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

Question

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°.

Accepted Answer

First, let us draw a free-body diagram of the arm.
In Figure 11.4, W_A is the weight of the arm, W is the weight that is being held, F_D is the force of the deltoid muscle, and F_S is the force exerted by the shoulder joint.

Writing the equations of motion in vector form, we have

[latex]\sum{F_{x}}= 0 = F_{s}\cos(\phi) - F_{D}\cos( heta)[/latex]

[latex]\sum{F_{y}}= 0 = - F_{s}\sin(\phi) + F_{D}\sin( heta) - W_{A} - W[/latex]

[latex]\sum{M_{0}}= 0 = (0.45 ft)F_{D}\sin( heta) - (1 ft)W_{A} - (2.1 ft)W[/latex]

Solving the third equation for F_D

[latex]F_{D}= \frac{(1 ft)(10 lbf) + (2.1 ft)(30 lbf)}{(0.45 ft)\sin(17^{\circ})}= 555 lbf[/latex]

Solving the first and second equations

[latex]F_{s}\cos(\phi)= F_{D}\cos( heta)= (555 lbf)\cos(17^{\circ})= 530 lbf[/latex]

[latex]F_{s}\sin(\phi)= F_{D}\sin( heta) - W_{A} - W= (555 lbf)\sin(17^{\circ}) - 10 lbf - 30 lbf= 122 lbf[/latex]

Solving for F_S,

[latex]F_{S}= \sqrt{(F_{s}\cos(\phi))^{2} + (F_{s}\sin(\phi))^{2}}= \sqrt{(530 lbf)^{2} + (122 lbf)^{2}}= 545 lbf[/latex]

To solve for [latex]\phi[/latex],

[latex]\phi= an^{-1} \left(\frac{122 lbf}{530 lbf} ight)= 13^{\circ}[/latex]

The force that is generated in the deltoid muscle is very large, which suggests that this is not a stable position for the arm.

Question 11.2: Calculate the force on the shoulder joint for an athlete who...

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer: