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Problem 4.37 The cable AB exerts a 290-kN force on
the building crane’s boom at B. The cable AC exerts a
148-kN force on the boom at C. Determine the sum of
the moments about Pdue to the forces the cables exert
on the boom.
8 m
16 m
38 m
56 m
P
G
A
BC
Boom
Solution:
p3200 ⊲290 kN⊳⊲56 m⊳8
p320 ⊲148 kN⊳⊲16 m⊳
D3.36 MNm
MPD3.36 MN-m CW
290 kN
148 kN
88
56
A
Problem 4.38 The mass of the building crane’s boom
in Problem 4.37 is 9000 kg. Its weight acts at G. The
sum of the moments about Pdue to the boom’s weight,
the force exerted at Bby the cable AB, and the force
exerted at Cby the cable AC is zero. Assume that the
tensions in cables AB and AC are equal. Determine the
tension in the cables.
Solution:
A
56
Problem 4.39 The mass of the luggage carrier and the
suitcase combined is 12 kg. Their weight acts at A. The
F
0.28 m 0.14 m
A
C
y
a
Solution: Ois the origin of the coordinate system
MODF⊲1.2 m cos ˛⊳
1.2 m cos ˛
Problem 4.40 The hydraulic cylinder BC exerts a
300-kN force on the boom of the crane at C. The force
is parallel to the cylinder. What is the moment of the
force about A?
Solution: The strategy is to resolve the force exerted by the hydra-
ulic cylinder into the normal component about the crane; determine the
distance; determine the sign of the action, and compute the moment.
1.2D63.43°.
617.15 kN-m Check: The force exerted by the actuator can be resolved
α
3 m
2.4 m
1.2 m
180
Problem 4.41 The hydraulic piston AB exerts a 400-lb
force on the ladder at Bin the direction parallel to the
piston. The sum of the moments about Cdue to the force
exerted on the ladder by the piston and the weight Wof
the ladder is zero. What is the weight of the ladder?
6 ft
W
Problem 4.42 The hydraulic cylinder exerts an 8-kN
force at Bthat is parallel to the cylinder and points from
Ctoward B. Determine the moments of the force about
points Aand D.
1 m
0.6 m Scoop
AB
D
C
0.15 m
0.6 m
1 m
Hydraulic
cylinder
Solution: Use x,ycoords with origin A. We need the unit vector
from Cto B,eCB. From the geometry,
eCB D0.780i0.625j
5.00 kN
6.25 kN
Problem 4.43 The structure shown in the diagram is
one of the two identical structures that support the scoop
of the excavator. The bar BC exerts a 700-N force at C
that points from Ctoward B. What is the moment of this
force about K?
H
LK
100
mm
160
mm
Shaft
260
mm
320
mm
380
mm
C
D
1040
mm 1120
mm
260
mm
Scoop
B
180
mm
J
Solution:
MKD353 Nm CW
520 mm
320
K
Problem 4.44 In the structure shown in Problem 4.43,
the bar BC exerts a force at Cthat points from C
toward B. The hydraulic cylinder DH exerts a 1550-N
force at Dthat points from Dtoward H. The sum of the
moments of these two forces about Kis zero. What is
the magnitude of the force that bar BC exerts at C?
Solving we find
FD796 N
1120
100
1550 N
K
182
c
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Problem 4.45 In Active Example 4.4, what is the
moment of Fabout the origin of the coordinate system?
(0, 6, 5) ft
A
C (7, 7, 0) ft
y
F
Solution: The vector from the origin to point Bis
Problem 4.46 Use Eq. (4.2) to determine the moment
of the 80-N force about the origin Oletting rbe the
vector (a) from Oto A; (b) from Oto B.
Solution:
(a) MODrOA ðF
Problem 4.47 A bioengineer studying an injury sus-
tained in throwing the javelin estimates that the magni-
Solution: The magnitude of the moment is jFj⊲0.55 m⊳D⊲360 N⊳
⊲0.55 m⊳D198 N-m. The moment vector is perpendicular to the xy
Problem 4.48 Use Eq. (4.2) to determine the moment
of the 100-kN force (a) about A, (b) about B.
A
y
100j (kN)
Solution: (a) The coordinates of Aare (0,6,0). The coordinates of
the point of application of the force are (8,0,0). The position vector
Problem 4.49 The cable AB exerts a 200-N force on
the support at Athat points from Atoward B. Use
Eq. (4.2) to determine the moment of this force about
point Pin two ways: (a) letting rbe the vector from P
to A; (b) letting rbe the vector from Pto B.
y
(0.3, 0.5) m
(0.9, 0.8) m
P
A
Solution: First we express the force as a vector. The force points
in the same direction as the position vector AB.
