A man holds a 179 N ball in his hand with the forearm horizontal as shown on the

Question

A man holds a 179 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, Fb 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 22 N and that the distance between the  elbow joint and the center of the ball is L= 0.328 m.  The distance between the point where the tendon connects the bicep to the forearm and the center of gravity of the arm is  x1 = 0.089 m. The distance between the point of connection of the  tendon and the elbow joint is x2 = 0.05 m.

1)

What is the magnitude of Fb?
Fb =

2)

What is the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint?
Fu =

Explanation / Answer

a) M*0.051 = 22 * ( 0.051+0.089) + 179*0.33
M = 1367.3 N

b) Sum of forces = 0

Fbone = M - 22 -179 = 1166.3 N (direction: downward)

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