0
Ans
Ez = 2.8125 kN
Ans
Ax = 0
Ans
Ay = 0
Ans
Az = 0.3125 kN
Ans
376
219
SM_CH04.indd
219
05b Ch05b 361-400.indd
376
4/8/11
11:48:29
AM AM
6/12/09
8:45:59
2011
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Inc.,
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Saddle
River,
All
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material
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under
copyright
laws
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currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
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form
any
means,
without
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writing
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publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
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4–43.
*5–72. Determine the components of reaction acting at the
smooth journal bearings A, B, and C.
z
450 N
C
0.6 m
300 N m 45
A
0.4 m
x
B
0.8 m
0.4 m
y
Equations of Equilibrium: From the free-body diagram of the shaft, Fig. a, Cy and
Cz can be obtained by writing the force equation of equilibrium along the y axis and the
moment equation of equilibrium about the y axis.
ΣFy = 0;
450 cos 45° + Cy = 0
Cy = –318.20 N = –318 N
ΣMy = 0;
Ans
Cz (0.6) – 300 = 0
Cz = 500 N
Ans
Using the above results and writing the moment equations of equilibrium about the x and z axes,
ΣMx = 0;
Bz (0.8) – 450 cos 45° (0.4) – 450 sin 45° (0.8 + 0.4) + (318.20)(0.4) + 500(0.8 + 0.4) = 0
Bz = –272.70 N = –273 N
ΣMz = 0;
Ans
Bx (0.8) – (–318.20)(0.6) = 0
Bx = –238.65 N = 239 N
Ans
Finally, using the above results and writing the force equations of equilibrium along the x and z axes,
ΣFx = 0;
Ax + 238.5 = 0
Ax = –238.65 N = –239 N
ΣFz = 0;
Ans
Az – (–272.70) + 500 – 450 sin 45° = 0
Az = 90.90 N = 90.9 N
Ans
The negative signs indicate that Cy, Bz and AX act in the opposite sense of that shown on the free-body
diagram.
377
220
220 377
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:46:04
11:48:30
6/12/09
AM AM
©©2011
rights
reserved.
This
material
is is
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under
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
*4–44. Determine the force components acting on the ball•5–73.
and-socket at A, the reaction at the roller B and the tension
on the cord CD needed for equilibrium of the quarter
circular plate.
z
D
350 N
200 N
2m
1m
A
C
60
200 N
3m
B
y
x
Equations of Equilibrium : The normal reaction NB and Az can be
obtained directly by summing moments about the x and y axes respectively.
ΣMx = 0; NB (3) – 200(3) – 200 (3 sin 60°) = 0
NB = –373.21 N = 373 N
Ans
ΣMy = 0; 350(2) + 200 (3 cos 60°) – Az (3) = 0
Az = 333.33 N = 333 N
Ans
ΣFx = 0;
TCD + 373.21 + 333.33 – 350 – 200 – 200 = 0
TCD = 43.5 N
Ans
ΣFx = 0;
Ax = 0
Ans
ΣFy = 0;
Ay = 0
Ans
378
221
SM_CH04.indd
221
05b Ch05b 361-400.indd
378
4/8/11
11:48:30
AM AM
6/12/09
8:46:07
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
or or
byby
any
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permission
in in
writing
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publisher.
5–74.
determine
thethe
x, y,
theload
loadhas
hasaaweight
weightofof200
200lb,
kN,
determine
x,
4–45. IfIfthe
zy,components
of reaction
at the
ball-and-socket
jointjoint
A and
z components
of reaction
at the
ball-and-socket
A
the
in each
of the
andtension
the tension
in each
ofwires.
the wires.
z
2m
ft
4m
ft
2 ft
m
D
F
ft
33 m
A
B
x
G
E
2m
ft
y
ft
2m
22 m
ft
C
Equations of Equilibrium : Expressing the forces indicated on the free-body diagram Fig. a,
in Cartesian vector form.
FA = Ax i + Ay j + Az k
W = [–200k] kN
kN
FBD = FBDk
FCD = FCD uCD = FCD
�
(4 – 4)i + (0 – 4)j + (3 – 0)k
(4 – 4)2 + (0 – 4)2 + (3 – 0)2
�
3
 4

= – FCD j + FCD k
5
 5

FEF = FEF k
Applying the force equation of equilibrium,
ΣF = 0;
FA + FBD + FCD + FEF + W = 0
 4

3
(Axi + Ay j + Az k) + FBD k + – FCD j + FCD k + FEF k + (–200k) = 0
 5

5




4
3
(Ax)i + Ay – FCD j + Az + FBD + FCD + FEF – 200 k = 0




5
5
Equating i, j, and k components,
Ax = 0
4
Ay – FCD = 0
5
3
Az + FBD + FCD + FEF – 200 = 0
5
(1)
(2)
(3)
In order to write the moment equation of equilibrium about point A, the position vectors rAB , rAG , rAC , and
rAE must be determined first.
rAB
rAG
rAC
rAE
= [4i] m
= [4i + 2j]m
m
= [4i + 4j] m
m
m
= [2i + 4j] m
Thus,
ΣMA = 0; (rAB * FBD) + (rAC * FCD) + (rAE * FEF) + (rAG * W) = 0
 4

3
(4i) * (FBD k) + (4i + 4j) * – FCD j + FCD k + (2i + 4j) * FEP k) + (4i + 2j) * (–200k)
 5

5
 12



 16

12
FCD – 2FEF + 800 j + –
FCD k = 0
 FCD + 4FEF – 400 i + –4FBD –

 5

5
5
379
222
222 379
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:46:10
11:48:31
6/12/09
AM AM
©©2011
rights
reserved.
This
material
is is
protected
under
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
Equating i, j, and k components,
12
F + 4FEF – 400 = 0
5 CD
12
– 4FBD –
F – 2FEF + 800 = 0
5 CD
16
–
F =0
5 CD
(4)
(5)
(6)
Solving Eqs. (1) through (6),
FCD = 0
Ans
FEF = 100 kNkN
Ans
FBD = 150 kN
kN
Ans
Ax = 0
Ans
Ay = 0
Ans
Az = 100 kN
kN
Ans
The negative signs indicate that Az acts in the opposite sense to that on the
free-body diagram.
F(–2, 0, 2) m
D(4, 0, 3) m
B(4, 0, 0) m
G(4, 2, 0) m
W = 200 kN
E(2, 4, 0) m
C(4, 4, 0) m
380
223
SM_CH04.indd
223
05b Ch05b 361-400.indd
380
4/8/11
11:48:31
AM AM
6/12/09
8:46:13
2011
Pearson
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Inc.,
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Saddle
River,
All
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material
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under
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laws
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currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
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in in
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4–46. If the cable can be subjected to a maximum tension
5–75.
300 kN,
lb, determine the maximum force F which may be
of 1.5
applied to the plate. Compute the x, y, z components of
reaction at the hinge A for this loading.
ΣMy = 0;
0.3 (F) – 1.5 (0.9) = 0
F = 4.5 kN
1.5 kN
Ax = 0
Ans
ΣFy = 0;
Ay = 0
Ans
ΣFz = 0;
–4.5 + 1.5 + Az = 0;
C
x
B
MAx = 0
0.9
9 ftm
Ans
Ans
*5–76. The
by by
a pin
at Aat
and
cablea
Themember
memberis issupported
supported
a pin
A aand
BC.
theIfload
at DatisD300
lb, kN,
determine
thethe
x, x,
y, y,
z
cableIfBC.
the load
is 1.5
determine
components
z componentsofofreaction
reactionatatthe
thepin
pinAA and
and the
the tension
tension in
cable B
C.
BC.
z
0.3 m
1 ft
C
3
6
2 
TBC = TBC  i – j + k 
7
7
7 

