In the human arm, the forearm and hand pivot about the elbow joint. Consider a s

Question

In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm

Now the person holds a 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow.

find the magnitude of the force that the elbow joint exerts on the forearm.

Explanation / Answer

The question states that the center of gravity of the weight is 33.0 (I assume cm) from the elbow. Use this number.

If F is the force created by the bicep and the loads are in Newtons
if the loads are in kg, then multiply kg by g (9.8) to get N

0 = F(3.8) - 15(15) - 80(33)

solve for F = [15(15) + 80(33)] / 3.8 => [225 + 2640] / 3.8 = 753.94

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