A man holds a 170 N ball in his hand with the forearm horizontal as shown on the

Question

A man holds a 170 N ball in his hand with the forearm horizontal as shown on the drawing to the left. He can support the ball in this position because of force, F_b from his bicep flexor muscule which is applied perpendicular to the forearm. This force applies a torque about the elbow joint. Let's assume the forearm weighs 20 N and that the distance between the elbow joint and the center of the ball is L = 0.332 m. The distance between the point where the tendon connects the bicep to the forearm and the center of gravity of the arm is x_1 = 0.087 m. The distance between the point of connection of the tendon and the elbow joint is x_2 = 0.049 m. What is the magnitude of F_b? F_b = N What is the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint? F_u = N

Explanation / Answer

Balancing torque about elbow joint,

Net torque = (x2 Fb) - ((x1 + x2) x 20 ) - (L x 170) = 0

0.087 Fb = (0.087 + 0.049)(20) + (0.332)(170)

Fb = 680 N ......Ans

2) Balancing forces in vertical direction,

Fb - 20 - 170 - Fu = 0

680 -20 - 170 = Fu

Fu = 490 N

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