ΣFy = 0;
3 ftm
A 0.3
Ans
Az = 3 kN
ΣMAz = 0; MAz = 0
ΣFx = 0;
F
20.2
ft m
10.1
ft m
y
0.3 m
ΣMAx = 0; MAx + 4.5 (0.1) – 0.3 (1.5) = 0;
3 ftm
0.3
0.9 m
Ans
ΣFx = 0;
z
0.6
2 ftm
 3
Ax +   TBC = 0
 7
0.6
2 ftm
1.8
6 ftm
0.6 m
B
0.6
m
2 ft
x
1.8 m
 6
Ay –   TBC = 0
 7
A
0.6
2 ftm
1.2 m
ΣFz = 0;
 2
Az – 1.5 +   TBC = 0
 7
ΣMx = 0;
 2
–1.5 (1.8) +   TBC (1.8) = 0
 7
ΣMy = 0;
 2
MAy – 1.5 (0.6) +   TBC (1.2) = 0
 7
ΣMz = 0;
 3
 6
MAz –   TBC (1.8) +   TBC (1.2) = 0
 7
 7
1.5 kN
y
D
Solving,
TBC = 5.25 kN
Ans
Ax = –2.25 kN
Ans
Ay = 4.5 kN
Ans
Az = 0
Ans
MAy = –0.9 kN · m
Ans
MAz = –1.35 kN · m
Ans
381
224
224 381
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:46:15
11:48:32
6/12/09
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2010Pearson
PearsonEducation,
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Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
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This
material
protected
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laws
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exist.No
Noportion
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thismaterial
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formororbybyany
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thepublisher.
publisher.
•8–1. Determine the minimum horizontal force P
•4–47.
required to hold the crate from sliding down the plane. The
crate has a mass of 50 kg and the coefficient of static friction
between the crate and the plane is ms = 0.25.
P
30
Free – Body Diagram. When the crate is on the verge of sliding down the plane, the frictional force F will act up the plane as indicated
on the free – body diagram of the crate shown in Fig. a.
Equations of Equilibrium.
©Fy¿ = 0;
N – P sin 30° – 50(9.81) cos 30° = 0
©Fx¿ = 0;
P cos 30° + 0.25 N – 50(9.81) sin 30° = 0
Solving
P = 140 N
N = 494.94 N
Ans
674
225
SM_CH04.indd
225
08a Ch08a 674-717.indd
674
4/8/11
11:48:32
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9:04:54
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River,
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laws
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currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
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form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
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permission
in in
writing
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thethe
publisher.
*4–48.
Determine
theminimum
minimumforce
forcePPrequired
required to
to push
8–2. Determine
the
the crate up the plane. The crate has a mass of 50 kg and the
coefficient of static friction between the crate and the plane
is ms = 0.25.
P
30
When the crate is on the verge of sliding down the plane, the frictional force F¿ will act down the plane as indicated on
the free – body diagram of the crate shown in Fig. b. Thus, F = msN = 0.25 N and F¿ = msN ¿ = 0.25 N¿.
By referring to Fig. b,
©Fy¿ = 0;
N ¿ – P sin 30° – 50(9.81) cos 30° = 0
©Fx¿ = 0;
P cos 30° – 0.25 N ¿ – 50(9.81) sin 30° = 0
Solving,
P = 474 N
Ans
N¿ = 661.92 N
675
226
226 675
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:04:57
11:48:33
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©©2011
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laws
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currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
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copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–49. A horizontal force of P = 100 N is just sufficient to
8–3.
hold the crate from sliding down the plane, and a horizontal
force of P = 350 N is required to just push the crate up the
plane. Determine the coefficient of static friction between
the plane and the crate, and find the mass of the crate.
P
30
Free – Body Diagram. When the crate is subjected to a force of P = 100 N, it is on the verge of slipping
down the plane. Thus, the frictional force F will act up the plane as indicated on the free – body diagram
of the crate shown in Fig. a. When P = 350 N, it will cause the crate to be on the verge of slipping up the
plane, and so the frictional force F acts down the plane as indicated on the free – body diagram of the
crate shown in Fig. a. Thus, F = msN and F ¿ = msN¿.
Equations of Equilibrium.
+a©Fy ¿ = 0;
N – 100 sin 30° – m(9.81) cos 30° = 0
+Q©Fx ¿ = 0;
msN + 100 cos 30° – m(9.81) sin 30° = 0
Eliminating N,
ms =
4.905 m – 86.603
8.496 m + 50
(1)
Also, by referring to Fig. b, we can write
+a©Fy ¿ = 0;
N ¿ – m(9.81) cos 30° – 350 sin 30° = 0
+Q©Fx ¿ = 0;
350 cos 30° – m(9.81) sin 30° – msN¿ = 0
Eliminating N ¿,
ms =
303.11 – 4.905 m
175 + 8.496 m
(2)
Solving Eqs. (1) and (2) yields
m = 36.46 kg
Ans
ms = 0.256
676
227
SM_CH04.indd
227
08a Ch08a 674-717.indd
676
4/8/11
11:48:33
AM AM
6/12/09
9:04:59
2011
Pearson
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Saddle
River,
Allrights
rightsreserved.
reserved.
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
This
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
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oror
byby
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inin
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from
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publisher.
*8–4.
4–50. If the coefficient of static friction at A is ms = 0.4
and the collar at B is smooth so it only exerts a horizontal
force on the pipe, determine the minimum distance x so
that the bracket can support the cylinder of any mass
without slipping. Neglect the mass of the bracket.
100 mm
x
B
C
200 mm
A
Free – Body Diagram. The weight of cylinder tends to cause the bracket to slide downward. Thus,
the frictional force FA must act upwards as indicated in the free – body diagram shown in Fig. a.
Here the bracket is required to be on the verge of slipping so that FA = msNA = 0.4 NA.
Equations of Equilibrium.
+ ↑©Fy = 0;
+©MB = 0;
0.4NA – mg = 0
NA = 2.5 mg
2.5 mg (0.2) + 0.4 (2.5 mg)(0.1) – m(g)(x + 0.1) = 0
Ans
x = 0.5 m
Note: Since x is independent of the mass of the cylinder, the bracket will not slip regardless of the
mass of the cylinder provided x > 0.5 m.
677
228
228 677
08aSM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:01
11:48:33
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laws
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2010Pearson
PearsonEducation,
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Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
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laws
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currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
•4–51. The
The 180-lb
90-kg man
•8–5.
man climbs
climbs up the
the ladder
ladder and stops at the
position shown after he senses that the ladder is on the verge
of slipping. Determine the inclination u of the ladder if the
coefficient of static friction between the friction pad A and the
ground is ms = 0.4.Assume the wall at B is smooth.The center
of gravity for the man is at G. Neglect the weight of the ladder.
B
G 10
3m
ft
u
0.9
3 ftm
A
Free – Body Diagram. Since the weight of the man tends to cause the friction pad A to slide to the right,
the frictional force FA must act to the left as indicated on the free – body diagram of the ladder, Fig. a.
Here, the ladder is on the verge of slipping. Thus, FA = s NA.
Equations of Equilibrium.
+↑ΣFy = 0;
+ ΣMB = 0;
NA – 90 (9.81) = 0
90 (9.81) (3 cos ) –
3 cos
NA = 882.9 N
s
(90) (9.81) (3 sin ) – 90 (9.81) (0.9) = 0
– 0.4 (3) sin = 0.9
Ans
= 52°
90 (9.81) N
3m
0.9 m
678
229
SM_CH04.indd
229
08a Ch08a 674-717.indd
678
4/8/11
11:48:34
AM AM
6/12/09
9:05:03
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
The
90-kg
man
climbs
8–6.
180-lb
man
climbs
the
ladder
*4–52.The
The
90-kg
man
climbsup
upthe
theladder
ladderand
andstops
stops at the
position shown after he senses that the ladder is on the verge
of slipping. Determine the coefficient of static friction between
the friction pad at A and ground if the inclination of the ladder
is u = 60° and the wall at B is smooth.The center of gravity for
the man is at G. Neglect the weight of the ladder.
B
G 10
3m
ft
u
0.9
3 ftm
A
Free – Body Diagram. Since the weight of the man tends to cause the friction pad A to slide to the right,
the frictional force FA must act to the left as indicated on the free – body diagram of the ladder, Fig. a.
Here, the ladder is on the verge of slipping. Thus, FA = s NA.
Equations of Equilibrium.
+↑ΣFy = 0;
+ ΣMB = 0;
NA – 90 (9.81) = 0
90 (9.81) (3 cos 60°) –
3 cos 60° –
s
NA = 882.9 N
s
(90) (9.81) (3 sin 60°) – 90 (9.81) (0.9) = 0
(3) sin 60° = 0.9
Ans
s = 0.231
90 (9.81)
3m
0.9 m
679
230
230 679
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:06
11:48:34
6/12/09
AM AM
©©2011
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reserved.
This
material
is is
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under
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–53. The uniform thin
N and a
8–7.
thin pole
pole has
has aa weight
weight of
of 150
30 lb
7.8ft.m.IfIf
placed
against
smooth
length of 26
it it
is is
placed
against
thethe
smooth
wallwall
andand
on
on the
rough
floor
in the
position
m,
will itit remain
remain in
the
rough
floor
in the
position
, will
d d= =103 ft
this position when it is released? The coefficient of static
µss ==0.3.
friction is m
0.3.
B
7.8ftm
26
A
d
+ ΣMA = 0;
150 (1.5) – NB (7.2) = 0
150 N
NB = 31.25 N
+
→ ΣFx
= 0;
31.25 – FA = 0
7.2 m
FA = 31.25 N
+↑ΣFy = 0;
NA – 150 = 0
NA = 150 N
(FA)max = 0.3 (150) = 45 N > 31.25 N
Yes, the pole will remain stationary.
1.5 m 1.5 m
Ans
680
231
SM_CH04.indd
231
08a Ch08a 674-717.indd
680
4/8/11
11:48:35
AM AM
6/12/09
9:05:09
© 2011
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Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
4–54.
N and
and a length
*8–8. The uniform pole
pole has
has aa weight
weight of
of150
30 lb
7.8ft.m.
Determine
maximum
distance
d be
it can
be
of 26
Determine
the the
maximum
distance
d it can
placed
placedthe
from
the smooth
wall
not
slip.
The coefficient
of
from
smooth
wall and
notand
slip.
The
coefficient
of static
static friction
between
the and
floorthe
and
theispole
µs =
friction
between
the floor
pole
ms =is 0.3
. 0.3.
B
26
7.8ftm
A
d
150 N
7.8 sin
+↑ΣFy = 0;
m
NA – 150 = 0
3.9 cos m
NA = 150 N
FA = (FA)max = 0.3 (150) = 45 N
+
→ ΣFx
= 0;
150 N
NB – 45 = 0
7.8 m
NB = 45 N
+ ΣMA = 0;
150 (3.9 cos ) – 45 (7.8 sin ) = 0
= 59.04°
d = 7.8 cos 59.04° = 4.013 m
Ans
681
232
232 681
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:13
11:48:36
6/12/09
AM AM
©©2011
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material
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under
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•4–55. If the coefficient of static friction at all contacting
•8–9.
surfaces is ms, determine the inclination u at which the
identical blocks, each of weight W, begin to slide.
A
B
u
Free – Body Diagram. Here, we will assume that the impending motion of the upper block is down the plane while the
impending motion of the lower block is up the plane. Thus, the frictional force F acting on the upper block acts up the
plane while the friction forces F and F¿ acting on the lower block act down the plane as indicated on the free – body
diagram of the upper and lower blocks shown in Figs. a and b, respectively. Since both block are required to be on the
verge of slipping, then F = msN and F ¿ = msN¿.
Equations of Equilibrium. Referring to Fig. a,
+a©Fy ¿ = 0;
N – W cos u = 0
N = W cos u
+Q©Fx ¿ = 0;
T + ms(W cos u) – W sin u = 0
T = W sin u – msW cos u
Using these results and referring to Fig. b,
+a©Fy ¿ = 0;
N¿ – W cos u – W cos u = 0
N¿ = 2W cos u
+Q©Fx ¿ = 0;
