Question

An athlete at the gym holds a 2.6 kg steel ball in his hand. His arm is 64 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?

(a) Straight out to his side, parallel to the floor

N m

(b) Straight, but 50

Answer 1

Weight of the ball:
Fb = 2.6 * 9.81 = 25.506 N
Weight of athlete's arm:
Fa= 4 * 9.81 = 39.24 N
We'll assume that weight of the arm acts in the center of the arm.

1) When hand is parallel to ground, torque is:
T = Fb * L + Fa * L/2
or
T = (Fb + Fa/2) * L
where L is length of the arm, L=0.64m
T = (25.506 + 39.24/2)*0.64 = 22.563 Nm

2) When hand is held at angle A below horizontal:
T = Fb * L * cosA + Fa * (L/2) * cos A
or
T = (Fb + Fa/2) * L * cos A
if A=35