Question

The moment arm of the shoulder abductor muscles is approximately 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.

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

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

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

Answer 1

1] The net force along the vertical direction is zero.

R_y + Tcos(150 - 90) - mg = 0

where R_y is the vertical component of the joint reaction force and T is the force from the muscle.

also, the horizontal net force must be zero.

=> R_x + Tcos(150) = 0

  where R_x is the horizontal component of the joint reaction force.

The net torque on the system must be zero for equilibrium. So, taking torque about the shoulder joint gives,

Tcos(150 - 90)L + R(0) + R(0) - mgd = 0

=> Tcos(150 - 90)(0.05) - 45(0.25) = 0

=> T = 22.5 N

this is the tension in the muscle.

Substitute this value in the above equations to give, R_y = 33.75 N and R_x = - 19.485 N

Orientation of joint reaction vector counter-clockwise from positive x axis is:

.