2(W sin u – msW cos u) – msW cos u – ms(2W cos u) – W sin u = 0
sin u – 5ms cos u = 0
Ans
u = tan–1 5ms
Since the analysis yields a positive u, the above assumption is correct.
682
233
SM_CH04.indd
233
08a Ch08a 674-717.indd
682
4/8/11
11:48:36
AM AM
6/12/09
9:05:16
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
or or
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
8–10.
20-lb ladder
*4–56. The uniform 10-kg
ladder rests
rests on the rough floor
for which the coefficient of static friction is ms = 0.8 and
against the smooth wall at B. Determine the horizontal
force P the man must exert on the ladder in order to cause
it to move.
B
1.5
5 ft m
ft
2.48 m
P
51.5
ft m
A
1.8
6 ftm
10 (9.81) N
Assume that the ladder tips about A :
10 (9.81) N
NB = 0;
+
→ ΣFx
= 0;
+↑ΣFy = 0;
P – FA = 0
1.2 m
+ ΣMA = 0;
1.2 m
–10 (9.81) + NA = 0
NA = 98.1 N
1.2 m
0.9 m
0.9 m
98.1 (0.9) – P (1.2) = 0
0.9 m
P = 73.6 N
Thus
FA = 73.6 N
(FA)max = 0.8 (10) (9.81) = 78.5 N > 73.6 N
OK
Ladder tips as assumed.
P = 73.6 N
Ans
683
234
234 683
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:18
11:48:36
6/12/09
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©©2011
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is is
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copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
8–11. The uniform 10-kg
20-lb ladder
ladder rests
rests on the rough floor
4–57.
for which the coefficient of static friction is ms = 0.4 and
against the smooth wall at B. Determine the horizontal
force P the man must exert on the ladder in order to cause
it to move.
B
51.5
ft m
ft
2.48 m
P
51.5
ft m
A
1.8
6 ftm
Assume that the ladder slips at A :
10 (9.81) N
FA = 0.4 NA
+↑ΣFy = 0;
1.2 m
NA – 10 (9.81) = 0
NA = 98.1 N
1.2 m
FA = 0.4 (98.1) = 39.24 N
+ ΣMB = 0;
P(1.2) – 98.1 (0.9) + 98.1 (1.8) – 39.24 (2.4) = 0
P = 4.905 N
+
→ ΣFx
= 0;
0.9 m
Ans
0.9 m
NB + 4.905 – 39.24 = 0
NB = 34.335 N > 0
OK
The ladder will remain in contact with the wall.
684
235
SM_CH04.indd
235
08a Ch08a 674-717.indd
684
4/8/11
11:48:37
AM AM
6/12/09
9:05:20
2011
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Saddle
River,
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rights
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This
material
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laws
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currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
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form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
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permission
in in
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4–58.
*8–12. The coefficients of static and kinetic friction
between the drum and brake bar are ms = 0.4 and mk = 0.3,
respectively. If M = 50 N # m and P = 85 N determine the
horizontal and vertical components of reaction at the pin O.
Neglect the weight and thickness of the brake. The drum has
a mass of 25 kg.
300 mm
700 mm
B
O
125 mm
500 mm
M
P
A
Equations of Equilibrium: From FBD (b),
+ΣMO = 0;
50 – FB (0.125) = 0
FB = 400 N
From FBD (a),
+ΣMA = 0;
85(1.00) + 400(0.5) – NB (0.7) = 0
NB = 407.14 N
Friction: Since FB > (FB)max = msNB = 0.4(407.14) = 162.86 N, the drum slips
at point B and rotates. Therefore, the coefficient of kinetic friction should
be used. Thus, FB = mkNB = 0.3 NB.
Equations of Equilibrium: From FBD (b),
+ΣMA = 0;
85(1.00) + 0.3 NB (0.5) – NB (0.7) = 0
NB = 154.54 N
From FBD (b),
+↑ΣFy = 0;
Oy – 245.25 – 154.54 = 0
Oy = 400 N
Ans
+
0.3(154.54) – Ox = 0
Ox = 46.4 N
Ans
S ΣFx = 0;
685
236
236 685
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:23
11:48:37
6/12/09
AM AM
©©2011
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laws
asas
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currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–59. The coefficient of static friction between the drum
•8–13.
and brake bar is ms = 0.4. If the moment M = 35 N # m,
determine the smallest force P that needs to be applied to
the brake bar in order to prevent the drum from rotating.
Also determine the corresponding horizontal and vertical
components of reaction at pin O. Neglect the weight and
thickness of the brake bar. The drum has a mass of 25 kg.
300 mm
700 mm
B
O
125 mm
P
35 – FB (0.125) = 0
FB = 280 N
+↑ΣFy = 0;
Oy – 245.25 – 700 = 0
Oy = 945 N
Ans
+
280 – Ox = 0
Ox = 280 N
Ans
S ΣFx = 0;
From FBD (a),
+ΣMA = 0;
A
Equations of Equilibrium: From FBD (b),
Equations of Equilibrium: From FBD (b),
+ΣMO = 0;
500 mm
M
P(1.00) + 280(0.5) – NB (0.7) = 0
Friction: When the drum is on the verge of rotating,
FB = msNB
280 = 0.4NB
NB = 700 N
Substituting NB = 700 N into Eq. (1) yields,
P = 350 N
Ans
686
237
SM_CH04.indd
237
08a Ch08a 674-717.indd
686
4/8/11
11:48:37
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6/12/09
9:05:25
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
Allrights
rightsreserved.
reserved.
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
This
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
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bebe
reproduced,
inin
any
form
oror
byby
any
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inin
writing
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publisher.
*4–60.
8–14. Determine the minimum coefficient of static
friction between the uniform 50-kg spool and the wall so
that the spool does not slip.
60
B
A
0.6 m
0.3 m
Free – Body Diagram. Here, the frictional force FA must act upwards to produce the counterclockwise moment
about the center of mass of the spool, opposing the impending clockwise rotational motion caused by force T as
indicated on the free – body diagram of the spool, Fig. a. Since the spool is required to be on the verge of slipping,
then FA = msNA.
Equations of Equilibrium. Referring to Fig. a,
+©MA = 0;
mg (0.6) – T cos 60° (0.3 cos 60° + 0.6) – T sin 60° (0.3 sin 60°) = 0
T = mg
+
S ©Fx = 0;
+↑©Fy = 0;
mg sin 60° – NA = 0
NA = 0.8660 mg
ms(0.8660 mg) + mg cos 60° – mg = 0
Ans
ms = 0.577
Note: Since ms is independent of the mass of the spool, it will not slip regardless of its mass provided ms > 0.577.
687
238
238 687
08aSM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:27
11:48:38
6/12/09
AM AM
©©2011
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laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
8–15.
4–61. The spool has a mass of 200 kg and rests against the
wall and on the floor. If the coefficient of static friction at B
is (ms)B = 0.3, the coefficient of kinetic friction is
(mk)B = 0.2, and the wall is smooth, determine the friction
force developed at B when the vertical force applied to the
cable is P = 800 N.
P
0.4 m
G
A
0.1 m
B
S ©Fx = 0;
+
FB – NA = 0
+c©Fy = 0;
800 – 200(9.81) + NB = 0
+©MO = 0;
–800(0.1) + FB (0.4) = 0
FB = 200 N
NB = 1162 N
(FB)max = 0.3(1162) = 348.6 N > 200 N
Thus,
FB = 200 N
Ans
688
239
SM_CH04.indd
239
08a Ch08a 674-717.indd
688
4/8/11
11:48:38
AM AM
6/12/09
9:05:29
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
or or
byby
any
means,
without
permission
in in
writing
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4–62. The 40-kg
boy stands
stands on
on the
the beam and pulls on the
*8–16.
80-lb boy
cord with a force large enough to just cause him to slip. If
the coefficient of static friction between his shoes and the
beam is (ms)D = 0.4, determine the reactions at A and B.
500 lb.
N. Neglect the
The beam is uniform and has a weight of 100
size of the pulleys and the thickness of the beam.
13
5
12
D
A
B
C
60
1.5
5 ftm
1 ftm
0.3
0.9
3 ftm
Equations of Equilibrium and Friction : When the boy is on the verge to
slipping, then FD = ( s)DND = 0.4ND. From FBD (a),
+↑ΣFy = 0;
 5
ND – T   – 40 (9.81) = 0
 13 
[1]
+
→ ΣFx
 12 
0.4ND – T   = 0
 13 
[2]
= 0;
1.2
m
4 ft
40 (9.81) N
ND = 470.88 N
500 N
1.95 m
FD = 188.352 N
Solving Eqs. [1] and [2] yields
T = 204.048 N
T = 204.048 N
ND = 470.88 N
Hence, FD = 0.4 (470.88) = 188.352 N. From FBD (b),
+ ΣMB = 0;
 5
500 (1.95) + 470.88 (2.4) – 204.048   (3.9)
 13 
1.5 m
T = 204.048 N
0.9 m
0.3 m
1.2 m
+ 204.048 (3.9) + 204.048 sin 30° (2.1) – Ay (1.2) = 0
Ans
Ay = 2340.9 N = 2.34 kN
+
→ ΣFx
= 0;
 12 
Bx + 204.048   – 188.352 – 204.048 cos 30° = 0
 13 
Bx = 176.7 N
+↑ΣFy = 0;
Ans
 5
2340.9 + 204.048   – 204.048
 13 
– 204.048 sin 30° – 470.88 – 500 – By = 0
By = 1142.4 N = 1.14 kN
Ans
689
240
240 689
08a SM_CH04.indd
Ch08a 674-717.indd
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11:48:38
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allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
No
ininany
without
exist.
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,
anyform
formororbybyany
anymeans,
means,
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•4–63. The 40-kg
boy stands
stands on
on the
the beam and pulls with a
•8–17.
80-lb boy
200lb.
N.IfIf(m
(µs)s)DD == 0.4
0.4,, determine the frictional force
force of 40
between his shoes and the beam and the reactions at A and
500 lb.
N. Neglect
B. The beam is uniform and has a weight of 100
the size of the pulleys and the thickness of the beam.
13
5
12
D
A
B
C
60
1.5
m
5 ft
1 ftm
0.3
0.9
3 ftm
1.2
m
4 ft
Equations of Equilibrium and Friction : From FBD (a),
40 (9.81) N
+↑ΣFy = 0;
 5
ND – 200   – 40 (9.81) = 0
 13 
ND = 469.32 N
+
→ ΣFx
 12 
FD – 200   = 0
 13 
FD = 184.62 N
= 0;
200 N
Since (FD)max = (µs) ND = 0.4 (469.32) = 187.73 N > FD, then the boy does
not slip. Therefore, the friction force developed is
Ans
FD = 184.62 N = 184.6 N
From FBD (b),
+ ΣMB = 0;
ND = 469.31 N 500 N
1.95 m
200 N
FD = 184.62 N
 5
500 (1.95) + 409.32 (2.4) – 200   (3.9)
 13 
+ 200 (3.9) + 200 sin 30° (2.1) – Ay (1.2) = 0
Ans
Ay = 2326.14 N = 2.326 kN
+
→ ΣFx
= 0;
1.2 m
 12 
Bx + 200   – 184.62 – 200 cos 30° = 0
 13 
Ans
Bx = 173.21 N
+↑ΣFy = 0;
0.9 m
1.5 m
0.3 m
200 N
200 N
 5
2326.14+ 200   – 200
 13 
– 200 sin 30° – 469.32 – 500 – By = 0
Ans
By = 1133.74 N = 1.133 kN
690
241
SM_CH04.indd
241
08a Ch08a 674-717.indd
690
4/8/11
11:48:39
AM AM
6/12/09
9:05:34
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
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©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
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form
any
means,
without
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writing
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exist.
NoNo
portion
of of
this
material
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reproduced,
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any
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oror
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in in
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*4–64. The tongs are used to lift the 150-kg crate, whose
8–18.
center of mass is at G. Determine the least coefficient of
static friction at the pivot blocks so that the crate can be
lifted.
P
275 mm
E
500 mm
C
30
F
H
D
500 mm
A
G
B
300 mm
Free – Body Diagram. Since the crate is suspended from the tongs, P must be equal to the weight of
the crate; i.e., P = 150(9.81) N as indicated on the free – body diagram of joint H shown in Fig. a.
Since the crate is required to be on the verge of slipping downward, FA and FB must act upward so
that FA = msNA and FB = msNB as indicated on the free – body diagram of the crate shown in Fig. c.
Equations of Equilibrium. Referring to Fig. a,
+
S ©Fx = 0;
+↑©Fy = 0;
FHE cos 30° – FHF cos 30° = 0
FHE = FHF = F
150(9.81) – 2F sin 30° = 0
F = 1471.5 N
Referring to Fig. b,
+©MC = 0;
1471.5 cos 30° (0.5) + 1471.5 sin 30° (0.275) – NA (0.5) – msNA (0.3) = 0
0.5 NA + 0.3msNA = 839.51
(1)
Due to the symmetry of the system and loading, NB = NA. Referring to Fig. c,
+↑©Fy = 0;
2msNA – 150(9.81) = 0
(2)
Solving Eqs. (1) and (2), yields
NA = 1237.57 N
Ans
ms = 0.595
691
242
242 691
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:38
