Question

An athlete at the gym holds a 3.5kg steel ball in his hand. His arm is 70cm long and has a mass of 4.0kg.What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55 degrees below horizontal?

Answer 1

Answer:

A. 38 nm

B. 22 nm

Explanation:

This question has given us m, mass of the ball as 3.5kg

M = The mass of arms = 4kg

R = length of arms as 70cm

70 cm = 0.7m

0.7m/2 = 0.35m

rmg + Rmg

g = 9.8m/s

= 0.7(3.5)(9.8)+(0.35)(4)(9.8)

= 24.01 + 13.72

= 37.72

~38

B.

55 degrees below horizontal

Cos 55⁰ = 0.5736

= 0.7(3.5)(9.8)(0.5736)+(0.35)(4)(9.8)(0.5736)

= 13.772136+7.869792

= 21.642

~22Nm

Answer 2

The magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm

The formula for calculating the magnitude of the torque about his arm is expressed as:

$\tau = (F_1r + F_2R) cos\theta$

F1 and F2 are the forces

$\theta$ is the given angle

F1 = 3.5 * 9.8 = 34.3 N

r = 70cm = 0.7m

F2 = 4.0 * 9.8 = 39.2N

R = 0.7/2 = 0.35m

Substitute the given values into the formula:

$\tau = (34.3(0.7) + (39.2)(0.35)) cos55^0\\\tau = (24.01+13.72)cos55\\\tau = 37.73cos55\\\tau=21.64Nm$

Hence the magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm

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