Question

IP Force to Hold a Baseball A person holds a 1.42-N baseball in his hand, a distance of 34.0 cm from the elbow joint, as shown in the figure(Figure 1) . The biceps, attached at a distance of 2.75 cm from the elbow, exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.50 kg . What is the net torque?

Answer 1

We will assume that the CM of the arm is at "L" from the elbow, and the ball is at 34cm. Then the net torque is computed by:

Net τ = 1.42 N * 34 cm + 1.50 kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm 

= 48.28 N*cm + 1.50kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm 

= 48.28 N*cm + 499.8 – 34.65 N*cm

τ  = 513.43 N*cm or 5.1343 N*m

Answer 2

Answer:

Explanation:

Net τ = 1.42 N * 34 cm + 1.50 kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm

= 48.28 N*cm + 1.50kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm

= 48.28 N*cm + 499.8 – 34.65 N*cm

τ  = 513.43 N*cm or 5.1343 N*m