11:48:39
6/12/09
AM AM
©©2011
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
8–19.
B have
have aa weight
weight of
of 50
10 N
lband
and30
6 lb,
N,
4–65. Two blocks A and B
respectively. They are resting on the incline for which the
coefficients of static friction are mA = 0.15 and mB = 0.25.
Determine the incline angle u for which both blocks begin
to slide. Also find the required stretch or compression in the
connecting spring for this to occur. The spring has a stiffness
of k == 40
2 lb>ft
N/m..
k k= 402N/m
lb/ft
A
u
50 N
Equations of Equilibrium : Using the spring force formula,
Fsp = kx = 40x. From FBD (a),
Fsp = 40x
ΣFx′ = 0;
40x + FA – 50 sin
+
ΣFy′ = 0;
NA – 50 cos = 0
[2]
↑
↑
+
=0
B
[1]
From FBD (b),
ΣFx′ = 0;
Fs – 40x – 30 sin = 0
[3]
+
ΣFy′ = 0;
NB – 30 cos = 0
[4]
↑
↑
+
Friction : If block A and B are on the verge to move, slipping
would have to occur at point A and B. Hence, FA = sA NA =
0.15NA and FB = s B NB = 0.25NB. Substituting these values into
Eqs. [1], [2], [3] and [4] and solving, we have
= 10.62°
NA = 49.14 N
x = 0.046 m
30 N
Fsp = 40x
Ans
NB = 29.49 N
692
243
SM_CH04.indd
243
08a Ch08a 674-717.indd
692
4/8/11
11:48:40
AM AM
6/12/09
9:05:40
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
4–66. Two blocks A and B
*8–20.
B have
have aa weight
weight of
of50
10N
lband
and30
6 lb,
N,
respectively. They are resting on the incline for which the
coefficients of static friction are mA = 0.15 and mB = 0.25.
Determine the angle u which will cause motion of one of
the blocks. What is the friction force under each of the
blocks when this occurs? The spring has a stiffness of
k == 40
2 lb>ft
N/m and is originally unstretched.
kk = 40
N/m
2 lb/ft
A
u
Equations of Equilibrium : Since Block A and B is either not moving or
on the verge of moving, the spring force Fsp = 0. From FBD (a),
ΣFx′ = 0;
FA – 50 sin = 0
[1]
+
ΣFy′ = 0;
NA – 50 cos
[2]
↑
↑
+
=0
B
50 N
From FBD (b),
ΣFx′ = 0;
FB – 30 sin = 0
[3]
+
ΣFy′ = 0;
NB – 30 cos = 0
[4]
↑
↑
+
30 N
Friction : Assuming block A is on the verge of slipping, then
FA =
sA
NA = 0.15NA
[5]
Solving Eqs. [1], [2], [3], [4] and [5] yields
= 8.531°
FA = 7.42 N
NA = 49.45 N
FB = 4.45 N
NB = 29.67 N
Since (FB)max = µsB NB = 0.25 (29.67) = 7.42 N > FB, block B does not slip.
Therefore, the above assumption is correct. Thus
= 8.53°
FB = 4.45 N
FA = 7.42 N
Ans
693
244
244 693
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:42
11:48:40
6/12/09
AM AM
©©2011
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•8–21.
weigh 1000
200 lb
N and
and 150
750 lb,
N,
•4–67. Crates A and B weigh
respectively. They are connected together with a cable and
placed on the inclined plane. If the angle u is gradually
increased, determine u when the crates begin to slide. The
coefficients of static friction between the crates and the
plane are mA = 0.25 and mB = 0.35.
B
D
A
C
u
Free – Body Diagram. Since both crates are required to be on the verge of sliding down the plane, the
frictional forces FA and FB must act up the plane so that FA = ANA = 0.25NA and FB = B NB = 0.35NB as
indicated on the free – body diagram of the crates shown in Figs. a and b.
+
ΣFy′ = 0;
NA – 1000 cos
+
ΣFx′ = 0;
FCD + 0.25 (1000 cos ) – 1000 sin
↑
↑
Equations of Equilibrium. Referring to Fig. a,
=0
NA = 1000 cos
=0
(1)
+
ΣFy′ = 0;
NB – 750 cos
+
ΣFx′ = 0;
0.35 (750 cos ) – FCD – 750 sin = 0
↑
↑
Also, by referring to Fig. b,
=0
NB = 750 cos
(2)
Solving Eqs. (1) and (2), yields
Ans
= 16.3°
FCD = 41.13 N
750 N
1000 N
694
245
SM_CH04.indd
245
08a Ch08a 674-717.indd
694
4/8/11
11:48:41
AM AM
6/12/09
9:05:44
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
byby
any
means,
without
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inin
writing
from
the
publisher.
8–30.
lb with center of
weight of
of 8000
40 kN
*4–68. The tractor has aa weight
gravity at G. Determine if it can push
push the
the 250-kg
550-lb log up the
incline. The coefficient of static friction between the log and
the ground is ms = 0.5, and between the rear wheels of the
tractor and the ground msœ = 0.8. The front wheels are free
to roll. Assume the engine can develop enough torque to
cause the rear wheels to slip.
C
G
1.25 ftm
0.375
B
10
A
2.5 ftm
0.75
2.1
7 ftm
30.9
ft m
40 kN
Log :
ΣFy = 0;
↑
+
P = 1633.5 N
NC – 250 (9.81) cos 10° = 0
0.75 m
NC = 2415.2 N
ΣFx = 0;
↑
+
0.9 m
0.375 m
2.1 m
–0.5 (2415.2) – 250 (9.81) sin 10° + P = 0
P = 1633.5 N
Tractor :
+ ΣMg = 0;
250 (9.81) N
3
1633.5 (0.375) + 40 (10 ) (cos 10°) (0.9)
+ 40 (103) (sin 10°) (0.75) – NA (3) = 0
NA = 13.758 kN
ΣFx = 0;
↑
+
FA – 40 (103) sin 10° – 1633.5 = 0
FA = 8.579 kN
(FA)max = 0.8 (13.758) = 11.006 kN > 8.579 kN
Tractor can move log.
Ans
702
246
SM_CH04.indd
246
08a Ch08a 674-717.indd
702
4/8/11
6/12/09 11:48:41
9:06:07 AM
AM
©©2011
This
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formororby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
weight of
of 8000
40 kN
8–31. The tractor has aa weight
lb with
with center of
4–69.
gravity at G. Determine the greatest weight of the log that
can be pushed up the incline. The coefficient of static
friction between the log and the ground is ms = 0.5, and
between the rear wheels of the tractor and the ground
msœ = 0.7. The front wheels are free to roll. Assume the
engine can develop enough torque to cause the rear wheels
to slip.
C
G
B
10
A
0.375
1.25 ftm
2.5 ftm
0.75
3 ft
72.1
ft m 0.9 m
40 kN
Tractor :
+ ΣMg = 0;
0.75 m
40 (cos 10°) (0.9) + 400 (sin 10°) (0.75)
+ P (0.375) – NA (3) = 0
0.9 m
0.375 m
2.1 m
NA – P (0.125) = 13.55417
ΣFx = 0;
↑
+
0.7 FA – 40 sin 10° – P = 0
0.7 FA – P = 6.94593
NA = 13.902 kN
P = 2.786 kN
P = 2.786 kN
+
ΣFy = 0;
NC – W cos 10° = 0
+
ΣFx = 0;
2.786 – W sin 10° – 0.5 NC = 0
↑
↑
Log :
NC = 4.119 kN
W = 4.183 kN
Ans
703
247
SM_CH04.indd
247
08a Ch08a 674-717.indd
703
4/8/11
6/12/09 11:48:42
9:06:09 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
byby
any
means,
without
permission
inin
writing
from
the
publisher.
8–26.
180 N
lb (and90
rests
4–70. The refrigerator has a weight of 900
kg)on
anda
tile
forfloor
which
If the
rests floor
on a tile
for which 0.25
If theman
man pushes
µs =. 0.25.
horizontally on the refrigerator in the direction shown,
determine the smallest magnitude of horizontal force
needed to move it. Also, if
if the
the man
man has
has aa weight
weight of
of 150
750 lb,
N
determine
smallest the
coefficient
friction between
his
( 75 kg),the
determine
smallestofcoefficient
of friction
shoes
andhis
theshoes
floorand
so that
he does
not slip.
between
the floor
so that
he does not slip.
0.9 m
G
0.45 m
1.2 m
0.9 m
A
Equations of Equilibrium : From FBD (a),
+↑ΣFy = 0;
N – 900 = 0
+
→ ΣFx
P–F=0
[1]
900 (x) – P (1.2) = 0
[2]
= 0;
+ ΣMA = 0;
N = 900 N
0.45 m
Friction : Assuming the refrigerator is on the verge of slipping, then F = N
= 0.25 (900) = 225 N. Substituting this value into Eqs. [1], and [2] and solving
yields
P = 225 N
900 N
x = 0.3 m
1.2 m
Since x < 0.45 m, the refrigerator does not tip. Therefore, the above assumption
is correct. Thus
Ans
P = 225 N
From FBD (b),
P = 225 N
+↑ΣFy = 0;
Nm – 750 = 0
Nm = 750 N
+
→ ΣFx
Fm – 225 = 0
Fm = 225 N
= 0;
750 N
When the man is on the verge of slipping, then
Fm =
225 =
′Nm
s
′ (750)
s
′ = 0.300
Ans
s
698
248
SM_CH04.indd
248
08a Ch08a 674-717.indd
698
4/8/11
6/12/09 11:48:42
9:05:57 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
900 lb
N (and90
kg)on
and
8–27.
rests
a
4–71. The refrigerator has a weight of 180
restsfloor
on afor
tile
floor for which
0.25.man
Also,
man has
µs = the
tile
which
hasthe
a weight
of
0.25. Also,
a weight
750
N ( 75ofkg)
andfriction
the coefficient
of static
150
lb andofthe
coefficient
static
between the
floor
friction
between
shoes
is µs = 0.6.onIfthe
he
and
his shoes
is the floor
he his
pushes
horizontally
0.6. Ifand
pushes horizontally
on the
refrigerator,
if he can
refrigerator,
determine
if he
can movedetermine
it. If so, does
the
move it. If so,
does
the refrigerator slip or tip?
refrigerator
slip
or tip?
0.9 m
G
0.45 m
1.2 m
0.9 m
A
Equations of Equilibrium : From FBD (a),
0.45 m
+↑ΣFy = 0;
N – 900 = 0
+
→ ΣFx
P–F=0
[1]
900 (x) – P (1.2) = 0
[2]
= 0;
+ ΣMA = 0;
N = 900 N
900 N
Friction : Assuming the refrigerator is on the verge of slipping, then F = N
= 0.25 (900) = 225 N. Substituting this value into Eqs. [1], and [2] and solving
yields
P = 225 N
1.2 m
x = 0.3 m
Since x < 0.45 m, the refrigerator does not tip. Therefore, the above assumption
is correct. Thus, the refrigerator slips.
Ans
750 N
P = 225 N
From FBD (b),
+↑ΣFy = 0;
Nm – 750 = 0
Nm = 750 N
+
→ ΣFx
Fm – 225 = 0
Fm = 225 N
= 0;
Since (Fm)max = s′Nm = 0.6 (750) = 450 N > Fm, then the man does not slip.
Thus, the man is capable of moving the refrigerator.
Ans
699
249
SM_CH04.indd
249
08a Ch08a 674-717.indd
699
4/8/11
6/12/09 11:48:42
9:06:01 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
*8–24.
The drum has a weight of 500
100 N
lb and rests on the
*4–72. The
floor for which the coefficient
coefficient of
ofstatic
staticfriction
frictionisismµss = 0.6.
0.6. If
a ==0.6
m and
and b == 0.9
m,determine
determinethe
the smallest
smallest magnitude of
2 ft
3 ft,
the force P that will cause impending motion of the drum.
P
5
3
a
4
Assume that the drum tips :
b
x = 0.3 m
+ ΣMO = 0;
 3
 4
500 (0.3) + P   (0.6) – P   (0.9) = 0
 5
 5
P = 416.67 N
+
→ ΣFx
= 0;
 4
–F – 416.67   = 0
 5
500 N
F = 333.33 kN
+↑ΣFy = 0;
 3
N – 500 – 416.67   = 0
 5
0.9 m
N = 750 N
0.3 m
Fmax = 0.6 (750) = 450 N > 333.33 N
0.3 m
OK
Drum tips as assumed,
P = 416.67 N
Ans
The drum has a weight of 500
100 N
lb and rests on the
•8–25.
•4–73. The
0.5. If
floor for which the coefficient
coefficient of
ofstatic
staticfriction
frictionisismµss = 0.5.
3 ft
4 ft,
a ==0.9
m and
and b == 1.2
m,determine
determinethe
the smallest
smallest magnitude of
the force P that will cause impending motion of the drum.
P
3
5
a
4
Assume that the drum slips :
b
F = 0.5 N
+
→ ΣFx
= 0;
+↑ΣFy = 0;
 4
–0.5 N + P   = 0
 5
 3
–P   – 500 + N = 0
 5
500 N
P = 500 N
N = 800 N
+ ΣMO = 0;
1.2 m
 3
 4
800 (x) + 500   (0.45) – 500   (1.2) = 0
 5
 5
x = 0.431 m < 0.45 m
0.45 m
0.45 m
OK
Drum slips as assumed,
P = 500 N
Ans
697
250
250 697
08a SM_CH04.indd
Ch08a 674-717.indd
4/8/119:05:52
11:48:43
6/12/09
AM AM
0.5 cos u – 1.06 sin u = 0
 0.5 
u = tan–1 
Ans
 = 25.3°
 1.06
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected
under
all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•4–74.
•8–77. The square threaded screw of the clamp has a
mean diameter of 14 mm and a lead of 6 mm. If ms = 0.2 for
the threads, and the torque applied to the handle is
1.5 N # m, determine the compressive force F on the block.
1.5 N m
 1 
 6 
Frictional Forces on Screw: Here, u = tan–1 
= tan–1 
 = 7.768°,
 2pr 
2p(7) 
W = F and f, = tan–1m, = tan–i (0.2) = 11.310°. Applying Eq. 8 – 3, we have
M = Wr tan (u + f)
F
1.5 = F (0.007) tan (7.768° + 11.310°)
F = 620 N
Ans
Note: Since f, > u, the screw is self-locking. It will not unscrew even if the
moment is removed.
F
745
08b Ch08b 718-761.indd 745
6/12/09 10:34:16 AM
251
SM_CH04.indd 251
4/8/11 11:48:44 AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
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exist.
No
portion
ofof
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inin
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inin
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8–78.
4–75. The device is used to pull the battery cable terminal
C from the post of a battery. If the required pulling force is
85
MM
that
must
be applied
to the
425lb,N,determine
determinethe
thetorque
torque
that
must
be applied
to
handle
on on
thethe
screw
to to
tighten
the handle
screw
tightenit.it.The
Thescrew
screw has
has square
threads, a mean
in., aalead
in., and the
mean diameter
diameterof
of0.2
5 mm,
leadofof0.08
2 mm,
coefficient of static friction is m
0.5.
µss ==0.5.
M


