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 15.0 N 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.

Answer 1

Answer:

In This Question there are three parts so answers of the three parts are in the following attachments .

Step-by-step explanation:

Answer 2

To answer this problem, we use moment balance, balancing the forces that tip the arm clockwise and those that tip the arm counterclockwise. In the balance, we multiply force by the perpendicular distance from the center.
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 
 F(3.8) - 15(15) = 80(33) 
solving for for F, F is equal to 753.98 Newtons