Problem 4.1 In Active Example 4.1, the 40-kN force
points 30°above the horizontal. Suppose that the force
points 30°below the horizontal instead. Draw a sketch
of the beam with the new orientation of the force. What
is the moment of the force about point A?
6 m
40 kN
30°
A
Solution: The perpendicular distance from Ato the line of action
of the force is unchanged
Problem 4.2 The mass m1D20 kg. The magnitude of
the total moment about Bdue to the forces exerted on
bar AB by the weights of the two suspended masses is
170 N-m. What is the magnitude of the total moment
due to the forces about point A?
0.35 m
A
B
0.35 m 0.35 m
Problem 4.3 The wheels of the overhead crane exert
downward forces on the horizontal I-beam at Band C.
If the force at Bis 40 kip and the force at Cis 44 kip,
determine the sum of the moments of the forces on the
beam about (a) point A, (b) point D.ABC
D
25 ft10 ft 15 ft
Solution: Use 2-dimensional moment strategy: determine normal
distance to line of action D; calculate magnitude DF; determine sign.
Add moments.
10 ft, and DAC D35 ft. The moments are clockwise (negative).
Hence,
MAD10⊲40⊳35⊲44⊳D1940 ft-kip .
(b) The normal distances from Dto the lines of action are DDB D
40 ft, and DDC D15 ft. The actions are positive; hence
MDDC⊲40⊳⊲40⊳C⊲15⊳⊲44⊳D2260 ft-kip
BC
10 ft 25 ft 15 ft
Problem 4.4 What force Fapplied to the pliers is
required to exert a 4 N-m moment about the center of
the bolt at P?
P
Solution:
MPD4 N-m DF⊲0.165 m sin 42°⊳)FD4 N-m
0.165 m sin 42°
D36.2N
160
Problem 4.5 Two forces of equal magnitude Fare
applied to the wrench as shown. If a 50 N-m moment is
required to loosen the nut, what is the necessary value
of F?
300 mm
Solution:
Problem 4.6 The force FD8 kN. What is the
moment of the force about point P?
y
(3, 7) m
Problem 4.7 If the magnitude of the moment due to
the force Fabout Qis 30 kN-m, what is F?
F
y
(3, 7) m
Q
Problem 4.8 The support at the left end of the beam
will fail if the moment about Aof the 15-kN force F
exceeds 18 kN-m. Based on this criterion, what is the
largest allowable length of the beam?
25°
A
B
F30°
Solution:
MAD7.5LkNÐm
set MADMAmax D18 kN ÐmD7.5L
max
Lmax D2.4m 25°
L
Problem 4.9 The length of the bar AP is 650 mm. The
radius of the pulley is 120 mm. Equal forces TD50 N
are applied to the ends of the cable. What is the sum of
the moments of the forces (a) about A; (b) about P.
T
T
A
45⬚30⬚
Solution:
(b) MPD⊲50 N⊳⊲0.12 m⊳
162
Problem 4.10 The force FD12 kN. A structural
engineer determines that the magnitude of the moment
due to Fabout P should not exceed 5 kN-m. What
is the acceptable range of the angle ˛? Assume that
0˛90°.
2 m
1 m
F
P
α
Solution: We have the moment about P
12 kN
Problem 4.11 The length of bar AB is 350 mm.
The moments exerted about points Band Cby the
vertical force Fare MBD1.75 kN-m and MCD
4.20 kN-m. Determine the force Fand the length of
bar AC.
B
C
20°
30°
A
Solution: We have
164
Problem 4.12 In Example 4.2, suppose that the 2-kN
force points upward instead of downward. Draw a sketch
of the machine part showing the orientations of the
forces. What is the sum of the moments of the forces
about the origin O?
300 mm
4 kN
2 kN
3 kNO
30⬚
5 kN
400 mm
300 mm
Solution: If the 2-kN force points upward, the magnitude of its
Problem 4.13 Two equal and opposite forces act on
the beam. Determine the sum of the moments of the two
forces (a) about point P; (b) about point Q; (c) about the
point with coordinates xD7m,y D5m.
2 m 2 m
40 N
30⬚
P
y
Q
40 N
30⬚
x
Solution:
MPD⊲40 N cos 30°⊳⊲2m⊳C⊲40 N cos 30°⊳⊲4m⊳
40 N 40 N
y
Problem 4.14 The moment exerted about point Eby
the weight is 299 in-lb. What moment does the weight
exert about point S?
S
30°
E40°
12 in.
13 in.
