Question

A woman lifts a 3.6-kg barbell in each hand with her arm in a horizontal position at the side of her body and holds it there for 3 s (see the figure below).

What force does the deltoid muscle in her shoulder exert on the humerus bone while holding the barbell? The deltoid attaches 13 cm from the shoulder joint and makes a 13? angle with the humerus. The barbell in her hand is 0.55 m from the shoulder joint, and the center of mass of her 4.0-kg arm is 0.24 m from the joint.

Answer 1

This is a torque question. The total weight of the arm and the object in the hand are supported by 2 forces. The 2 forces are the vertical component of the tension in the deltoid muscle and the upward force which the scapula exerts on the shoulder joint. If we choose the shoulder joint as the pivot point, the upward force which the scapula exerts on the shoulder joint does not produce torque.

So we start by choosing a pivot and assigning sign of torque to each force.
In placing the pivot at the scapula the deltoid has a counterclockwise (positive) force which is balanced by the arm and the weight both having clockwise (negative) forces

T(arm) = m*g*x = 4.0kg*9.81m/s^2*0.24m = 9.4176N
T(weight) = m*g*x = 3.6kg*9.81m/s^2*0.55m = 19.4238N
T(delt) = F*x = sin(13)*.13m

Now balance the forces:
T(delt) = T(weight) + T(arm)
T*0.13sin(13) = 9.4176N + 19.4238N
T = 986 N