Question 11.2: Calculate the force on the shoulder joint for an athlete who...
Calculate the force on the shoulder joint for an athlete who is holding a weight with the arm perfectly horizontal (see Figure 11.3). To calculate the reaction forces within the shoulder joint, consider that the weight of the arm is equal to 10 lbf and is located at a distance 1 ft from the shoulder joint. The weight that the athlete is holding is equal to 40 lbf, and it is held at a distance of 2.1 ft from the shoulder joint. To simplify the problem, let us consider that the deltoid muscle is the primary muscle holding the arm in place and that it attaches to the humerus at a distance of 0.45 ft from the shoulder joint at an angle of 17°.
First, let us draw a free-body diagram of the arm.
In Figure 11.4, WA is the weight of the arm, W is the weight that is being held, FD is the force of the deltoid muscle, and FS is the force exerted by the shoulder joint.
Writing the equations of motion in vector form, we have
\sum{F_{x}}= 0 = F_{s}\cos(\phi) – F_{D}\cos(\theta)
\sum{F_{y}}= 0 = – F_{s}\sin(\phi) + F_{D}\sin(\theta) – W_{A} – W
\sum{M_{0}}= 0 = (0.45 ft)F_{D}\sin(\theta) – (1 ft)W_{A} – (2.1 ft)W
Solving the third equation for FD
F_{D}= \frac{(1 ft)(10 lbf) + (2.1 ft)(30 lbf)}{(0.45 ft)\sin(17^{\circ})}= 555 lbf
Solving the first and second equations
F_{s}\cos(\phi)= F_{D}\cos(\theta)= (555 lbf)\cos(17^{\circ})= 530 lbf
F_{s}\sin(\phi)= F_{D}\sin(\theta) – W_{A} – W= (555 lbf)\sin(17^{\circ}) – 10 lbf – 30 lbf= 122 lbf
Solving for FS,
F_{S}= \sqrt{(F_{s}\cos(\phi))^{2} + (F_{s}\sin(\phi))^{2}}= \sqrt{(530 lbf)^{2} + (122 lbf)^{2}}= 545 lbf
To solve for \phi,
\phi= \tan^{-1} \left(\frac{122 lbf}{530 lbf}\right)= 13^{\circ}
The force that is generated in the deltoid muscle is very large, which suggests that this is not a stable position for the arm.
