An athlete at the gym holds a 4.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg.

1) What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?

Known Variables:

Mass of Arm: 4 kg

Mass of Ball: 4 kg

arm length: 70 cm

First you have to find the weight of both the ball and the arm. Multiply them by the force of gravity.

weight of arm: 4kg x 9.8 = 39.2

weight of ball: 4kg x 9.8 = 39.2

We are looking for the Torque

Torque, τ = r₁ F₁ sin θ +r₂ F₂ sin θ

We need to find r1 and r2

r1= 70/2= 35 cm or .35m (arm: to find the center of gravity, divide by 2)

r2= 70cm or .70m (ball)

Since the arms are pointed straight out, the angle is 180°

τ=m1g(L/2) + m2gL.

L is the length of the system.

τ =4 * 9.8 *(0.7/2) + 4 * 9.8 * 0.7

= 40 Nm

What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55° below horizontal?

Now the arms are pointed out at an 55° angle.

you would be using cosine since its the horizontal distance.

τ = .35 x 39.2cos(55) + .70 x 39.2cos(55)

τ = 7.86 + 15.73

τ = 24 N*m