Question

A bowler holds a bowling ball with mass M = 6.8 kg in the palm of his hand. Lower arm has mass m = 1.7 kg. As the figure shows, his upper arm is vertical and his lower arm is horizontal.What is the magnitude of (a) the force of the biceps muscle on the lower arm and (b) the force between the bony structures at the elbow contact point?

Answer 1

Solution:

Given:

M = 6.8 kg

m = 1.7 kg

Part (A) Solution:

At equilibrium condition,the net torque acting on the system is :

Net Torque = ((0.02) F - (0.14) mg - (0.32) Mg

Therefore, the magnitude of the force on the biceps muscle on the lower arm is given by :

F = { (0.14) mg + (0.32) Mg } / (0.02)

F = { (0.14)(1.7)(9.81) + (0.32)(6.8)(9.81) } / (0.02)

F = 1148.067 N = 1.18 x 10³ N

Part (B) Solution:

At the equilibrium condition, the net force acting on the system is :

Net Force = F - F' - (m+M)g = 0

Then the magnitude of the force between the bone structure at the elbow contact point is :

F' = F - (m+M)g

F' = 1148.067 N - (1.7 + 6.8)(9.81)

F' = 1100.68 N = 1.1 x 10³ N

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