 1 
2
Frictional Forces on Screw : Here, = tan–1 
= tan–1 
 = 7.256°,

 2π r 
 2π (2.5) 
W = 425 N and
s
= tan–1
s
= tan–1 (0.5) = 26.565°. Applying Eq. 8–3, we have
M = Wr tan ( + )
= 425 (2.5) tan (7.256° + 26.565°)
= 711.85 N · m
= 0.712 N · m
Note: Since
s
A
Ans
C
B
> , the screw is self-locking. It will not unscrew even if the moment is removed.
8–79.
The jacking mechanism
mechanism consists
consistsofofaalink
linkthat
thathas
has
9
a
*4–76.
a square-threaded
screw
with
a mean
diameter
ofin.
12and
mm
square-threaded
screw
with
a mean
diameter
of 0.5
a
and aoflead
of in.,
5 mm,
coefficientofofstatic
static friction
friction is
lead
0.20
andand
thethe
coefficient
µss ==0.4.
m
Determinethe
thetorque
torqueM
M that
that should
should be applied to
0.4.Determine
to start
start lifting
lifting the
the6000-lb
30-kN load
the screw to
load acting
acting at
at the end of
member ABC.
30 kNlb
6000
C
B
M
187.5
mm
7.5 in.
250
10mm
in.
D
A
20 imm
n.
500
15 mm
in.
375
10 in.
250 mm
 250 
= tan–1 
= 21.80°
 625 
+ ΣMA = 0;
–30 (875) + FBD cos 21.80° (250) + FBD sin 21.80° (500) = 0
30 kN
FBD = 62.828 kN
z
= tan–1 (0.4) = 21.80°
 5 
= tan–1 
 = 7.555°
 2π (6) 
250 mm
M = Wr tan ( + )
M = 62.828 (6) tan (7.555° + 21.80°)
500 mm
375 mm
M = 212.02 kN · mm = 212.0 N · m
746
252
SM_CH04.indd
252
08b Ch08b 718-761.indd
746
4/8/11 11:48:45
6/12/09
10:34:18 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
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formor
orby
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anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
4–77. Determine the magnitude of the horizontal force P
*8–80.
that must be applied to the handle of the bench vise in order
to produce a clamping force of 600 N on the block. The
single square-threaded screw has a mean diameter of
25 mm and a lead of 7.5 mm. The coefficient of static
friction is ms = 0.25.
100 mm
Here, M = P (0.1)
P
 7.5 
 L 
= tan–1 
u = tan 
 = 5.455°

2pr


2p(12.5)
–1
fs = tan–1ms = tan–1(0.25) = 14.036°
W = 600 N
Thus
M = Wr tan (fs + u)
P(0.1) = 600(0.0125) tan (14.036° + 5.455°)
P = 26.5 N
Ans
Note: Since fs > u, the screw is self-locking.
•4–78. Determine the clamping force exerted on the
•8–81.
block if a force of P = 30 N is applied to the lever of the
bench vise. The single square-threaded screw has a mean
diameter of 25 mm and a lead of 7.5 mm. The coefficient of
static friction is ms = 0.25.
100 mm
P
Here, M = 30(0.1) = 3 N · m
 7.5 
 L 
u = tan–1 
= tan–1 
 = 5.455°

 2pr 
2p(12.5)
fs = tan–1ms = tan–1(0.25) = 14.036°
W=F
Thus
M = Wr tan (fs + u)
3 = F (0.0125) tan (14.036° + 5.455°)
F = 678 N
Ans
Note: Since fs > u, the screw is self-locking.
747
253
SM_CH04.indd
253
08b
Ch08b 718-761.indd
747
4/8/11 11:48:46
6/12/09
10:34:21 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
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material
may
bebe
reproduced,
inin
any
form
oror
byby
any
means,
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inin
writing
from
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publisher.
8–82.
4–79. Determine the required horizontal force that must
be applied perpendicular to the handle in order to develop
a 900-N clamping force on the pipe. The single squarethreaded screw has a mean diameter of 25 mm and a lead of
5 mm. The coefficient of static friction is ms = 0.4. Note: The
screw is a two-force member since it is contained within
pinned collars at A and B.
E
C
200 mm
150 mm
A
B
200 mm
D
Referring to the free-body diagram of member ED shown in Fig. a,
+©MD = 0;
FAB(0.2) – 900(0.4) = 0
FAB = 1800 N

 L 
5 
Here, u = tan–1 
= tan–1 
 = 3.643°

 2pr 
2p(12.5)
fs = tan–1ms = tan–1(0.4) = 21.801°
M = F (0.15); and W = FAB = 1800 N
M = Wr tan (fs + u)
F(0.15) = 1800(0.0125) tan (21.801° + 3.643°)
F = 71.4 N
Ans
Note: Since fs > u, the screw is self-locking.
748
254
SM_CH04.indd
254
08b Ch08b 718-761.indd
748
4/8/11 11:48:47
6/12/09
10:34:23 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
8–83.
*4–80. If the clamping force on the pipe is 900 N,
determine the horizontal force that must be applied
perpendicular to the handle in order to loosen the screw.
The single square-threaded screw has a mean diameter of
25 mm and a lead of 5 mm. The coefficient of static friction
is ms = 0.4. Note: The screw is a two-force member since it
is contained within pinned collars at A and B.
E
C
200 mm
150 mm
A
B
200 mm
D
Referring to the free-body diagram of member ED shown in Fig. a,
+©MD = 0;
FAB(0.2) – 900(0.4) = 0
FAB = 1800 N

 L 
5 
Here, u = tan–1 
= tan–1 
 = 3.643°

2p(12.5)
2pr




fs = tan–1ms = tan–1(0.4) = 21.801°
M = F (0.15); and W = FAB = 1800 N
M = Wr tan (fs – u)
F(0.15) = 1800(0.0125) tan (21.801° + 3.643°)
F = 49.2 N
Ans
749
255
SM_CH04.indd
255
08b
Ch08b 718-761.indd
749
4/8/11 11:48:48
6/12/09
10:34:25 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
byby
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inin
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*8–84.
4–81. The clamp provides pressure from several directions
on the edges of the board. If the square-threaded screw has a
lead of 3 mm, mean radius of 10 mm, and the coefficient of
static friction is ms = 0.4, determine the horizontal force
developed on the board at A and the vertical forces
developed at B and C if a torque of M = 1.5 N # m is applied
to the handle to tighten it further. The blocks at B and C are
pin connected to the board.
B
45
D
A
M
45
C
fz = tan–1(0.4) = 21.801°

3 
up = tan–1 
 = 2.734°
 2p(10) 
M = W(r) tan (fs + up)
1.5 = Az (0.01) tan (21.801° + 2.734°)
Ans
Az = 328.6 N
+
S ©Fx = 0;
328.6 – 2T cos 45° = 0
T = 232.36 N
By = Cy = 232.36 sin 45° = 164 N
Ans
750
256
SM_CH04.indd
256
08b Ch08b 718-761.indd
750
4/8/11 11:48:50
6/12/09
10:34:27 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•4–82. If the jack supports the 200-kg crate, determine the
•8–85.
horizontal force that must be applied perpendicular to the
handle at E to lower the crate. Each single square-threaded
screw has a mean diameter of 25 mm and a lead of 7.5 mm.
The coefficient of static friction is ms = 0.25.
C
A
E
45
45
45
45
100 mm
B
D
The force in rod AB can be obtained by first analyzing the equilibrium of joint C followed by joint B. Referring
to the free-body diagram of joint C shown in Fig. a,
S ©Fx = 0;
+
FCA sin 45° – FCB sin 45° = 0
FCA = FCB = F
+c©Fy = 0;
2F cos 45° – 200(9.81) = 0
F = 1387.34 N
Using the result of F and referring to the free-body diagram of joint B shown in Fig. b,
+c©Fy = 0;
FBD sin 45° – 1387.34 sin 45° = 0
FBD = 1387.34 N
+
1387.34 cos 45° +1387.34 cos 45° – FAB = 0
FAB = 1962 N
S ©Fx = 0;

 L 
5 
= tan–1 
Here, u = tan–1 
 = 5.455°

2p(12.5)
2pr




fs = tan–1ms = tan–1(0.25) = 14.036°
M = F (0.1) and W = FAB = 1962 N
Since M must overcome the friction of two screws,
M = 2[Wr tan (fs – u)]
F(0.1) = 2[1962(0.0125) tan (14.036° – 5.455°)]
Ans
F = 74.0 N
Note: Since fs > u, the screw are self-locking.
fz = tan–1(0.4) = 21.801°

3 
up = tan–1 
 = 2.734°
2p(10)


M = W(r) tan (fs + up)
1.5 = Az (0.01) tan (21.801° + 2.734°)
Ans
Az = 328.6 N
+
S ©Fx = 0;
328.6 – 2T cos 45° = 0
T = 232.36 N
By = Cy = 232.36 sin 45° = 164 N
Ans
751
257
SM_CH04.indd
257
08b
Ch08b 718-761.indd
751
4/8/11 11:48:51
6/12/09
10:34:30 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
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reproduced,
inin
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form
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inin
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8–86. If the jack is required to lift the 200-kg crate,
4–83.
determine the horizontal force that must be applied
perpendicular to the handle at E. Each single squarethreaded screw has a mean diameter of 25 mm and a lead of
7.5 mm. The coefficient of static friction is ms = 0.25.
C
A
E
45
45
45
45
100 mm
B
D
The force in rod AB can be obtained by first analyzing the equilibrium of joint C followed by joint B. Referring
to the free-body diagram of joint C shown in Fig. a,
S ©Fx = 0;
+
FCA sin 45° – FCB sin 45° = 0
FCA = FCB = F
+c©Fy = 0;
2F cos 45° – 200(9.81) = 0
F = 1387.34 N
Using the result of F and referring to the free-body diagram of joint B shown in Fig. b,
+c©Fy = 0;
FBD sin 45° – 1387.34 sin 45° = 0
FBD = 1387.34 N
+
1387.34 cos 45° +1387.34 cos 45° – FAB = 0
FAB = 1962 N
S ©Fx = 0;

