torque and angular

Question

Mary is doing a lateral raise exercise using a 100N dumbbell. She is performing an isometric contraction and holds her arm still in a position in which the horizontal distance from the dumbbell to her shoulder joint is 0.70m.

a) for this position, what torque is created by the dumbbell about her shoulder joint axis?

b) if the moment arm of her shoulder abductor muscles is .04m about the shoulder joint axis, what amount of force must these muscles produce to hold the arm in the same position described? ignore the weight of her arm.

Answer 1

a) The torque created by the dumbbell about Mary's shoulder joint axis is given by:

Torque = Force x Moment Arm

where the force is the weight of the dumbbell, which is 100N, and the moment arm is the horizontal distance between the dumbbell and her shoulder joint, which is 0.70m. Thus, the torque is:

Torque = 100N x 0.70m = 70 N·m

Therefore, the torque created by the dumbbell about her shoulder joint axis is 70 N·m.

b) The force that Mary's shoulder abductor muscles must produce to hold her arm in the same position is given by:

Force = Torque / Moment Arm

where the torque is the same as before, which is 70 N·m, and the moment arm is the distance between the shoulder joint axis and the line of action of the shoulder abductor muscles, which is given as 0.04m. Thus, the force is:

Force = 70 N·m / 0.04m = 1750 N

Therefore, the amount of force that Mary's shoulder abductor muscles must produce to hold her arm in the same position is 1750 N.