Question

Consider the case of the bicepts muscle acting on a forearm holding a 43 N weight. The forearm is 30° below the horizontal and the bicepts muscle inserts at an angle of 45° with the forearm. The muscle insertion is at 5 cm from the elbow joint. The forearm has a weight of 21 N and has its center of gravity 15 cm from the elbow. The 43 N weight is held in the hand 30 cm from the joint center.

Determine the muscle force (M) needed to maintain this position.

M. 5cm 15 cm 30 cm B

Answer 1

As the forearm is in equilibrium net force and net torque acting on it must be zero.

Apply net torque about joint = 0

M*5*sin(45) - W*15*sin(90 - 30) - L*30*sin(90 - 30)

M*5*sin(45) = W*15*sin(60) + L*30*sin(60)

M = (W*15*sin(60) + L*30*sin(60))/(5*sin(45))

= (21*15*sin(60) + 43*30*sin(60))/(5*sin(45))

= 393 N <<<<<<<<<<-------------------Answer