Physics, 6th Edition
Torque and Rotational Equilibrium
*5-49. An car has a distance of 3.4 m between front and rear axles. If 60 percent of the weight
rests on the front wheels, how far is the center of gravity located from the front axle?
F
Axis
x
= 0.6

Physics, 6th Edition
Torque and Rotational Equilibrium
5-19. A uniform meter stick is balanced at its midpoint with a single support. A 60-N weight is
suspended at the 30 cm mark. At what point must a 40-N weight be hung to balance the
system?
20 cm
(The

Physics, 6th Edition
Torque and Rotational Equilibrium
5-16. Determine the resultant torque about the corner A for Fig. 5-15.
80 N
B
= +(160 N)(0.60 m) sin 400 - (80 N)(0.20 m)
20 cm
= 61.7 N m 16.0 N m = 45.7 N m
A
60 cm
400
r
R = 45.7 N m
160 N
400
C

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-18. Find the resultant torque about axis B in Fig. 5-15.
Fx = 160 cos 400;
Fy = 160 sin 400
= (123 N)(0.2 m) + (103 N)(0.6 m) = 37.2 N m
B
20 cm
160 N
80 N
60 cm
Fy
400
Fx

Physics, 6th Edition
Torque and Rotational Equilibrium
5-14. Two wheels of diameters 60 cm and 20 cm are fastened together and turn on the same axis
as in Fig. 5-14. What is the resultant torque about a central axis for the shown weights?
r1 = (60 cm) = 0

Physics, 6th Edition
Torque and Rotational Equilibrium
5-13. What horizontal force must be exerted at point A in Fig 5-11b to make the resultant torque
about point B equal to zero when the force F = 80 N?
F = 80 N
= P (2 m) (80 N)(5 m) (sin 300) = 0
P =

Physics, 6th Edition
Torque and Rotational Equilibrium
5-7. A leather belt is wrapped around a pulley 20 cm in diameter. A force of 60 N is applied to
the belt. What is the torque at the center of the shaft?
r = D = 10 cm;
= (60 N)(0.10 m) = +6.00 N m
F

Physics, 6th Edition
Torque and Rotational Equilibrium
5-1. Draw and label the moment arm of the force F about an axis at point A in Fig. 5-11a. What
is the magnitude of the moment arm?
Moment arms are drawn perpendicular to action line:
rA = (2 ft) sin 2

Physics, 6th Edition
Torque and Rotational Equilibrium
5-11. What is the resultant torque about point A in Fig. 5-13. Neglect weight of bar.
= +(30 N)(6 m) - (15 N)(2 m) - (20 N)(3 m)
15 N
4m
= 90.0 N m, Counterclockwise.
30 N
3m
2m
A
20 N

Physics, 6th Edition
Torque and Rotational Equilibrium
5-9. A person who weighs 650 N rides a bicycle. The pedals move in a circle of radius 40 cm. If
the entire weight acts on each downward moving pedal, what is the maximum torque?
= (250 N)(0.40 m)
=

Torque and Rotational Equilibrium
Physics, 6th Edition
5-10. A single belt is wrapped around two pulleys. The drive pulley has a diameter of 10 cm,
and the output pulley has a diameter of 20 cm. If the top belt tension is essentially 50 N at
the edge of e

Physics, 6th Edition
Torque and Rotational Equilibrium
5-8. The light rod in Fig. 5-12 is 60 cm long and pivoted about point A. Find the magnitude and
sign of the torque due to the 200 N force if is (a) 900, (b) 600, (c) 300, and (d) 00.
= (200 N) (0.60

Physics, 6th Edition
Torque and Rotational Equilibrium
5-6. The force F in Fig. 5-11b is 400 N and the angle iron is of negligible weight. What is the
resultant torque about axis A and about axis B?
Counterclockwise torques are positive, so that A is + an

Physics, 6th Edition
Torque and Rotational Equilibrium
5-5. If the force F in Fig. 5-11a is equal to 80 lb, what is the resultant torque about axis A
neglecting the weight of the rod. What is the resultant torque about axis B?
Counterclockwise torques are

Physics, 6th Edition
Torque and Rotational Equilibrium
5-3. Determine the moment arm if the axis of rotation is at point A in Fig. 5-11b. What is the
magnitude of the moment arm?
rB = (2 m) sin 60
0
F
600
rB = 1.73 m
5m
B
300
2m
A
rA
rB