Solution: The key is the geometry
From trigonometry,
We are given that
Now,
MsD⊲d1Cd2⊳W
MsD⊲20.35⊳⊲30.0⊳
MsD611 in-lb clockwise
Problem 4.15 The magnitudes of the forces exerted on
the pillar at Dby the cables A,B, and Care equal: FAD
FBDFC. The magnitude of the total moment about E
due to the forces exerted by the three cables at Dis
1350kN-m. What is FA?FA
FB
FC
A
E
D
D
BC
6 m
Solution: The angles between the three cables and the pillar are
The vertical components of each force at point Dexert no moment
about E. Noting that FADFBDFC, the magnitude of the moment
about Edue to the horizontal components is
166
Problem 4.16 Three forces act on the piping. Deter-
mine the sum of the moments of the three forces about
0.2 m
P
0.2 m 0.2 m
Solution:
2 kN
4 kN
0.2 m
0.2 m0.2 m0.2 m
P
Problem 4.17 The forces F1D30 N, F2D80 N, and
F3D40 N. What is the sum of the moments of the
forces about point A?
F3
30⬚
A
C
x
y
8 m
Problem 4.18 The force F1D30 N. The vector sum
of the forces is zero. What is the sum of the moments
of the forces about point A?
F1
F3
30⬚
45⬚
A
B
C
y
2 m
Solution: The sums of the forces in the xand ydirections equal
Problem 4.19 The forces FAD30 lb, FBD40 lb, FCD
20 lb, and FDD30 lb. What is the sum of the
moments of the forces about the origin of the coordinate
system? FD
30⬚
FBFC
FA
y
x
Solution: The moment about the origin due to FAand FDis
zero. Treating counterclockwise moments as positive, the sum of the
Problem 4.20 The force FAD30 lb. The vector sum
of the forces on the beam is zero, and the sum of the
moments of the forces about the origin of the coordinate
system is zero.
(a) Determine the forces FB,FC, and FD.
(b) Determine the sum of the moments of the forces
about the right end of the beam.
FD
30⬚
FBFC
FA
y
x
6 ft 4 ft
Solution:
(a) The sum of the forces and the sum of the moments equals zero
(b) The sum of the moments about the right end is
168
Problem 4.21 Three forces act on the car. The sum of
the forces is zero and the sum of the moments of the
forces about point Pis zero.
(a) Determine the forces Aand B.
(b) Determine the sum of the moments of the forces
about point Q.
BPQ
A
y
x
2800 lb
6 ft 3 ft
Solution:
6 ft 3 ft
Problem 4.22 Five forces act on the piping. The vector
sum of the forces is zero and the sum of the moments
of the forces about point Pis zero.
(a) Determine the forces A, B, and C.
(b) Determine the sum of the moments of the forces
about point Q.
2 ft
P
80 lb
2 ft 2 ft
20 lb
2 ft
45⬚
A
B
y
x
Q
C
Solution:
Problem 4.23 In Example 4.3, suppose that the attach-
ment point Bis moved upward and the cable is length-
ened so that the vertical distance from Cto Bis 9 ft.
(the positions of points Cand Aare unchanged.) Draw a
sketch of the system with the cable in its new position.
What is the tension in the cable?
2 ft 2 ft
4 ft
C
W
7 ft
B
A
Solution: The angle ˛between the cable AB and the horizontal is.
Problem 4.24 The tension in the cable is the same on
both sides of the pulley. The sum of the moments about
point Adue to the 800-lb force and the forces exerted
on the bar by the cable at Band Cis zero. What is the
tension in the cable?
A
BC
30
170
Problem 4.25 The 160-N weights of the arms AB and
BC of the robotic manipulator act at their midpoints.
Determine the sum of the moments of the three weights
about A.
600 mm
600 mm
40°
20°
160 N
40 N
C
150 mm
B
Solution: The strategy is to find the perpendicular distance from
the points to the line of action of the forces, and determine the sum
of the moments, using the appropriate sign of the action.
Problem 4.26 The space shuttle’s attitude thrusters
exert two forces of magnitude FD7.70 kN. What
moment do the thrusters exert about the center of
mass G?
18 m 12 m
5°6°
2.2 m
2.2 m
G
FF
Problem 4.27 The force Fexerts a 200 ft-lb counter-
clockwise moment about Aand a 100 ft-lb
clockwise moment about B. What are Fand ?
A
(–5, 5) ft
B
(3, –4) ft
(4, 3) ft
F
θ
y
x
Solution: The strategy is to resolve Finto x– and y-components,
and compute the perpendicular distance to each component from A
rAF D⊲4⊲5⊳⊳iC⊲35⊳jD9i2j.
(4, 3) ft
y
172
Problem 4.28 Five forces act on a link in the gear-
shifting mechanism of a lawn mower. The vector sum
of the five forces on the bar is zero. The sum of their
moments about the point where the forces Axand Ayact
is zero.
(a) Determine the forces Ax,Ay, and B.
(b) Determine the sum of the moments of the forces
about the point where the force Bacts.
650 mm
650 mm 350 mm
450 mm
30 kN
45°
20°
B
Ax
Ay
25 kN
Solution: The strategy is to resolve the forces into x– and
y-components, determine the perpendicular distances from Bto the line
of action, determine the sign of the action, and compute the moments.
The angles are measured counterclockwise from the xaxis. The
forces are
F2D30⊲icos 135°Cjsin 135°⊳D21.21iC21.21j
F1D25⊲icos 20°Cjsin 20°⊳D23.50iC8.55j.
y-component of F2is ⊲0.650 C0.350⊳D1 m, and the action is
positive. The distance to the line of action of the x-component
of F2is ⊲0.650 0.450⊳D0.200 m, and the action is positive.
The moment about Ais
MAD⊲8.55⊳⊲1⊳C⊲23.5⊳⊲0.2⊳C⊲BX⊳⊲0.65⊳D0.
Problem 4.29 Five forces act on a model truss built by
a civil engineering student as part of a design project.
The dimensions are bD300 mm and hD400 mm; FD
100 N. The sum of the moments of the forces about the
point where Axand Ayact is zero. If the weight of the
truss is negligible, what is the force B?
h
60°
F60°
F
Ax
Problem 4.30 Consider the truss shown in Problem
4.29. The dimensions are bD3 ft and hD4 ft; FD
300 lb. The vector sum of the forces acting on the truss
is zero, and the sum of the moments of the forces about
the point where Axand Ayact is zero.
(a) Determine the forces Ax,Ay, and B.
(b) Determine the sum of the moments of the forces
about the point where the force Bacts.
174
Problem 4.31 The mass mD70 kg. What is the mom-
ent about Adue to the force exerted on the beam at B
by the cable?
3 m
30°
m
B
A
45°
Solution: The strategy is to resolve the force at Binto components
parallel to and normal to the beam, and solve for the moment using
FOC DjFOCj⊲icos 45°Cjsin 45°⊳DjFOCj⊲0.707iC0.707j⊳.
WD⊲70⊳⊲9.81⊳⊲0i1j⊳D686.7j(N).
The equilibrium conditions are:
FxD⊲0.866jFOBjC0.7070jFOCj⊳iD0
FYD⊲0.500jFOBjC⊲.707jFOCj⊳686.7⊳jD0.
Solve: jFOBjD502.70 N. This is used to resolve the cable tension at B:
FBD502.7⊲icos 330°Cjsin 330°⊳D435.4i251.4j. The distance
from Ato the action line of the y-component at Bis 3 m, and the
action is negative. The x-component at passes through A, so that the
action line distance is zero. The moment at Ais MAD3⊲251.4⊳D
754.0 N-m
Problem 4.32 The weights W1and W2are suspended
by the cable system shown. The weight W1D12 lb. The
cable BC is horizontal. Determine the moment about
point Pdue to the force exerted on the vertical post at
Dby the cable CD.
AD
BC
P
W2
W1
6 ft
50⬚
Solution: Isolate part of the cable system near point B. The equi-
librium equations are
Problem 4.33 The bar AB exerts a force at Bthat helps
support the vertical retaining wall. The force is parallel
to the bar. The civil engineer wants the bar to exert a
38 kN-m moment about O. What is the magnitude of
the force the bar must exert?
B
A
O
4 m
1 m
1 m 3 m
Solution: The strategy is to resolve the force at Binto components
parallel to and normal to the wall, determine the perpendicular distance
By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes
with the horizontal is cos D3
5D0.600, and sin D4
5D0.800. The
force at Bis FBDjFBj⊲0.600iC0.800j⊳. The perpendicular distance
the y-component is 1 m, and the action is positive. The moment about
Ois MOD5⊲0.600⊳jFBjC1⊲0.800⊳jFBjD3.8jFBjD38 kN, from
which jFBjD10 kN
FB
O
4 m
1 m
1 m 3 m
Problem 4.34 A contestant in a fly-casting contest
snags his line in some grass. If the tension in the line is
5 lb, what moment does the force exerted on the rod by
the line exert about point H, where he holds the rod?
4 ft 6 ft
H
Solution: The strategy is to resolve the line tension into a compo-
nent normal to the rod; use the length from Hto tip as the perpen-
176
Problem 4.35 The cables AB and AC help support the
tower. The tension in cable AB is 5 kN. The points A,
B,C, and Oare contained in the same vertical plane.
(a) What is the moment about Odue to the force exerted
on the tower by cable AB?
(b) If the sum of the moments about Odue to the forces
exerted on the tower by the two cables is zero, what
is the tension in cable AC?
20 m
45°60°
A
BOC
Solution: The strategy is to resolve the cable tensions into compo-
nents normal to the vertical line through OA; use the height of the
tower as the perpendicular distance; determine the sign of the action,
and compute the moments.
AA
FNFN
60°45°
Problem 4.36 The cable from Bto A(the sailboat’s
forestay) exerts a 230-N force at B. The cable from Bto
C(the backstay) exerts a 660-N force at B. The bottom
of the sailboat’s mast is located at xD4m,yD0. What
x
Solution: Triangle ABP
tan ˛D4
11.8,˛D18.73°
P
178
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