Chapter 2, Problem 118P

(a)

To determine

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

Answer to Problem 118P

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

Explanation of Solution

Draw the free body diagram.

Write the equilibrium condition in x-direction.

    $F_m\cos \theta =F_c\cos \varphi$

Here, $\theta$ is the angle made by ${\overrightarrow{F}}_m$ with horizontal.

Rewrite the expression for $F_c$.

  $F_c=\frac{\cos \theta }{\cos \varphi }{F}_m$        (I)

Write the equilibrium condition in y-direction.

    $F_c\sin \varphi =F_m\sin \theta +W$

Here, $W$ is the weight of head.

Rewrite the expression for $F_c$.

  $F_c=\frac{F_m\sin \theta +W}{\sin \varphi }$        (II)

Rewrite the above equation by substituting equation (I).

    $\begin{array}{c}\frac{\cos \theta }{\cos \varphi }{F}_m=\frac{F_m\sin \theta +W}{\sin \varphi }\\ \frac{\sin \varphi \cos \theta }{\cos \varphi }{F}_m=F_m\sin \theta +W\\ \tan \varphi =F_m\sin \theta +W\\ \tan \varphi =\frac{F_m\sin \theta +W}{F_m\cos \theta }\end{array}$

Simplify and rewrite the equation for $\varphi$.

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

Conclusion:

Substitute 50 N for $W$, $35°$ for $\theta$ and 60 N for $F_m$ in the above equation to find $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\tan 35°+\frac{50\text{N}}{\left(60\text{N}\right)\left(\cos 35°\right)}\right)\\ ={\tan }^{-1}\left(0.70+1.01\right)\\ ={\tan }^{-1}\left(1.71\right)\\ =59.7°\end{array}$

Substitute 50 N for $W$, $35°$ for $\theta$, $59.7°$ for $\varphi$ and 60 N for $F_m$ in the above equation to find $\varphi$.

    $\begin{array}{c}{F}_c=\frac{\left(60\text{N}\right)\left(\sin 35°\right)+50\text{ N}}{\sin 59.7°}\\ =\frac{84.4\text{N}}{0.86}\\ =98\text{N}\end{array}$

Thus, the magnitude of force ${\overrightarrow{F}}_c$ is 98 N.

(a)

To determine

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

Answer to Problem 118P

${\overrightarrow{F}}_c$ is directed at $59.7°$ above the horizontal.

Explanation of Solution

Substitute 50 N for $W$, $35°$ for $\theta$ and 60 N for $F_m$ in equation (III) to find $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\tan 35°+\frac{50\text{N}}{\left(60\text{N}\right)\left(\cos 35°\right)}\right)\\ ={\tan }^{-1}\left(0.70+1.01\right)\\ ={\tan }^{-1}\left(1.71\right)\\ =59.7°\end{array}$

Conclusion:

Thus, ${\overrightarrow{F}}_c$ is directed at $59.7°$ above the horizontal.