 L 
5 
= tan–1 
Here, u = tan–1 
 = 5.455°

 2pr 
2p(12.5)
fs = tan–1ms = tan–1(0.25) = 14.036°
M = F (0.1) and W = FAB = 1962 N
Since M must overcome the friction of two screws,
M = 2[Wr tan (fs + u)]
F(0.1) = 2[1962(0.0125) tan (14.036° + 5.455°)]
F = 174 N
Ans
Note: Since fs > u, the screws are self-locking.
752
258
SM_CH04.indd
258
08b Ch08b 718-761.indd
752
4/8/11 11:48:52
6/12/09
10:34:33 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formororby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
*4–84. The machine part is held in place using the
8–87.
double-end clamp. The bolt at B has square threads with a
mean radius of 4 mm and a lead of 2 mm, and the
coefficient of static friction with the nut is ms = 0.5. If a
torque of M = 0.4 N # m is applied to the nut to tighten it,
determine the normal force of the clamp at the smooth
contacts A and C.
90 mm
260 mm
B
A
C
f = tan–1(0.5) = 26.565°
 2 
= 4.550°
u = tan–1 
 2p(4)
M = W(r) tan (u + f)
0.4 = W (0.004) tan (4.550° + 26.565°)
W = 165.67 N
+©MA = 0;
NC (350) – 165.67 (260) = 0
NC = 123.1 = 123 N
+c©Fy = 0;
Ans
NA – 165.67 + 123.1 = 0
NA = 42.6 N
Ans
4–85. Blocks A and
*8–88.
andBBweigh
weigh250
50 lb
30 lb,
N and 150
N, respectively.
Using the coefficients of static friction indicated, determine
the greatest weight of block D without causing motion.
m
B
0.5
mBA
20
0.6
A
C
D
mAC
0.4
For block A and B : Assuming block B does not slip
+↑ΣFy = 0;
NC – (250 + 150)= 0
NC = 400 N
+
→ ΣFx
0.4 (400) – TB = 0
TB = 160 N
= 0;
(250 + 150) N
For block B :
+↑ΣFy = 0;
NB cos 20° + FB sin 20° – 150 = 0
[1]
+
→ ΣFx
FB cos 20° – NB sin 20° – 160 = 0
[2]
= 0;
Solving Eqs. [1] and [2] yields :
FB = 201.65 N
150 N
NB = 86.23 N
TB = 160 N
Since FB = 201.65 N > NB = 0.6 (86.23) = 51.74 N, slipping does occur
between A and B. Therefore, the assumption is no good.
Since slipping occurs, FB = 0.6 NB.
+↑ΣFy = 0;
NB cos 20° + 0.6NB sin 20° – 150 = 0
NB = 131.0 N
+
→ ΣFx
0.6 (131.0) cos 20° – 131.0 sin 20° – TB = 0
TB = 29.055 N
= 0;
T2 = T1e
Where T2 = WD, T1 = TB = 5.812 N,
150 N
= 0.5π rad
WD = 29.055e0.5(0.5π)
= 63.73 N
Ans
259
753
SM_CH04.indd
259
08b Ch08b 718-761.indd
753
4/8/11 11:48:53
6/12/09
10:34:36 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
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publisher.
exist.
No
portion
ofof
this
material
may
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reproduced,
inin
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oror
byby
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inin
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•4–86. Blocks A and B
•8–89.
N each,
each, and D weighs
B weigh
weigh 375
75 lb
150
N. Using
Using the
the coefficients
coefficients of
of static
static friction indicated,
30 lb.
determine the frictional force between blocks A and B and
between block A and the floor C.
For the rope, T3 = T1 e , where T2 = 150 N, T1 = TB, and
m
B
0.5
mBA
20
0.6
A
= 0.5π rad.
C
D
mAC
0.4
150 = TB e0.5(0.5π)
TB = 68.391 N
Ans
FC = 68.4 N
For block B :
+↑ΣFy = 0;
NB cos 20° + FB sin 20° – 3.75 = 0
[1]
+
→ ΣFx
FB cos 20° – NB sin 20° – 68.391 = 0
[2]
= 0;
Solving Eqs. [1] and [2] yields :
NB = 329.0 N
Ans
FB = 192.5 N
Since FB = 192.5 N > NB = 0.6 (329.0) = 197.4 N, slipping between
A and B does not occur.
8–90.
4–87. A cylinder having a mass of 250 kg is to be
supported by the cord which wraps over the pipe.
Determine the smallest vertical force F needed to support
the load if the cord passes (a) once over the pipe, b = 180°,
and (b) two times over the pipe, b = 540°. Take ms = 0.2.
F
Frictional Force on Flat Belt: Here, T1 = F and T2 = 250(9.81) = 2452.5 N
Applying Eq. 8 – 6, we have
a) If b = 180° = p rad
T2 = T1emb
2452.5 = Fe0.2p
Ans
F = 1308.38 N = 1.31 kN
b) If b = 540° = 3p rad
T2 = T1emb
2452.5 = Fe0.2(3p)
Ans
F = 372.38 N = 372 kN
260
754
SM_CH04.indd
260
08b Ch08b 718-761.indd
754
4/8/11 11:48:53
6/12/09
10:34:39 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
*4–88. A cylinder having a mass of 250 kg is to be
8–91.
supported by the cord which wraps over the pipe.
Determine the largest vertical force F that can be applied
to the cord without moving the cylinder. The cord passes
(a) once over the pipe, b = 180°, and (b) two times over the
pipe, b = 540°. Take ms = 0.2.
F
Frictional Force on Flat Belt: Here, T1 = 250(9.81) = 2452.5 N and T2 = F.
Applying Eq. 8 – 6, we have
a) If b = 180° = p rad
T2 = T1emb
F = 2452.5e0.2p
Ans
F = 4597.10 N = 4.60 kN
b) If b = 540° = 3p rad
T2 = T1emb
F = 2452.5e0.2(3p)
Ans
F = 152.32 N = 16.2 kN
755
261
SM_CH04.indd
261
08b
Ch08b 718-761.indd
755
4/8/11 11:48:54
6/12/09
10:34:41 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
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portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
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means,
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inin
writing
from
the
publisher.
4–89. The boat
*8–92.
boat has
hasa aweight
weightof of
500N lb
2500
( and
250 is
kg)held
and in
is
position
off the side
of aside
shipofby
the spars
atspars
A andatB.
man
held in position
off the
a ship
by the
AA
and
B.
having
a weight
of 130 oflb650
getsN in
boat,
wraps
a boat,
rope
A man having
a weight
( the
65 kg)
gets
in the
around
overhead
at C, and
ties itattoC,the
end
wraps aan
rope
aroundboom
an overhead
boom
and
tiesofitthe
to
boat
as shown.
If theasboat
is disconnected
from
the spars,
the end
of the boat
shown.
If the boat is
disconnected
determine
the minimum
number
of half turns
the rope
from the spars,
determine
the minimum
number
of must
half
make
around
the boom
so thatthe
theboom
boatsocan
turns the
rope must
make around
thatbe
thesafely
boat
lowered
into the
water at
constant
velocity.
Also, what
is the
can be safely
lowered
into
the water
at constant
velocity.
normal
forceisbetween
the boat
the man?
coefficient
Also, what
the normal
forceand
between
theThe
boat
and the
of
kinetic
friction between
thefriction
rope and
the boom
is
man?
The coefficient
of kinetic
between
the rope
m
. Hint:isThe
requires
that the requires
normal force
0.15
and
boom
µs =problem
0.15. Hint:
The problem
that
s =the
between
theforce
man’s
feet andthe
theman’s
boat be
small
possible.
the normal
between
feetasand
theasboat
be as
small as possible.
C
A
B
Frictional Force on Flat Belt : If the normal force between the man and the boat
is equal to zero, then, T1 = 650 N and T2 = 2500 N. Applying Eq. 8–6, we have
T 2 = T1 e
2500 = 650e0.15
= 8.980 rad
2500 N
The least number of half turns of the rope required is d = 2.86 turns. Thus
Use
Ans
n = 3 half turns
650 N
Equations of Equilibrium : From FBD (a),
+↑ΣFy = 0;
T2 – Nm – 2500 = 0
T2 = Nm + 2500
T1 + Nm – 650 = 0
T1 = 650 – Nm
From FBD (b),
+
→ ΣFx
= 0;
Frictional Force on Flat Belt : Hence,
= 3π rad. Applying Eq. 8–6, we have
T 2 = T1 e
Nm + 2500 = (650 – Nm)e0.15
Nm = 33.71 N
Ans
756
262
SM_CH04.indd
262
08b Ch08b 718-761.indd
756
4/8/11 11:48:54
6/12/09
10:34:42 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formor
orby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•8–93. The 50-kg
100-lb boy
boy at
at A
A is
is suspended from the cable
•4–90.
that passes over the quarter circular cliff rock. Determine if
it is possible for
for the
the 92.5-kg
185-lb woman
woman to
to hoist
hoist him
him up; and if
this is possible, what smallest force must she exert on the
horizontal cable? The coefficient of static friction between
the cable and the rock is ms = 0.2, and between the shoes of
the woman and the ground msœ = 0.8.
=
π
2
92.5 (9.81) N
T2 = T1 e = 50 (9.81) e0.2
+↑ΣFy = 0;
π
2
= 671.55 N
T2 = 671.55 N
A
N = 92.5 (9.81) = 0
N = 907.425 N
+
→ ΣFx
= 0;
671.55 – F = 0
F = 671.55 N
Fmax = 0.8 (907.425) = 725.9 N > 671.55 N
Yes, just barely.
Ans
8–94. The 50-kg
100-lb boy
boy at
at A
A is
is suspended from the cable
4–91.
that passes over the quarter circular cliff rock. What
horizontal force must the woman at A exert on the cable in
order to let the boy descend at constant velocity? The
coefficients of static and kinetic friction between the cable
and the rock are ms = 0.4 and mk = 0.35, respectively.
=
π
2
T2 = T1 e ;
50 (9.81) = T1 e0.35
T1 = 283.1 N
A
π
2
Ans
757
263
SM_CH04.indd
263
08b Ch08b 718-761.indd
757
4/8/11 11:48:55
6/12/09
10:34:44 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
byby
any
means,
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inin
writing
from
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publisher.
8–95.
*4–92. A 10-kg cylinder D, which is attached to a small
pulley B, is placed on the cord as shown. Determine the
smallest angle u so that the cord does not slip over the peg at
C. The cylinder at E has a mass of 10 kg, and the coefficient
of static friction between the cord and the peg is ms = 0.1.
A
u
u
C
B
E
D
Since pulley B is smooth, the tension in the cord between pegs A and C remains constant. Referring to the free-body
diagram of the joint B shown in Fig. a, we have
+c©Fy = 0;
2T sin u – 10(9.81) = 0
T=
49.05
sin u
In the case where cylinder E is on the verge of ascending, T2 = T =
49.05
p
and T1 = 10(9.81) N. Here,
+ u, Fig. b. Thus,
sin u
2
T2 = T1emsb
 p
49.05
= 10(9.81)e0.1  2
sin u
In

+ u

p

0.5
= 0.1  + u
sin u
2

Solving by trial and error, yields
Ans
u = 0.4221 rad = 24.2°
4–93.
*8–96. A 10-kg cylinder , which is attached to a small
pulley B, is placed on the cord as shown. Determine the
largest angle u so that the cord does not slip over the peg at
C. The cylinder at E has a mass of 10 kg, and the coefficient
of static friction between the cord and the peg is ms = 0.1.
A
u
u
C
B
E
D
In the case where cylinder E is on the verge of descending, T2 = 10(9.81) N and T1 =
49.05
p
. Here,
+ u. Thus,
sin u
2
T2 = T1emsb
10(9.81) =
49.05 0.1  p
e 2
sin u

