Question

A person weighing 845N is holding a 111N dumbbell horizontally at arm's length. The distance from the center of the dumbbell to the shoulder joints is 0.61m. The upper extremity weighs 53.4N and the center of gravity is located 0.23m from the proximal end. Assume that a muscle inserts 0.15m from the axis of rotation. How much vertical muscle force is needed to keep the siutation in equilibrium?

Answer 1

How much vertical muscle force is needed to keep the siutation in equilibrium?

For an equilibrium of torque, we have

= 0

F_muscle d - w_dumbbell L - w_extremity r = 0

where, w_dumbbell = weight of dumbbell = 111 N

L = distance from the center of dumbbell to the shoulder joints = 0.61 m

r = distance from the proximal end to the center of gravity = 0.23 m

w_extremity = weight of an upper extremity = 53.4 N

d = distance from the axis of rotation = 0.15 m

then, we get

F_muscle (0.15 m) - (111 N) (0.61 m) - (53.4 N) (0.23 m) = 0

F_muscle (0.15 m) = [(67.71 N.m) + (12.282 N.m)]

F_muscle = [(79.992 N.m) / (0.15 m)]

F_muscle = 533.2 N