Pise

Chapter 5. Torque and Rotational Equilibrium

Unit Conversions
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 250 rA = 0.845 ft
5-2. Find the moment arm about axis B in Fig. 11a. (See figure above.)
rB = (3ft) sin 250 rB = 1.27 ft
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 600 rB = 1.73 m
5-4. Find the moment arm about axis B in Fig. 5-11b.
rB = (5 m) sin 300 rB = 2.50 m

Torque
5-5. If the force F in Fig. 5-11a is equal to 80 lb, what is the resultant torque aboutaxis A neglecting the weight of the rod. What is the resultant torque about axis B?
Counterclockwise torques are positive, so that (A is - and (B is +.
(a) (A = (80 lb)(0.845 ft) = -67.6 lb ft (b) (B = (80 lb)(1.27 ft) = +101 lb ft
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 andabout axis B?
Counterclockwise torques are positive, so that (A is + and (B is -.
(a) (A = (400 N)(1.732 m) = +693 N m; (b) (B = (400 N)(2.50 m) = -1000 N m
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 = ½D = 10 cm; ( = (60 N)(0.10 m) = +6.00 N m5-8. 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 m) sin ( for all angles:
(a) ( = 120 N m (b) ( = 104 N m
(b) ( = 60 N m (d) ( = 0
5-9. A person who weighs 650 N rides a bicycle. The pedals move in acircle of radius 40 cm. If the entire weight acts on each downward moving pedal, what is the maximum torque?
( = (250 N)(0.40 m) ( = 260 N m
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 each pulley, whatare the input and output torques?
Input torque = (50 N)(0.10 m) = 5 N m
Output torque = (50 N)(0.20 m) = 10 N m

Resultant Torque
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)
τ = 90.0 N m, Counterclockwise.


5-12. Find the resultant torque in Fig. 5-13, if the axisis moved to the left end of the bar.
(( = +(30 N)(0) + (15 N)(4 m) - (20 N)(9 m)
τ = -120 N m, counterclockwise.

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?
τ = P (2 m) – (80 N)(5 m) (sin 300) = 0
2 P = 200 N; P = 100 N

5-14. Two wheels ofdiameters 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 = ½(60 cm) = 0.30 m ; r2 = ½(30 cm) = 0.15 m
τ = (200 N)(0.30 m) – (150 N)(0.15 m) = 37.5 N m; τ = 37.5 N m, ccw
5-15. Suppose you remove the 150-N weight from the small wheel inFig. 5-14. What new weight can you hang to produce zero resultant torque?
τ = (200 N)(0.30 m) – W (0.15 m) = 0; W = 400 N
5-16. Determine the resultant torque about the corner A for Fig. 5-15.
(( = +(160 N)(0.60 m) sin 400 - (80 N)(0.20 m)
(( = 61.7 N m – 16.0 N m = 45.7 N m
τR = 45.7 N m
5-17. Find the...