Angular Kinetics

PEP 294: Solutions

XIII. Angular Kinetics

	Read questions in the textbook first before you start reading these solutions.
	Then, compare your interpretation of the questions and the solutions provided.

Introductory Problems

3. m₁ = 23 kg, d₁ = 1.5 m, m₂ = 21 kg Þ d₂ (d₁ & d₂ = moment arms)

Torque 1: T₁ = (23)(9.81)(1.5) = 338.45 Nm

Torque 2: T₂ = (21)(9.81)(d₂)

T₁ = T₂: (21)(9.81)(d₂) = 338.45 Nm

d₂ = 1.64 m

4. FA = 3 cm, R = 200 N, RA = 25 cm Þ F

T_F = (F)(FA) = (F)(0.03)

T_R = (R)(RA) = (200)(0.25) = 50 Nm

T_F = T_R: (F)(0.03) = 50

F = 1666.67 N

5. F_A = 40 N, FA_A = 20 cm, F_B = 30 N, FA_B = 25 cm Þ T_net (resultant torque)

Assuming that A is trying to rotate the door in counterclockwise direction:

T_A (counterclockwise, positive) = (F_A)(FA_A) = (40)(0.20) = 8 Nm

T_B (clockwise, negative) = -(F_B)(FA_B) = -(30)(0.25) = -7.5 Nm

T_net = T_A + T_B = 8 - 7.5 = 0.5 Nm

0.5 Nm in the direction that A pushes

7. Since mechanical advantage = R/F = FA/RA, it depends on the lengths of the force arm and resistance arm.

8. F = 100 N, angle of pull = 20° Þ F_rot, F_stab

Rotary component (perpendicular to the bone): F_rot = (F)(sin 20°) = 34.20 N

Stabilizing component (along the bone): F_stab = (F)(cos 20°) = 93.97 N

9. m = 10 kg, F = 2 N Þ N, F_f

Normal reaction force: N = weight of the block = (m)(g) = (10)(9.81) = 98.1 N

External forces acting on the block horizontally: F & F_f

In order for the block not to move at all the net force should be 0. In order words, F_f must be equal but opposite to F:

F_f = 2 N

Additional Problems

3. See the figure on p. 432:

(a) W_a = 35 N, d_m = 3 cm, d_a = 15 cm, joint angle = 45°

Torque due to the arm weight: T_a = (W_a)(d_a)(sin 45°) = (35)(0.15)(sin 45°) = 3.71 Nm

Torque due to muscle force: T_m = (F_m)(d_m)(sin 135°) = (F_m)(0.03)(sin 135°)

Since T_a = T_m: (F_m)(0.03)(sin 135°) = 3.71

F_m = 175 N

(b) W_w = 50 N, d_w = 25 cm

T_net = T_a + T_w = 3.71 + (50)(0.25)(sin 45°) = 3.71 + 8.84 = 12.55 Nm

Since T_m = T_a + T_w: (F_m)(0.03)(sin 135°) = 12.55

F_m = 591.61 N

4. See the figure on p. 432:

Torque due to muscle force: T_m (clockwise, negative) = -(F_m)(0.03)

Torque due to arm weight: T_a (clockwise, negative) = -(40)(0.17) = -6.8 Nm

Torque measured at the scale: T_s (counterclockwise, positive) = (90)(0.32) = 28.8 Nm

Resultant torque = 0 (equilibrium): T_net = T_s + T_m + T_a = 0

28.8 - 6.8 - (F_m)(0.03) = 0 or (F_m)(0.03) = 28.8 - 6.8

F_m = 733.33 N

7. See the figure on p. 433:

(90)(0.7) = (F)(0.06)

F = (90)(0.7) / (0.06) = 1050 N

9. See the figure on p. 433:

(a) (80)(0.25) = (F_m)(0.03)

F_m = (80)(0.25)/(0.03) = 666.7 N (to the left)

(b) Linear equilibrium: F_fulcrum + F + R = 0

F_fulcrum = F_r

F = F_m = -666.7 N

R = 80 N

Therefore, F_r = -(F_m + R) = -(-666.7 + 80) = 586.67 N

10. See the figure on p. 433:

In order for this system to stay in equilibrium, the sum of F and T_a should be equal but opposite to T_b so that the F_net of the system becomes 0. The resultant force is 80 N. (Here, T means tension, or force, not torque.)

This problem then reduces to a simple triangular problem we did in vector algebra earlier. Use the law of sines to compute the magnitudes of F and T_a. See the figure below.