Chapter 5, Problem 63P

(a)

To determine

To Calculate: The maximum and minimum values of applied force for which the block does not slip.

Answer to Problem 63P

  $\begin{array}{l}{F}_{\max }=84\text{N}\\ F_{\min }=1.6\text{N}\end{array}$

Explanation of Solution

Given data:

Mass of the block, $m=0.50\text{kg}$

Mass of the wedge, $m\text{'}=2.0\text{kg}$

Coefficient of static friction, ${\mu }_s=0.80$

Angle of wedge, $\theta =35°$

Formula Used:

Newton’s second law of motion:

  ${\displaystyle \sum F=ma}$

Where, m is the mass and a is the acceleration.

Calculation:

Free-body diagram:

  

Maximum force, $F_{\max }=m_t{a}_{\max }$

Minimum force, $F_{\min }=m_t{a}_{\min }$

Where, $m_t$ is the total mass of the block and the wedge.

For the block,

  $\begin{array}{l}{\displaystyle \sum F_x=N\sin \theta -f\cos \theta =m{a}_x}\mathrm{...}\left(1\right)\\ {\displaystyle \sum F_y=N\cos \theta +f\sin \theta -mg=0}\mathrm{...}\left(2\right)\\ f_{\max }={\mu }_{s}N\\ N\cos \theta +{\mu }_{s}N\sin \theta -mg=0\\ N=\frac{mg}{\cos \theta +{\mu }_s\sin \theta }\mathrm{...}\left(3\right)\end{array}$

Substitute (3) in (1) to find the minimum acceleration:

  $\begin{array}{l}\frac{mg}{\cos \theta +{\mu }_s\sin \theta }\sin \theta -{\mu }_s\frac{mg}{\cos \theta +{\mu }_s\sin \theta }\cos \theta =m{a}_{\min }\\ a_{\min }=\frac{g\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }\\ F_{\min }=\frac{m_{t}g\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }\end{array}$

Substitute the values and solve:

  $\begin{array}{l}{F}_{\min }=\left|\frac{\left(2.0+0.5\right)\left(9.8\right)\left(\sin 35°-0.80\cos 35°\right)}{\cos 35°+0.80\sin 35°}\right|\\ F_{\min }=1.6\text{N}\end{array}$

For maximum force, reverse the direction of f

  $\begin{array}{l}{a}_{\max }=\frac{g\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }\\ F_{\max }=\frac{m_{t}g\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }\end{array}$

Substitute the values and solve:

  $\begin{array}{l}{F}_{\max }=\left|\frac{\left(2.0+0.5\right)\left(9.8\right)\left(\sin 35°+0.80\cos 35°\right)}{\cos 35°-0.80\sin 35°}\right|\\ F_{\max }=84\text{N}\end{array}$

Conclusion:

The maximum and minimum values of applied force for which the block does not slip are 84 N and 1.6 N respectively.

(b)

To determine

To Calculate: The maximum and minimum values of applied force for which the block does not slip.

Answer to Problem 63P

  $\begin{array}{l}{F}_{\max }=37.5\text{N}\\ F_{\min }=5.8\text{N}\end{array}$

Explanation of Solution

Given data:

Mass of the block, $m=0.50\text{kg}$

Mass of the wedge, $m\text{'}=2.0\text{kg}$

Coefficient of static friction, ${\mu }_s=0.40$

Angle of wedge, $\theta =35°$

Formula Used:

From previous part:

  $F_{\min }=\frac{m_{t}g\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }$

  $F_{\max }=\frac{m_{t}g\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }$

Calculation:

Substitute the values and solve for minimum force:

  $F_{\min }=\frac{m_{t}g\left(\sin \theta -{\mu }_s\cos \theta \right)}{\cos \theta +{\mu }_s\sin \theta }$

  $\begin{array}{l}{F}_{\min }=\left|\frac{\left(2.0+0.5\right)\left(9.8\right)\left(\sin 35°-0.40\cos 35°\right)}{\cos 35°+0.40\sin 35°}\right|\\ F_{\min }=5.8\text{N}\end{array}$

For maximum force, reverse the direction of f

  $F_{\max }=\frac{m_{t}g\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }$

Substitute the values and solve:

  $\begin{array}{l}{F}_{\max }=\left|\frac{\left(2.0+0.5\right)\left(9.8\right)\left(\sin 35°+0.40\cos 35°\right)}{\cos 35°-0.40\sin 35°}\right|\\ F_{\max }=37.5\text{N}\end{array}$

Conclusion:

The maximum and minimum values of applied force for which the block does not slip are 37.5 N and 5.8 N respectively.