184
Problem 4.50 The line of action of Fis contained in
the xyplane. The moment of Fabout Ois 140k(N-
m), and the moment of Fabout Ais 280k(N-m). What
O
x
(5, 3, 0) m
Solution: The strategy is to find the moments in terms of the
components of Fand solve the resulting simultaneous equations. The
position vector from Oto the point of application is rOF D5iC3j.
Take the dot product of both sides with kto eliminate k. The simul-
taneous equations are:
O
Problem 4.51 Use Eq. (4.2) to determine the sum of
the moments of the three forces (a) about A, (b) about B.
6 kN
y
Problem 4.52 Three forces are applied to the plate.
Use Eq. (4.2) to determine the sum of the moments of
the three forces about the origin O.
3 ft
200 lb
y
Solution: The position vectors from Oto the points of applica-
Problem 4.53 Three forces act on the plate. Use
Eq. (4.2) to determine the sum of the moments of the
three forces about point P.
3 kN
4 kN
30⬚
45⬚
y
Solution:
Problem 4.54 (a) Determine the magnitude of the
moment of the 150-N force about Aby calculating the
perpendicular distance from Ato the line of action of
the force.
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 150-N force about A.
A
y
x
150k (N)
(0, 6, 0) m
(6, 0, 0) m
Solution:
(a) The perpendicular from Ato the line of action of the force lies
186
Problem 4.55 (a) Determine the magnitude of the
moment of the 600-N force about Aby calculating the
perpendicular distance from Ato the line of action of
the force.
(b) Use Eq. (4.2) to determine the magnitude of the
moment of the 600-N force about A.
y
x
(0.6, 0.5, 0.4) m
0.8 m
A
600i (N)
Solution:
(a) Choose some point P⊲x, 0,0.8m⊳. on the line of action of the
force. The distance from Ato Pis then
Problem 4.56 what is the magnitude of the moment of
Fabout point B?
y
A
(4, 4, 2) ft
F ⫽ 20i ⫹ 10j ⫺ 10k (lb)
Solution: The position vector from Bto Ais
Problem 4.57 In Example 4.5, suppose that the attach-
ment point Cis moved to the location (8,2,0) m and the
tension in cable AC changes to 25 kN. What is the sum
of the moments about Odue to the forces excerted on
the attachment point Aby the two cables?
y
x
(6, 3, 0) m
(0, 4, 8) m O
B
C
Solution: The position vector from Ato Cis
The moment about Ois
ij k
Problem 4.58 The rope exerts a force of magnitude
jFjD200 lb on the top of the pole at B. Determine the
magnitude of the moment of Fabout A.
yB (5, 6, 1) ft
F
Solution: The position vector from Bto Cis
188
Problem 4.59 The force FD30iC20j10k(N).
(a) Determine the magnitude of the moment of F
about A.
(b) Suppose that you can change the direction of Fwhile
keeping its magnitude constant, and you want to
choose a direction that maximizes the moment of
Fabout A. What is the magnitude of the resulting
maximum moment?
y
x
z
(4, 3, 3) m
A(8, 2, – 4) m
F
Solution: The vector from Ato the point of application of Fis
rD4i1j7km
and
(b) The maximum moment occurs when r?F. In this case
Problem 4.60 The direction cosines of the force Fare
cos xD0.818, cos yD0.182, and cos zD0.545.
The support of the beam at Owill fail if the magnitude of
the moment of Fabout Oexceeds 100 kN-m. Determine
the magnitude of the largest force Fthat can safely be
applied to the beam. z
y
O
x
F
3 m
Solution: The strategy is to determine the perpendicular distance
Problem 4.61 The force Fexerted on the grip of the
exercise machine points in the direction of the unit vector
eD2
3i2
3jC1
3kand its magnitude is 120 N. Determine
the magnitude of the moment of Fabout the origin O.
y
O
F
150 mm
200 mm
Solution: The vector from Oto the point of application of the
force is
or
Problem 4.62 The force Fin Problem 4.61 points in
the direction of the unit vector eD2
3i2
3jC1
3k. The
support at Owill safely support a moment of 560 N-m
magnitude.
(a) Based on this criterion, what is the largest safe
magnitude of F?
(b) If the force Fmay be exerted in any direction, what
is its largest safe magnitude?
Solution: See the figure of Problem 4.61.
