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Problem 4.174 A plumber exerts the two forces shown
to loosen a pipe.
(a) What total moment does he exert about the axis of
the pipe?
(b) If you represent the two forces by a force Facting
at Oand a couple M, what are Fand M?
(c) If you represent the two forces by a wrench
consisting of the force Fand a parallel couple Mp,
what is Mp, and where does the line of action of F
intersect the xyplane?
x
–70 k (lb)
12 in
6 in
16 in
16 in
y
z
O
Solution: The sum of the forces is
(a) The total moment exerted on the pipe is
254
Problem 4.175 The Leaning Tower of Pisa is approx-
imately 55 m tall and 7 m in diamteter. The horizontal
displacement of the top of the tower from the vertical
is approximately 5 m. Its mass is approximately 3.2ð
106kg. If you model the tower as a cylinder and assume
that its weight acts at the center, what is the magnitude
of the moment exerted by the weight about the point at
the center of the tower’s base?
5m
41.5 101.463
42.5 101.472
Problem 4.176 The cable AB exerts a 300-N force on
the support Athat points from Atoward B. Determine the
magnitude of the moment the force exerts about point P.
x
y
(0.3, 0.6) m
A
B
(⫺0.4, 0.3) m
Solution:
Problem 4.177 Three forces act on the structure. The
sum of the moments due to the forces about Ais zero.
Determine the magnitude of the force F.
b
45⬚30⬚
4 kN 2 kN
Solution:
Problem 4.178 Determine the moment of the 400-N
force (a) about A, (b) about B.
220 mm
30°
A
400 N
256
Problem 4.179 Determine the sum of the moments
exerted about Aby the three forces and the couple.
300 lb
5 ft
A
Problem 4.180 In Problem 4.179, if you represent the
three forces and the couple by an equivalent system
consisting of a force Facting at Aand a couple M,
what are the magnitudes of Fand M?
Problem 4.181 The vector sum of the forces acting on
the beam is zero, and the sum of the moments about A
is zero.
(a) What are the forces Ax,Ay, and B?
(b) What is the sum of the moments about B?
30°
Ax
Ay400 N
220 mm
260 mm
Solution: The vertical and horizontal components of the 400 N
force are:
Problem 4.182 The hydraulic piston BC exerts a 970-
lb force on the boom at Cin the direction parallel to the
piston. The angle ˛D40°. The sum of the moment about
Adue to the force exerted on the boom by the piston
and the weight of the suspended load is zero. What is
the weight of the suspended load?
9 ft 6 ft
AB
C
a
Solution: The horizontal (x) and vertical (y) coordinates of point
Crelative to point Bare
Problem 4.183 The force FD60iC60j(lb).
(a) Determine the moment of Fabout point A.
(b) What is the perpendicular distance from point Ato
the line of action of F?
y
x
F
(4, –4, 2) ft
A
(8, 2, 12) ft
z
Solution: The position vector of Aand the point of action are
(b) The magnitude of the moment is
258
Problem 4.184 The 20-kg mass is suspended by cables
attached to three vertical 2-m posts. Point Ais at (0, 1.2,
0) m. Determine the moment about the base Edue to the
force exerted on the post BE by the cable AB.B
A
C
D
y
z
x
0.3 m
Solution: The strategy is to develop the simultaneous equations in
The equilibrium conditions are TAB CTAC CTAD DW. Collect like
Problem 4.185 What is the total moment due to the
two couples?
(a) Express the answer by giving the magnitude and
stating whether the moment is clockwise or coun-
terclockwise.
(b) Express the answer as a vector.
y
100 N
100 N
100 N
100 N
4 m
2 m
2 m
4 m
x
Solution:
Problem 4.186 The bar AB supporting the lid of the
grand piano exerts a force FD6iC35j12k(lb) at
Solution: The position vector of point Bis rOB D3iC4jC3k.
The moment about the xaxis due to the force is
260
Problem 4.187 Determine the moment of the vertical
800-lb force about point C.
800 lb
B
A (4, 3, 4) ft
y
Problem 4.188 In Problem 4.187, determine the mo-
ment of the vertical 800-lb force about the straight line
through points Cand D.