+ u

p

In (2 sin u) = 0.1  + u
2


Solving by trial and error, yields
u = 0.6764 rad = 38.8°
Ans
758
264
SM_CH04.indd
264
08b Ch08b 718-761.indd
758
4/8/11 11:48:56
6/12/09
10:34:47 AM
AM
©©2011
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybe
bereproduced,
reproduced,ininany
anyform
formororby
byany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
•4–94. Determine the smallest lever force P needed to
•8–97.
prevent the wheel from rotating if it is subjected to a torque
of M = 250 N # m. The coefficient of static friction between
the belt and the wheel is ms = 0.3. The wheel is pin
connected at its center, B.
400 mm
B
M
A
200 mm
750 mm
P
+©MA = 0;
–F(200) + P(950) = 0
F = 4.75 P
T2 = T1emb
 3p
2 
F¿ = 4.75 Pe0.3 
+©MB = 0;
= 19.53 P
–19.53 P (0.4) + 250 + 4.75 P (0.4) = 0
P = 42.3 N
Ans
759
265
SM_CH04.indd
265
08b Ch08b 718-761.indd
759
4/8/11 11:48:56
6/12/09
10:34:49 AM
AM
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
this
material
may
bebe
reproduced,
inin
any
form
oror
byby
any
means,
without
permission
inin
writing
from
the
publisher.
8–98.
•4–95. If a force of P = 200 N is applied to the handle of
the bell crank, determine the maximum torque M that can
be resisted so that the flywheel is not on the verge of
rotating clockwise. The coefficient of static friction between
the brake band and the rim of the wheel is ms = 0.3.
P
900 mm
400 mm
C
A
B
100 mm
O
M
300 mm
Referring to the free-body diagram of the bell crank shown in Fig. a and the flywheel shown in Fig. b,
+©MB = 0;
TA (0.3) + TC (0.1) – 200(1) = 0
(1)
+©MO = 0;
TA (0.4) – TC (0.4) – M = 0
(2)
By considering the friction between the brake band and the rim of wheel where b =
270°
p = 1.5p rad and
180°
TA > TC, we can write
TA = TCemsb
TA = TCe0.3(1.5p)
TA = 4.1112 TC
(3)
Solving Eqs. (1), (2), and (3) yields
M = 187 N · m
TA = 616.67 N
Ans
TC = 150.00 N
760
266
SM_CH04.indd
266
08b Ch08b 718-761.indd
760
4/8/11 11:48:57
6/12/09
10:34:51 AM
AM
©©2011
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
*4–96. The uniform
*8–104.
beam
is supported
by the
250-N
uniform50-lb
25-kg
beam
is supported
byrope
the
which
is attached
to thetoend
wraps
overover
the
rope which
is attached
the of
endthe
of beam,
the beam,
wraps
rough
peg,peg,
andand
is then
connected
totothe
500-N
the rough
is then
connected
the100-lb
50-kg block.
block. If
the coefficient of static friction between the beam and the
block, and between the rope and the peg, is ms = 0.4,
determine the maximum distance that the block can be
placed from A and still remain in equilibrium. Assume the
block will not tip.
d
1 ftm
0.3
A
3m
10
ft
50 (9.81) N
Block :
+↑ΣFy = 0;
N – 50 (9.81) = 0
N = 490.5 N
+
→ ΣFx
= 0;
T1 – 0.4 (490.5) = 0
50 (9.81) N
T1 = 196.2 N
T2 = T1 e ;
T2 = 196.2 e
196.2 N
0.4( π )
2
367.8 N
= 367.8 N
1.5 m
System :
+ ΣMA = 0;
0.3 m
3 m 25 (9.81) N
–490.5 (d) – 196.2 (0.3) – 25 (9.81) (1.5) + 367.8 (3) = 0
d = 1.380 m
Ans
766
267
SM_CH04.indd
267
08c Ch08c 762-805.indd
766
4/8/11
11:48:57
AM AM
6/15/09
10:48:45
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
rights
reserved.
This
material
is protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
AllAll
rights
reserved.
This
material
is protected
under
allall
copyright
laws
as as
they
currently
exist.
portion
material
may
reproduced,
form
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
thisthis
material
may
bebe
reproduced,
in in
anyany
form
or or
byby
anyany
means,
without
permission
in in
writing
from
thethe
publisher.
•8–105.
•4–97. The 80-kg man tries to lower the 150-kg crate
using a rope that passes over the rough peg. Determine the
least number of full turns in addition to the basic wrap
(165°) around the peg to do the job. The coefficients of
static friction between the rope and the peg and between
the man’s shoes and the ground are ms = 0.1 and msœ = 0.4,
respectively.
15
If the man is on the verge of slipping, F = ms¿ N = 0.4 N. Referring to the free-body diagram of the man shown in Fig. a,
+
→ ©Fx
= 0;
0.4N – T sin 15° = 0
+↑©Fy = 0;
N + T cos 15° – 80(9.81) = 0
Solving,
T = 486.55 N
N = 314.82 N
Using the result for T and considering the friction between the rope and the peg, where T2 = 150(9.81) N, T1 = T = 486.55 N
�
�
 90° + 75°
and b1 = n(2p) + 
p = (2n + 0.9167)p rad, Fig. b,
 180° 
T2 = T1emsb
150(9.81) = 486.55e0.1(2n + 0.9167)p
In 3.024 = 0.1(2n + 0.9167)p
n = 1.303
Thus, the required number of full turns is
Ans
n=2
767
268
SM_CH04.indd
268 767
08c Ch08c
762-805.indd
4/8/11
11:48:58
6/15/09
10:48:47
AM AM
©©2011
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
8–106.
4–98. If the rope wraps three full turns plus the basic
wrap (165°) around the peg, determine if the 80-kg man can
keep the 300-kg crate from moving. The coefficients of
static friction between the rope and the peg and between
the man’s shoes and the ground are ms = 0.1 and msœ = 0.4,
respectively.
15
If the man is on the verge of slipping, F = ms¿ N = 0.4 N. Referring to the free-body diagram of the man shown in Fig. a,
+
→ ©Fx
= 0;
0.4N – T sin 15° = 0
+↑©Fy = 0;
N + T cos 15° – 80(9.81) = 0
Solving,
T = 486.55 N
N = 314.82 N
Using the result for T and considering the friction between the rope and the peg, where T2 = 300(9.81) N, T1 = T = 486.55 N
�
�
 90° + 75°
and b = n(2p) + 
p = (2n + 0.9167)p rad, Fig. b,
 180° 
T2 = T1emsb
300(9.81) = 486.55e0.1(2n + 0.9167)p
In 6.049 = 0.1(2n + 0.9167)p
n = 2.406
Ans
Since n > 3, the man can hold the crate in equilibrium.
768
269
SM_CH04.indd
269
08c Ch08c 762-805.indd
768
4/8/11
11:48:58
AM AM
6/15/09
10:48:48
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
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form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
any
means,
without
permission
in in
writing
from
thethe
publisher.
8–107. The drive pulley B in a video tape recorder is on
4–99.
the verge of slipping when it is subjected to a torque of
0.005 N m. If the coefficient of static friction between
the tape and the drive wheel and between the tape and the
fixed shafts A and C is
0.1, determine the tensions 1
and 2 developed in the tape for equilibrium.
10 mm
T1
A
M
0.005 N m
10 mm
B
10 mm
C
Here T3 must overcome T4 and M, so T3 > T4. Also b = p rad. Thus,
T2
T3 = T4 emsb
T3 = T4 e0.1(p)
T3 = 1.3691T4
(1)
Referring to the free-body diagram of pulley B in Fig. a,
+©M0 = 0;
0.005 + T4 (0.01) – T3 (0.01) = 0
(2)
Solving Eqs. (1) and (2), yields
T4 = 1.3546 N
T3 = 1.8546 N
Using the result of T4 and considering the friction on the fixed shaft A, where T1 > T4 and b = p rad,
T1 = T4 e msb
= 1.3546e0.1p
Ans
= 1.85N
Using the result of T3 and considering the friction on the fixed shaft C, where T3 > T2 and b =
P
rad,
2
T3 = T2emsb
1.8546 = T2e0.1(p/2)
Ans
T2 = 1.59 N
769
270
270 769
08c SM_CH04.indd
Ch08c 762-805.indd
4/8/11
11:48:59
6/15/09
10:48:50
AM AM
©©2011
rights
reserved.
This
material
is is
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under
allall
copyright
laws
asas
they
currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
means,without
withoutpermission
permissionininwriting
writingfrom
fromthe
thepublisher.
publisher.
*4–100.
*8–108. Determine
Determinethethe
maximum
number
25-kg
*8–108.
maximum
number
of 50-lbof
packages
packages
that
can be
the belt
without
that
can be
placed
on placed
the belton
without
causing
the causing
belt to
the belt
to drive
slip atwheel
the drive
wheel
which with
is rotating
with
slip
at the
A which
is A
rotating
a constant
a constant
angular
velocity.
B is
free to
rotate.
angular
velocity.
Wheel
B isWheel
free to
rotate.
Also,
findAlso,
the
find the corresponding
M that
must be
corresponding
torsional torsional
moment moment
M that must
be supplied
supplied
A. The conveyor
belt is pre-tensioned
to
wheel to
A. wheel
The conveyor
belt is pre-tensioned
with the
with the
1500-N horizontal
The coefficient
of friction
kinetic
300-lb
horizontal
force. Theforce.
coefficient
of kinetic
friction between
theand
beltplatform
and platform
between
the belt
P isP misk =k =0.2
, and
0.2,
and the
coefficient of static friction between the belt and the rim of
each wheel is ms == 0.35.
0.35.
0.5 ftm
0.15
0.5m
ft
0.15
B
P
A
P = 1500
300 lb
N
M
The maximum tension T2 of the conveyor belt can be obtained by considering the equilibrium of the free
– body diagram of the top belt shown in Fig. a,
+↑ΣFy = 0;
n(25 (9.81)) – N = 0
N = 245.25
(1)
+
→ ΣFx
750 + 0.2 (245.25n) – T2 = 0
T2 = 750 + 49.05n
(2)
= 0;
By considering the case when the drive wheel A is on the verge of slipping, where
and T1 = 750 N,
= π rad, T2 = 750 + 49.05n
T2 = T1 e
750 + 49.05n = 750 e0.35(π)
n = 30.62
Thus, the maximum allowable number of boxes on the belt is
Ans
n = 30
Substituting n = 30 into Eq. (2) gives T2 = 2221.5 N. Referring to the free – body diagram of the wheel A shown
in Fig. b,
+ ΣMO = 0;
M + 750 (0.15) – 2221.5 (0.15) = 0
Ans
M = 220.7 N · m
n (245.25) N
0.15 m
T2 = 2221.5 N
750 N
T1 = 750 N
770
271
SM_CH04.indd
271
08c Ch08c 762-805.indd
770
4/8/11
11:48:59
AM AM
6/15/09
10:48:52
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.All
Allrights
rightsreserved.
reserved.This
Thismaterial
materialisisprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.
exist.
No
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
No
portion
ofof
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material
may
bebe
reproduced,
inin
any
form
oror
byby
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inin
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8–151. A roofer, having a mass of 70 kg, walks slowly in an
4–101.
upright position down along the surface of a dome that has
a radius of curvature of r = 20 m. If the coefficient of static
friction between his shoes and the dome is ms = 0.7,
determine the angle u at which he first begins to slip.
20 m
u
60
+QΣFy' = 0;
N – 70(9.81) cos u = 0
(1)
R+ΣFx' = 0;
70(9.81) sin u – 0.7 N = 0
(2)
Solving Eqs. (1) and (2) yields:
Ans
u = 35.0°
N = 562.6 N
*8–152. Column D is subjected to a vertical load of
40 kN.
8000
lb.ItItisissupported
supportedon
ontwo
two identical
identical wedges
wedges A
A and B for
which the coefficient of static friction at the contacting
surfaces between A and B and between B and C is s = 0.4.
Determine the force P needed to raise the column and the
equilibrium force P¿ needed to hold wedge A stationary.
The contacting surface between A and D is smooth.
40 kN
8000
lb
Wedge A :
+↑ΣFy = 0;
D
N cos 10° – 0.4N sin 10° – 40 = 0
P
N = 43.70 kN
+
→ ΣFx
= 0;
10
10
A
P¿
C
40 kN
0.4 (43.70) cos 10° + 43.70 sin 10° – Pʹ = 0
Pʹ = 24.80 kN
B
Ans
Wedge B :
+↑ΣFy = 0;
NC + 0.4 (43.70) sin 10° – 43.70 cos 10° = 0
N = 43.70 kN
0.4 (43.70) kN
NC = 40 kN
+
→ ΣFx
= 0;
P – 0.4 (40) – 43.70 sin 10° – 0.4 (43.70) cos 10° = 0
P = 40.80 kN
Ans
804
272
SM_CH04.indd
272
08c Ch08c 762-805.indd
804
4/8/11 11:49:00
6/15/09
