Chapter 12, Problem 20P

(a)

To determine

The magnitude of the force $\overrightarrow{F}$ applied on the board.

Answer to Problem 20P

  $181\text{N}$

Explanation of Solution

Given:

Length of the board, $L=3.0$ m

Mass of the board, $m=5$ kg

The angle between the board and the horizontal surface, $\theta ={30}^0$

Mass of the block rests on the board, $M=60$ kg

Distance between the block and the hinge = $80$ cm = $0.8$ m

Formula used:

Applying rotational equilibrium condition to the board, that is $\Sigma \overrightarrow{\tau }=0$ about an axis through the hinge,

  $F\left[L\cos \theta \right]-mg\left[\frac{L}{2}\cos \theta \right]-Mg\left[\frac{L}{4}\cos \theta \right]=0$

  $\Rightarrow F=\frac{m\left[\frac{L}{2}\cos \theta \right]+M\left[\frac{L}{4}\cos \theta \right]}{\left[L\cos \theta \right]}g$

  $\Rightarrow F=\frac{m\left(\frac{L}{2}\right)+M\left(\frac{L}{4}\right)}{\left(L\right)}g$  $\left(1\right)$

Calculation:

In figure 1,

  $m\overrightarrow{g}$ is the weight of the board

  ${\overrightarrow{F}}_{hinge}$ is the force exerted by the hinge

  $M\overrightarrow{g}$ is the weight of the block

  $\overrightarrow{F}$ is the force acting vertically on the right end of the board

  

FIGURE: 1

Substituting the numerical values in equation $\left(1\right)$ ,

  $\Rightarrow F\text{=}\frac{\text{5}\text{.0 kg × }\frac{\text{3m}}{\text{2}}\text{ + 60 kg × }\frac{\text{3 m}}{\text{4}}}{\text{3 m}}\left(\text{9}{\text{.8 m/s}}^{\text{2}}\right)$

  $\text{=}\frac{\text{5}\text{.0 kg ×1}\text{.5 m +60 kg × 0}\text{.8 m}}{\text{3 m}}\left(\text{9}{\text{.8 m/s}}^{\text{2}}\right)$

  $=181\text{N}$

Conclusion:

The magnitude of the force $\overrightarrow{F}$ applied on the board is $181\text{N}$ .

(b)

To determine

The force exerted by the hinge.

Answer to Problem 20P

  $\text{456}\text{N}$

Explanation of Solution

Given:

Length of the board, $L=3.0$ m

Mass of the board, $m=5$ kg

The angle between the board and the horizontal surface, $\theta ={30}^0$

Mass of the block rests on the board, $M=60$ kg

Distance between the block and the hinge = $80$ cm = $0.8$ m

Formula used:

Applying $\Sigma F_y=0$ to the board,

  $F_{hinge}-Mg-mg+F=0$

  $\Rightarrow F_{hinge}=Mg+mg-F=\left(M+m\right)g-F$  $\left(2\right)$

Calculation:

Substituting the numerical values in equation $\left(2\right)$ ,

  $\begin{array}{l}{F}_{hinge}=\left(\text{60kg+5kg}\right)\left(\text{9}{\text{.8m/s}}^{\text{2}}\right)\text{-181N}\\ F_{hinge}\text{= 456}\text{N}\end{array}$

Conclusion:

The force exerted by the hinge is $\text{456}\text{N}$ .

(c)

To determine

The magnitude of the force $\overrightarrow{F}$ as well as the force exerted by the hinge, if $\overrightarrow{F}$ is exerted at right angles to the board.

Answer to Problem 20P

The magnitude of the force $\overrightarrow{F}$ is $0.16\text{kN}$

The magnitude of the force exerted by the hinge $F_{hinge}$ is $\text{0}\text{.51}\text{kN}$ .

Explanation of Solution

Given:

Length of the board, $L=3.0$ m

Mass of the board, $m=5$ kg

The angle between the board and the horizontal surface, $\theta ={30}^0$

Mass of the block rests on the board, $M=60$ kg

Distance between the block and the hinge = $80$ cm = $0.8$ m

Formula used:

Applying rotational equilibrium condition to the board, that is $\Sigma \overrightarrow{\tau }=0$ about the hinge,

  $F\left[L\right]-mg\left[\frac{L}{2}\cos \theta \right]-Mg\left[\frac{L}{4}\cos \theta \right]=0$

  $\Rightarrow F\left[L\right]=mg\left[\frac{L}{2}\cos \theta \right]+Mg\left[\frac{L}{4}\cos \theta \right]$

  $\Rightarrow F=\frac{mg\left[\frac{L}{2}\cos \theta \right]+Mg\left[\frac{L}{4}\cos \theta \right]}{L}=\frac{m\frac{L}{2}+M\frac{L}{4}}{L}g\cos \theta$  $\left(3\right)$

Applying $\Sigma F_y=0$ to the board,

  $F_{hinge}\sin \theta -Mg-mg+F\cos {30}^0=0$

  $\Rightarrow F_{hinge}\sin \theta =\left(M+m\right)g-F\cos {30}^0$  $\left(4\right)$

Applying $\Sigma F_x=0$ to the board,

  $F_{hinge}\cos \theta -F\sin {30}^0=0$

  $\Rightarrow F_{hinge}\cos \theta =F\sin {30}^0$  $\left(5\right)$

  $\Rightarrow F_{hinge}=\frac{F\sin {30}^0}{\cos \theta }$  $\left(6\right)$

Dividing equation $\left(4\right)$ by equation $\left(5\right)$ ,

  $\frac{F_{hinge}\sin \theta }{F_{hinge}\cos \theta }=\frac{\left(M+m\right)g-F\cos {30}^0}{F\sin {30}^0}$

  $\Rightarrow \tan \theta =\frac{\left(M+m\right)g+F\cos {30}^0}{F\sin {30}^0}$

  $\Rightarrow \theta ={\tan }^{-1}\left[\frac{\left(M+m\right)g+F\cos {30}^0}{F\sin {30}^0}\right]$  $\left(7\right)$

Calculation:

  

FIGURE: 2

Substituting the numerical values in equation $\left(3\right)$ ,

  $\begin{array}{l}F\text{=}\frac{\text{5 kg×}\frac{\text{3}}{\text{2}}\text{+60 kg ×}\frac{\text{3}}{\text{4}}}{\text{3}}\text{×9}{\text{.8 m/s}}^{\text{2}}{\text{ cos30}}^{\text{0}}\\ F=0.16\text{kN}\end{array}$

Substituting numerical values in equation $\left(7\right)$ ,

  $\theta ={\tan }^{-1}\left[\frac{\left(\text{60+5}\right)\text{kg×9}{\text{.8 m/s}}^{\text{2}}{\text{-157 N×cos 30}}^{\text{0}}}{{\text{157N×sin 30}}^{\text{0}}}\right]$ = ${81.1}^0$

Now substituting the numerical values in equation $\left(6\right)$ ,

  $\begin{array}{l}{F}_{hinge}=\frac{157N\times \sin {30}^0}{\cos {81.1}^0}\\ =507.39\text{N}\\ \text{= 0}\text{.51}\text{kN}\end{array}$

Conclusion:

The magnitude of the force $\overrightarrow{F}$ is $0.16\text{kN}$ .

The magnitude of the force exerted by the hinge $F_{hinge}$ is $\text{0}\text{.51}\text{kN}$ .