Chapter 4, Problem 154P

(a)

To determine

The magnitude of ${\overrightarrow{F}}_c$ .

Answer to Problem 154P

The magnitude of ${\overrightarrow{F}}_c$ is $98\text{ N}$ .

Explanation of Solution

Take $+y$ direction to be upward and $+x$ direction to be the right.

Write the equation for the net force in $x$ direction.

$\Sigma F_x=F_m\cos \theta -F_c\cos \varphi$

Here, $\Sigma Fx$ is the magnitude of the sum of the forces in $x$ direction, $F_m$ is the magnitude of the muscle force, $\theta$ is the angle ${\overrightarrow{F}}_m$ makes with the horizontal, $F_C$ is the contact force and $\varphi$ is the angle ${\overrightarrow{F}}_c$ makes with the horizontal

The sum of the forces in $x$ direction must be equal to zero. Equate the above equation to zero and rewrite it for $F_C$ .

$\begin{array}{c}{F}_m\cos \theta -F_c\cos \varphi =0\\ F_c=F_m\frac{\cos \theta }{\cos \varphi }\end{array}$ (I)

Write the equation for the net force in $y$ direction.

$\Sigma F_y=-F_m\sin \theta +F_c\sin \varphi -W$

Here, $W$ is the weight of the head

The sum of the forces in $y$ direction must be equal to zero. Equate the above equation to zero and rewrite it for $F_C$ .

$\begin{array}{c}-F_m\sin \theta +F_c\sin \varphi -W=0\\ F_c=\frac{W+F_m\sin \theta }{\sin \varphi }\end{array}$ (II)

Equate equations (I) and (II) and solve for $\varphi$ .

$\begin{array}{c}{F}_m\frac{\cos \theta }{\cos \varphi }=\frac{W+F_m\sin \theta }{\sin \varphi }\\ \frac{\sin \varphi }{\cos \varphi }=\frac{W+F_m\sin \theta }{F_m\cos \theta }\\ \tan \varphi =\frac{W}{F_m\cos \theta }+\tan \theta \\ \varphi ={\tan }^{-1}\left(\frac{W}{F_m\cos \theta }+\tan \theta \right)\end{array}$ (III)

Conclusion:

Given that the weight of the head is $50.0\text{ N}$ , value of $F_m$ is $60.0\text{ N}$ and that of $\theta$ is $35°$ .

Substitute $50.0\text{ N}$ for $W$ , $60.0\text{ N}$ for $F_m$ and $35°$ for $\theta$ in equation (III) to find $\varphi$ .

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{50.0\text{ N}}{\left(60.0\text{ N}\right)\cos 35°}+\tan 35°\right)\\ =60°\end{array}$

Substitute $60.0\text{ N}$ for $F_m$ , $35°$ for $\theta$ and $60°$ for $\varphi$ in equation (I) to find $F_c$ .

$\begin{array}{c}{F}_c=\left(60.0\text{ N}\right)\frac{\cos 35°}{\cos 60°}\\ =98\text{ N}\end{array}$

Therefore, the magnitude of ${\overrightarrow{F}}_c$ is $98\text{ N}$ .

(b)

To determine

The direction of ${\overrightarrow{F}}_c$ .

Answer to Problem 154P

The force ${\overrightarrow{F}}_c$ is $60°\text{ above the horizontal}$ .

Explanation of Solution

The force ${\overrightarrow{F}}_c$ is at an angle of $\varphi$ above the horizontal.

Conclusion:

Substitute $50.0\text{ N}$ for $W$ , $60.0\text{ N}$ for $F_m$ and $35°$ for $\theta$ in equation (III) to find $\varphi$ .

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{50.0\text{ N}}{\left(60.0\text{ N}\right)\cos 35°}+\tan 35°\right)\\ =60°\end{array}$

Therefore, the force ${\overrightarrow{F}}_c$ is $60°\text{ above the horizontal}$ .