10:49:51 AM
AM
©©2011
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This
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is is
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under
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copyright
laws
asas
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currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
•5–89. Determine the horizontal and vertical components
•4–102.
of reaction at the pin A and the reaction at the roller B
required to support the truss. Set F = 600 N.
A
2m
45
2m
F
2m
F
B
2m
F
Equations of Equilibrium : The normal reaction NB can be obtained
directly by summing moments about point A.
+ ΣMA = 0;
600(6) + 600(4) + 600(2) – NB cos 45° (2) = 0
Ans
NB = 5091.17 N = 5.09 kN
+
S ΣFx = 0;
Ax – 5091.17 cos 45° = 0
Ans
Ax = 3600 N = 3.60 kN
+ ↑ΣFy = 0;
5091.17 sin 45° – 3(600) – Ay = 0
Ans
Ay = 1800 N = 1.80 kN
5–90. If the roller at B can sustain a maximum load of
4–103.
3 kN, determine the largest magnitude of each of the three
forces F that can be supported by the truss.
A
2m
45
2m
F
2m
F
B
2m
F
Equations of Equilibrium : The unknowns Az and Ay can be
eliminated by summing moments about point A.
+ ΣMA = 0;
F(6) + F(4) + F(2) – 3 cos 45° (2) = 0
F = 0.3536 kN = 354 N
Ans
394
273
SM_CH04.indd
273
05b Ch05b 361-400.indd
394
4/8/11
11:49:01
AM AM
6/12/09
8:46:46
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4–105.
274
SM_CH04.indd 274
4/8/11 11:49:01 AM
©©2011
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asas
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2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
maybebereproduced,
reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
4–106.
*5–92. The shaft assembly is supported by two smooth
journal bearings A and B and a short link DC. If a couple
moment is applied to the shaft as shown, determine the
components of force reaction at the journal bearings and the
force in the link. The link lies in a plane parallel to the y–z
plane and the bearings are properly aligned on the shaft.
z
D
30
20
120 mm
C
250 mm
A
x
©Mx = 0;
Ans
Ans
–Ay(0.7) – 1015.43 cos 20° (0.42) = 0
Ay = 572.51 = 573 N
©Fy = 0;
Ans
–208.38 + 1015.43 sin 20° + Bz = 0
Bc = –139 N
©(MB)z = 0;
y
–Az(0.7) – 1015.43 sin 20° (0.42) = 0
Az = –208.38 = –208 N
©Fz = 0;
300 mm
– 250 + FCD cos 20° (0.25 cos 30°) + FCD sin 20° (0.25 sin 30°) = 0
FCD = 1015.43 N = 1.02 kN
©(MB)y = 0;
250 N m
400 mm
B
Ans
572.51 – 1015.43 cos 20° + By = 0
By = 382 N
Ans
396
275
SM_CH04.indd
275
05b Ch05b 361-400.indd
396
4/8/11
11:49:01
AM AM
6/12/09
8:46:50
2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
Allrights
rightsreserved.
reserved.
Thismaterial
materialis isprotected
protectedunder
underallallcopyright
copyrightlaws
lawsasasthey
theycurrently
currently
©©
2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
This
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
the
publisher.
exist.
NoNo
portion
ofof
this
material
may
bebe
reproduced,
in in
any
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oror
byby
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in in
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•4–107.
•5–93. Determine the reactions at the supports A and B of
the frame.
50
10 kN
kip
35
kN
7 kip
25
kN
5 kip
2.4
8 ftm
1.8
6 ftm
1.8
m
6 ft
10
kN
2 kip
A
8 ft
2.4
m
0.5
2.5 kip
kN
6 ft
1.8
m
B
25 kN
35 kN
2.4 m
50 kN
1.8 m
10 kN
1.8 m
2.4 m
2.5 kN
+ ΣMB = 0;
+
→ ΣFx
= 0;
+↑ΣFy = 0;
25 (4.2) + 35 (1.8) + 2.5 (1.8) – 10 (1.8) – Ay (4.2) = 0
Ay = 36.786 kN = 36.8 kN
Ans
Bx – 2.5 = 0
Ans
Bx = 2.5 kN
1.8 m
By + 36.786 – 25 – 35 – 50 – 10 = 0
By = 83.2 kN
Ans
397
276
276 397
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/118:46:51
11:49:02
6/12/09
AM AM
©©2011
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This
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is is
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copyright
laws
asas
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currently
2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
*4–108.A A
skeletal
diagramofofthe
thelower
lowerleg
legisisshown
shown in the
5–94.
skeletal
diagram
lower figure. Here it can be noted that this portion of the leg
is lifted by the quadriceps muscle attached to the hip at A
and to the patella bone at B. This bone slides freely over
cartilage at the knee joint. The quadriceps is further
extended and attached to the tibia at C. Using the
mechanical system shown in the upper figure to model the
lower leg, determine the tension in the quadriceps at C and
the magnitude of the resultant force at the femur (pin), D,
in order to hold the lower leg in the position shown. The
lower leg has a mass of 3.2 kg and a mass center at G1; the
foot has a mass of 1.6 kg and a mass center at G2.
75 mm
25 mm
A
350 mm
B
C
75
A
300 mm
D
B
G1
G2
C
D
+ ΣMD = 0; T sin 18.43° (75) – 3.2(9.81)(425 sin 75°)
–1.6(9.81)(725 sin 75°) = 0
T = 1006.82 N = 1.01 kN
+ ↑ΣFy = 0;
Ans
Dy + 1006.82 sin 33.43° – 3.2(9.81) – 1.6(9.81) = 0
Dy = –507.66 N
+
S ΣFx = 0;
Dx – 1006.82 cos 33.43° = 0
Dz = 840.20 N
FD =
D2x + D2y = (–507.66)2 + 840.202 = 982 N
Ans
398
277
SM_CH04.indd
277
05b Ch05b 361-400.indd
398
4/8/11
11:49:02
AM AM
6/12/09
8:46:53
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
any
form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
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reproduced,
in in
any
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or or
byby
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N acts
acts on
on the
the crankshaft.
4–109. A vertical force
5–95.
force of
of 400
80 lb
Determine the horizontal equilibrium force P that must be
applied to the handle and the x, y, z components of force at
the smooth journal bearing A and the thrust bearing B. The
bearings are properly aligned and exert only force reactions
on the shaft.
z
80 lb
400
N
B
250in.mm
10
y
350
14 mm
in.
A
350
14 mm
in.
6150
in.mm
x
8 in.
200
mm
4100
in. mm
P
400 N
ΣMy = 0;
P(200) – 400 (250) = 0
P = 500 N
Ans
ΣMz = 0;
Bz(700) – 400 (350) = 0
Bz = 200 N
Ans
ΣMz = 0;
–Bx(700) – 500 (250) = 0
Bx = –178.6 N
Ans
ΣFz = 0;
As + (–178.6) – 500 = 0
As = 678.6 N
Ans
ΣFy = 0;
By = 0
ΣFx = 0;
As + 200 – 400 = 0
250
350 mm
350 mm
150 mm
Ans
As = 200 N
mm
200 mm
100 mm
Ans
Negative sign indicates that Bx acts in the opposite sense to that shown on the FBD.
399
278
278 399
05b SM_CH04.indd
Ch05b 361-400.indd
4/8/11
11:49:03
6/12/09
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©©2011
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asas
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2010Pearson
PearsonEducation,
Education,Inc.,
Inc.,Upper
UpperSaddle
SaddleRiver,
River,NJ.
NJ.All
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
exist.
exist.No
Noportion
portionofofthis
thismaterial
materialmay
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reproduced,ininany
anyform
formororbybyany
anymeans,
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withoutpermission
permissionininwriting
writingfrom
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thepublisher.
publisher.
•4–110. The truck has a mass of 1.25 Mg and a center of
•8–145.
mass at G. Determine the greatest load it can pull if (a) the
truck has rear-wheel drive while the front wheels are free to
roll, and (b) the truck has four-wheel drive. The coefficient of
static friction
friction between
betweenthe
thewheels
wheelsand
andthe
theground
groundisisms s== 0.5
0.5,,
œ
and between the
the crate
crate and
andthe
theground,
ground itisis m
′s =
0.4.
0.4.
s =
800 mm
G
600 mm
A
B
1.5 m
1m
a) The truck with rear wheel drive.
Equations of Equilibrium and Friction: It is required that rear wheels
of the truck slip. Hence FA = ms NA = 0.5 NA. From FBD (a)
+©MB = 0; 1.25(103)(9.81)(1) + T (0.6) – NA (2.5) = 0
[1]
+
S ©Fx = 0; 0.5 NA – T = 0
[2]
Solving Eqs. [1] and [2] yields
NA = 5573.86 N
T = 2786.93 N
Since the crate moves, Fc = ms¿Nc = 0.4 Nc. From FBD (c),
+c©Fy = 0;
Nc – W = 0
+
2786.93 – 0.4 W = 0
S ©Fx = 0;
Nc = W
W = 6967.33 N = 6.97 kN
Ans
b) The truck with four wheel drive.
Equations of Equilibrium and Friction: It is required that rear wheels
and front wheels of the truck slip. Hence FA = msNA = 0.5 NA and FB
= msNB = 0.5 NB. From FBD (b),
+©MB = 0; 1.25(103)(9.81)(1) + T (0.6) – NA (2.5) = 0
[3]
+©MA = 0; NB (2.5) + T (0.6) – 1.25(103)(9.81)(1.5)= 0
[4]
+
S ©Fx = 0; 0.5 NA + 0.5 NB – T = 0
[5]
Solving Eqs. [3], [4] and [5] yields
NA = 6376.5 N
NB = 5886.0 N
T = 6131.25 N
Since the crate moves, Fc = ms¿Nc = 0.4 Nc. From FBD (c),
+c©Fy = 0;
Nc – W = 0
+
6131.25 – 0.4 W = 0
S ©Fx = 0;
Nc = W
W = 15328.125 N = 15.3 kN
Ans
798
279
SM_CH04.indd
279
08c Ch08c 762-805.indd
798
4/8/11
11:49:04
AM AM
6/15/09
10:49:43
© 2011
Pearson
Education,
Inc.,
Upper
Saddle
River,
All
rights
reserved.
This
material
protected
under
copyright
laws
they
currently
© 2010
Pearson
Education,
Inc.,
Upper
Saddle
River,
NJ.NJ.
All
rights
reserved.
This
material
is is
protected
under
allall
copyright
laws
asas
they
currently
exist.
portion
this
material
may
reproduced,
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form
any
means,
without
permission
writing
from
publisher.
exist.
NoNo
portion
of of
this
material
may
bebe
reproduced,
in in
any
form
oror
byby
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4–111. Solve Prob. 8–145 if the truck and crate are
8–146.
traveling up a 10° incline.
800 mm
G
600 mm
A
B
1.5 m
1m
a) The truck with rear wheel drive.
Equations of Equilibrium and Friction: It is required that the rear wheel
of the truck slip hence FA = msNA = 0.5 NA. From FBD (a).
+©MB = 0;
+Q©Fx¿ = 0;
1.25(103)(9.81) cos 10°(1)
+ 1.25(103)(9.81) sin 10°(0.8)
+ T (0.6) – NA (2.5) = 0
[1]
0.5 NA – 1.25(103)(9.81) sin 10° – T = 0
[2]
Solving Eqs. [1] and [2] yields
NA = 5682.76 N
T = 712.02 N
Since the crate moves, FC = ms¿NC = 0.4NC. From FBD (c),
a+©Fy¿ = 0;
NC – W cos 10° = 0
+Q©Fx = 0;
712.02 – W sin 10° – 0.4(0.9848 W) = 0
NC = 0.9848 W
W = 1254.50 N = 1.25 kN
Ans
b) The truck with four wheel drive.
Equations of Equilibrium and Friction: It is required that rear wheels
of the truck slip hence FA = msNA = 0.5 NA. From FBD (b).
+©MB = 0;
+©MA = 0;
+Q©Fx = 0;
1.25(103)(9.81) cos 10°(1)
+ 1.25(103)(9.81) sin 10°(0.8)
+ T (0.6) – NA (2.5) = 0
[3]
–1.25(103)(9.81) cos 10°(1.5)
+ 1.25(103)(9.81) sin 10°(0.8)
+ T (0.6) + NB (2.5) = 0
[4]
0.5 NA + 0.5 NB – 1.25(103)(9.81) sin 10° – T = 0
[5]
Solving Eqs. [3], [4] and [5] yields
NA = 6449.98 N
NB = 5626.23 N
T = 3908.74 N
Since the crate moves, FC = ms¿NC = 0.4 NC . From FBD (c),
a+©Fy¿ = 0;
NC – W cos 10° = 0 NC = 0.9848 W
+
3908.74 – W sin 10° – 0.4(0.9848 W) = 0
S ©Fx = 0;
Ans
W = 6886.79 N = 6.89 kN
799
280
280 799
08c SM_CH04.indd
Ch08c 762-805.indd
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