Physics, 6th Edition
Torque and Rotational Equilibrium
5-12. Find the resultant torque in Fig. 5-13, if the axis is moved to the left end of the bar.
= +(30 N)(0) + (15 N)(4 m) - (20 N)(9 m)
15 N
A 4m
2m
3m
= -120 N m, counterclockwise.
30 N
20 N

Physics, 6th Edition
Torque and Rotational Equilibrium
5-21. An 8-m board of negligible weight is supported at a point 2 m from the right end where a
50-N weight is attached. What downward force at the must be exerted at the left end to
produce equilibriu

Physics, 6th Edition
Torque and Rotational Equilibrium
5-22. A 4-m pole is supported at each end by hunters carrying an 800-N deer which is hung at a
point 1.5 m from the left end. What are the upward forces required by each hunter?
= A (0) (800 N)(1.5 m

Physics, 6th Edition
Torque and Rotational Equilibrium
5-24. For equilibrium, what are the forces F1 and F2 in Fig. 5-17. (Neglect weight of bar.)
F1
= (90 lb)(5 ft) F2 (4 ft) (20 lb)(5 ft) = 0;
5 ft
F2 = 87.5 lb Fy = F1 F2 20 lb 90 lb = 0
F1 = F2 +110 l

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-46. For the conditions set in Problem 5-5, what are the horizontal and vertical components
of the force exerted by the floor hinge on the base of the boom?
Fx = H 1169 N = 0;
or
H = 1169 N
Fy = V 1

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-45. Suppose the boom in Fig. 5-24 has a weight of 100 N and the suspended weight W is
300
equal to 400 N. What is the tension in the cord?
m) sin 300) (400 N)(6 m) cos 300
(100 N)(3 m) cos 300 = 0

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-44. (a) What weight W will produce a tension of 400 N in the rope attached to the boom in
Fig. 5-24?. (b) What would be the tension in the rope if W = 400 N? Neglect the weight
300
of the boom in e

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-47. What is the tension in the cable for Fig. 5-25. The weight of the boom is 300 N but its
length is unknown. (Select axis at wall, L cancels.)
450
L
TL sin 75 (300 N ) sin 300 546 L sin 300 0
2

Physics, 6th Edition
Torque and Rotational Equilibrium
*5-40. Find the resultant torque about point B in Fig. 5-22.
= (70 N)(0) (50 N)(a + b) ;
First find a and b.
a = (0.05 m) cos 500 = 0.0231 m; b = (0.16 m) sin 550 = 0.131 m
= (50 N)(0.0231 m + 0.131

Physics, 6th Edition
Torque and Rotational Equilibrium
B
*5-39. Find the resultant torque about point A in Fig. 5-22.
500
= (70 N)(0.05 m) sin 500 (50 N)(0.16 m) sin 550
5 cm
70 N
A
r
= 2.68 N m 6.55 N m = 3.87 N m
16 cm
= 3.87 N m
r
550
50 N
B
500
5 c

Torque and Rotational Equilibrium
Physics, 6th Edition
5-42. On a lab bench you have a small rock, a 4-N meterstick and a single knife-edge support.
Explain how you can use these three items to find the weight of the small rock.
a
Measure distances a and

Physics, 6th Edition
Torque and Rotational Equilibrium
5-23. Assume that the bar in Fig. 5-16 is of negligible weight. Find the forces F and A provided
the system is in equilibrium.
= (80 N)(1.20 m) F (0.90 m) = 0;
F
F = 107 N
Fy = F A 80 N = 0; A = 107

Physics, 6th Edition
Torque and Rotational Equilibrium
5-35. What is the resultant torque about the hinge in Fig. 4-20? Neglect weight of the curved bar.
= (80 N)(0.6 m) (200 N)(0.4 m) sin 400
60 cm
80 N
= 48.0 N m 51.4 N m;
= 3.42 N m
r
400
40
0
200 N

Physics, 6th Edition
Torque and Rotational Equilibrium
5-36. What horizontal force applied to the left end of the
bar in Fig. 4-20 will produce rotational equilibrium?
F
From Prob. 5-33: = - 3.42 N m.
Thus, if = 0, then torque of +3.42 N m must be added.

Torque and Rotational Equilibrium
Physics, 6th Edition
5-26. A V-belt is wrapped around a pulley 16 in. in diameter. If a resultant torque of 4 lb ft is
required, what force must be applied along the belt?
R = (16 in.) = 8 in. R = (8/12 ft) = 0.667 ft
F (