Chapter 9, Problem 34P

Part (a)

To determine

The magnitude of the force $F_m$ requiredof the "deltoid" muscle to hold up the outstretched arm.

Answer to Problem 34P

Solution:

  $F_m$=249.9N

Explanation of Solution

Given:

Total mass of the arm, $M=3.3\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum \tau =0}\\ {\displaystyle \sum F=0}\end{array}$

Calculation:

Take torque about the joint:

  $\begin{array}{l}{\displaystyle \sum \tau =0}\\ F_m\sin 15°(0.12)-3.3(9.8)(0.24)=0\\ 0.0311F_m=7.762\\ F_m=249.9\text{N}\end{array}$

Conclusion:

Thus, the force exerted by the deltoid muscle=249.9N.

Part (b)

To determine

The magnitude of the force F_j exerted by the shoulder joint on the upper arm and the angle (to the horizontal) at which it acts.

Answer to Problem 34P

Solution:

  $F_j=\text{243}.5\text{ N},\theta =\text{7}.{\text{62}}^{\text{o}}$

Explanation of Solution

Given:

Total mass of the arm, $M=3.3\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum \tau =0}\\ {\displaystyle \sum F=0}\end{array}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_m\sin 15°-mg-F_{jy}=0\\ 249.9\sin 15°-3.3(9.8)-F_{jy}=0\\ F_{jy}=32.3\text{N}\\ {\displaystyle \sum F_y}=0\\ F_m\cos 15°-F_{jx}=0\\ 249.9\cos 15°-F_{jx}=0\\ F_{jx}=241.4\text{N}\\ F_j=\sqrt{{(241.4)}^2+{(32.3)}^2}\\ F_j=243.5\text{N}\\ \text{tan}\theta \text{=}\frac{F_{jy}}{F_{jx}}\\ \theta ={\tan }^{-1}\left(\frac{32.3}{241.4}\right)\\ \theta =7.62°\end{array}$

Conclusion:

Thus, the magnitude of the force exerted by the shoulder joint on the upper arm is 243.5 N at an angle of 7.62^o.