If we set jMOjD560 N-m, we can solve for jFmaxj
190
Problem 4.63 A civil engineer in Boulder, Colorado
estimates that under the severest expected Chinook
winds, the total force on the highway sign will be
FD2.8i1.8j(kN). Let MObe the moment due to F
about the base Oof the cylindrical column supporting
the sign. The ycomponent of MOis called the torsion
exerted on the cylindrical column at the base, and the
component of MOparallel to the xzplane is called
the bending moment. Determine the magnitudes of the
torsion and bending moment.
y
O
F
x
8 m
8 m
Solution: The total moment is
Problem 4.64 The weights of the arms OA and AB of
the robotic manipulator act at their midpoints. The direc-
y
600 mm
B
Solution: By definition, the direction cosines are the scalar compo-
nents of the unit vectors. Thus the unit vectors are e1D0.5iC0.866j,
The sum of moments is
MDr1ðW1Cr2ðW2
ijk
ij k
Problem 4.65 The tension in cable AB is 100 lb. If
you want the magnitude of the moment about the base
Oof the tree due to the forces exerted on the tree by the
two ropes to be 1500 ft-lb, what is the necessary tension
in rope AC ?
B
O
C
z
(0, 0, 10) ft (14, 0, 14) ft
Solution: We have the forces
F1D100 lb
p164 ⊲8jC10k⊳, F2DTAC
p456 ⊲14i8jC14k⊳
Thus the total moment is
Problem 4.66* A force Facts at the top end Aof the
pole. Its magnitude is jFjD6 kN and its xcomponent
is FxD4 kN. The coordinates of point Aare shown.
Determine the components of Fso that the magnitude
Solution: The force is given by FD⊲4kNiCFyjCFzk⊳.
Since the magnitude is constrained we must have
192
Problem 4.67 The force FD5i(kN) acts on the ring
Awhere the cables AB,AC, and AD are joined. What is
the sum of the moments about point Ddue to the force
Fand the three forces exerted on the ring by the cables?
(0, 6, 0) m
y
D
AF
Problem 4.68 In Problem 4.67, determine the mo-
ment about point Ddue to the force exerted on the ring
Aby the cable AB.
Solution: We need to write the forces as magnitudes times the
D(0, 6, 0)
194
Problem 4.69 The tower is 70 m tall. The tensions
in cables AB,AC, and AD are 4 kN, 2 kN, and 2 kN,
respectively. Determine the sum of the moments about
the origin Odue to the forces exerted by the cables at
point A.
35 m
35 m
40 m 40 m
40 m
A
y
x
z
B
O
C
D
Solution: The coordinates of the points are A(0, 70, 0), B(40, 0,
0), C(40, 0, 40) D(35, 0, 35). The position vectors corresponding
to the cables are:
The sum of the forces acting at Aare
TAD0.2792i6.6615jC0.07239k(kN-m)
The position vector of Ais rOA D70j. The moment about Ois MD
Problem 4.70 Consider the 70-m tower in Prob-
lem 4.69. Suppose that the tension in cable AB is 4 kN,
and you want to adjust the tensions in cables AC and
AD so that the sum of the moments about the origin O
due to the forces exerted by the cables at point Ais zero.
Determine the tensions.
Solution: From Varignon’s theorem, the moment is zero only if
the resultant of the forces normal to the vector rOA is zero. From
Problem 4.69 the unit vectors are:
eAD DrAD
jrADjD35
85.73 i70
85.73 j35
85.73
The tensions are TAB D4eAB,TAC DjTACjeAC, and TAD DjTADjeAD.
The components normal to rOA are
FXD⊲0.4082jTADj0.4444jTACjC1.9846⊳iD0
Problem 4.71 The tension in cable AB is 150 N. The
tension in cable AC is 100 N. Determine the sum of the
moments about Ddue to the forces exerted on the wall
by the cables.
C
y
x
D
z
A
B
5 m
5 m
8 m
8 m
4 m
Solution: The coordinates of the points A,B,Care A(8, 0, 0),
B(0, 4, 5), C(0, 8, 5), D(0, 0, 5). The point Ais the intersection of
the lines of action of the forces. The position vector DA is
rDA D8iC0j5k.
The position vectors AB and AC are
The unit vectors parallel to the cables are:
ijk
⊲⊲8⊳⊲C32.77⊳⊲5⊳⊲181.79⊳⊳jC⊲8⊳⊲123.24⊳k
196
Problem 4.72 Consider the wall shown in Prob-
lem 4.71. The total force exerted by the two cables in
the direction perpendicular to the wall is 2 kN. The
magnitude of the sum of the moments about Ddue to
the forces exerted on the wall by the cables is 18 kN-m.
What are the tensions in the cables?
Problem 4.73 The tension in the cable BD is 1 kN. As
a result, cable BD exerts a 1-kN force on the “ball” at
z
Solution: We have the force and position vectors
Problem 4.74* Suppose that the mass of the sus-
pended object Ein Problem 4.73 is 100 kg and the mass
of the bar AB is 20 kg. Assume that the weight of the
bar acts at its midpoint. By using the fact that the sum
Solution: We have the following forces applied at point B.
F1D⊲100 kg⊳⊲9.81 m/s2⊳j,F2DTBC
p33 ⊲4iCj4k⊳,
MyD2.089TBC 3.333TBD D0
198
