Chapter 9, Problem 33P

To determine

The magnitude of $F_v$ (the vertebra).

Answer to Problem 33P

Solution: $F=2.4w$

Explanation of Solution

Explanation: Given: Weight of the head, $w_H=0.07w$

Weight of the two arms, $w_A=0.12w$

Weight of the trunk, $w_T=0.46w$

Formula Used:

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

  ${\displaystyle \sum {\tau }_A}=0$

  ${{\displaystyle \sum F}}_{\gamma }=0$

  $\theta ={\tan }^{-1}(F_{v\gamma }/F_{vx})$

Calculation:

From the force diagram taking torques about point A

  $\begin{array}{l}{\displaystyle \sum {\tau }_A}=0\\ (0.72W_H+0.48W_A+0.36W_r)\cos 45-0.48F_M\sin B=0\end{array}$

  $(0.72\times 0.07w+0.48\times 0.12w+0.36\times 0.46w)\cos 45°-0.48\sin 12°F_M=0$

  $\begin{array}{l}{F}_M=\frac{\left[\left(0.72\right)(0.07)+\left(01.48\times 0.12\right)+\left(0.36\right)\left(0.46\right)\right]w\cos 45°}{0.48\times \sin 12°}\\ F_M=1.94w\end{array}$

  $\begin{array}{l}{{\displaystyle \sum F}}_{\gamma }=0\\ F_{v\gamma }-F_M\sin 18-W_H-W_A-W_{\gamma }=0\\ F_{v\gamma }=1.94w\sin (45°-12°)+0.07w+0.12w+0.46w\\ F_{v\gamma }=1.71w\end{array}$

Taking x components ${\displaystyle \sum F_x}=0$

  $\begin{array}{l}{F}_{vx}-F_M\cos {33}^{\text{o}}=0\\ F_{vx}=1.94w\cos 33°\end{array}$

  $\begin{array}{c}{F}_v=1.63w\widehat{i}+1.71w\widehat{j}\\ \left|F_v\right|=\sqrt{{\left(1.63w\right)}^2+{\left(1.71w\right)}^2}\\ =2.36w\end{array}$

The inclination $\theta ={\tan }^{-1}(F_{v\gamma }/F_{vx})$

  $\begin{array}{l}={\tan }^{-1}\left(\frac{1.71}{1.63}\right)\\ \theta ={46.4}^{\text{o}}\end{array}$

Conclusion: Therefore,the force on vertebra is 2.4 w.