Question

The moment arm of the shoulder abductor muscles is approxi- mately 5 cm from the axis of rotation of the shoulder joint, and the force vector representing shoulder abductor muscle force is oriented at approximately 150° relative to the positive x-axis.The weight of the upper extremity of an 80-kg male is approximately 45 N. If the person is 1.78 m in height, then the center of gravity of the upper-extremity is approximately 0.25 m from the shoulder joint.

3. Calculate the vertical component of the joint reaction force.

4. Calculate the horizontal component of the joint reaction force.

5. Calculate the orientation of the vector representing joint reaction force.

Answer 1

muscle force f

the force vector makes 150 deg with the +ve x-axis

arm length of the force = 5 cm

extremity weight = 45 N - vertically down

arm length = 0.25 m

only vertical component of the muscle force fSin(150) contributes for the torque. horizontal force do not contribute any torque

M_J = 0.25 * 45 - 0.05 fSin(150) =0 , moments about the shoulder joint

f = 450 N

R - reaction at the joint

F_y = R_y + 450 sin(150) - 45 = 0

R_y = -180 N

F_x = R_x + 450 Cos(150) = 0 ( only muscle force acts in horizontal direction)

R_x = 389.7 N

R = 429 N , joint reaction force.

orientation of the reaction Tan (p) = -180/390

p = -24.77 deg , below the horizontal