Solution: In Problem 4.197, we found the moment of the 800 lb
Problem 4.189 The system of cables and pulleys sup-
line of action intersect the xaxis?
y
F
B
H
G
D
E
Solution: The cable-pulley combination does not produce a mo-
Problem 4.190 Consider the system in Problem 4.189.
F, and where does its line of action intersect the
xaxis?
Solution: The vertical component of the tension is each cable must
equal half the weight supported.
The single force must equal the sum of the vertical components; since
262
Problem 4.191 The two systems are equivalent. Deter-
mine the forces Axand Ay, and the couple MA.y
400 mm
20 N
System 1
Solution: The sum of the forces for System 1 is
Problem 4.193 If you represent the equivalent systems
in Problem 4.191 by a force F, what is F, and where does
its line of action intersect the xaxis?
Problem 4.194 The two systems are equivalent. If
FD100iC40jC30k(lb),
x
x
6 in
6 in
zz
Solution: The sum of forces in the two systems must be equal,
thus F0DFD100iC40jC30k(lb).
The moment for the unprimed system is MTDrðFCM.
Problem 4.195 The tugboats Aand Bexert forces
FAD1 kN and FBD1.2 kN on the ship. The angle
D30°. If you represent the two forces by a force F
acting at the origin Oand a couple M, what are F
and M?
FA
y
x
O
A
25 m
F
B
60 m
60 m
Solution: The sums of the forces are:
FXD⊲1C1.2 cos 30°⊳iD2.0392i(kN)
FYD⊲1.2 sin 30°⊳jD0.6j(kN).
The equivalent force at the origin is
FEQ D2.04iC0.6j
tions are
264
Problem 4.196 The tugboats Aand Bin Problem 4.195
exert forces FAD600 N and FBD800 N on the ship.
The angle D45°. If you represent the two forces by
a force F, what is F, and where does its line of action
intersect the yaxis?
MODrAðFACrBðFB.
The vector positions are
0.600
0.5656 0.5656 0 D16.20k(kN-m)
The single force must produce this moment.
ijk
Problem 4.197 The tugboats Aand Bin Problem 4.195
want to exert two forces on the ship that are equivalent
to a force Facting at the origin Oof 2-kN magnitude. If
FAD800 N, determine the necessary values of FBand
angle .
Solution: The equivalent force at the origin is ⊲FACFBcos ⊳2C
These are two equations in two unknowns FBsin and FBcos . For
25 FAC60
25 x2
Reduce to obtain the quadratic in x:
25 2F2
Substitute FAD800 N to obtain 6.76x27616xC326400 D0. In
FBDp44.62C1812.92D1813.4N
FBDp676.82C1082.02D1276.2N,
Problem 4.198 If you represent the forces exerted by
the floor on the table legs by a force Facting at the
origin Oand a couple M, what are Fand M?
y
1 m 2 m
Solution: The sum of the forces is the equivalent force at the
The sum of the moments about the origin is
ijk
C
ijk
C
ijk
Problem 4.199 If you represent the forces exerted by
the floor on the table legs in Problem 4.198 by a force
F, what is F, and where does its line of action intersect
the xzplane?
Solution: From the solution to Problem 4.198 the equivalent force
is FD190j. This force must produce the moment MD98iC184k
obtained in Problem 4.198.
ijk
266
Problem 4.200 Two forces are exerted on the crank-
0.545, and its magnitude is 4 kN. The direction cosines
of FBare cos xD0.182, cos yD0.818, and cos zD
0.545, and its magnitude is 2 kN. If you represent the
two forces by a force Facting at the origin Oand a
couple M, what are Fand M?
Solution: The equivalent force is the sum of the forces:
D0.728iC3.272jC2.18k(kN)
FBD2⊲0.182iC0.818j0.545k⊳D0.364iC1.636j1.09k(kN).
The sum of the moments:
Problem 4.201 If you represent the two forces exerted
on the crankshaft in Problem 4.200 by a wrench consis-
ting of a force Fand a parallel couple Mp, what are F
and Mp, and where does the line of action of Fintersect
the xzplane?
Solution: From the solution to Problem 4.200,
FD0.364iC4.908jC1.09k(kN) and
MD0.1309i0.0438jC1.1125k(kN-m).
This is the equivalent couple parallel to F.
The component of the moment perpendicular to Fis
4.908 D0.2178 m,
4.908 D0.0236 m
