Chapter 5, Problem 50P

(a)

To determine

The minimum acceleration so that box will not fall to the ground.

Explanation of Solution

Given:

Mass of box is $2.0\text{kg}$ .

Coefficient of friction between the box and the wall is $0.60$ .

Formula used:

Draw free body diagram of the box.

  

Write expression for net force in x -direction.

  ${\displaystyle \sum F_x}=F_n$

Here, $F_n$ is the normal force on the box.

Substitute $m{a}_{\min }$ for $F_n$ in above expression.

  ${\displaystyle \sum F_x}=m{a}_{\min }$

Here, $m$ is mass of the box and $a_{\min }$ is the minimum acceleration.

Write expression for net force in y -direction.

  ${\displaystyle \sum F_y}=0$

Substitute $f_{s,\max }-mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $f_{s,\max }-mg=0$

Here, $f_{s,\max }$ is the maximum friction force by the wall on the box.

Here, $g$ is acceleration due to gravity.

Substitute ${\mu }_s{F}_n$ for $f_{s,\max }$ in above expression.

  ${\mu }_s{F}_n-mg=0$

Here, ${\mu }_s$ is the coefficient of friction between box and wall.

Substitute $m{a}_{\min }$ for $F_n$ in above expression.

  ${\mu }_s\left(m{a}_{\min }\right)-mg=0$

Solve above expression for $a_{\min }$ .

  $a_{\min }=\frac{g}{{\mu }_s}$.......(1)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.60$ for ${\mu }_s$ in equation (1).

  $\begin{array}{c}{a}_{\min }=\frac{9.81\text{m}/{\text{s}}^{\text{2}}}{0.60}\\ =16.35\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the minimum acceleration is $16.35\text{m}/{\text{s}}^{\text{2}}$ .

(b)

To determine

The magnitude of frictional force.

Explanation of Solution

Given:

Mass of box is $2.0\text{kg}$ .

Coefficient of friction between the box and the wall is $0.60$ .

Formula used:

Draw free body diagram of the box.

  

Write expression for net force in y -direction.

  ${\displaystyle \sum F_y}=0$

Substitute $f_{s,\max }-mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $f_{s,\max }-mg=0$

Here, $f_{s,\max }$ is the maximum friction force by the wall on the box.

Rearrange above expression for $f_{s,\max }$ .

  $f_{s,\max }=mg$.......(1)

Calculation:

Substitute $2.0\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ .

  $\begin{array}{c}{f}_{s,\max }=\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =19.62\text{N}\end{array}$

Conclusion:

Thus, the magnitude of frictional force is $19.62\text{N}$ .

(c)

To determine

The force of friction on box if the acceleration of box is twice the minimum acceleration.

Explanation of Solution

Given:

Mass of box is $2.0\text{kg}$ .

Coefficient of friction between the box and the wall is $0.60$ .

Formula used:

Draw free body diagram of the box.

  

Write expression for net force in y -direction.

  ${\displaystyle \sum F_y}=0$

Substitute $f_{s,\max }-mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $f_{s,\max }-mg=0$

Here, $f_{s,\max }$ is the maximum friction force by the wall on the box.

Rearrange above expression for $f_{s,\max }$ .

  $f_{s,\max }=mg$.......(1)

Calculation:

Substitute $2.0\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ .

  $\begin{array}{c}{f}_{s,\max }=\left(2.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =19.62\text{N}\end{array}$

Conclusion:

The friction force is independent of minimum acceleration. Thus, the magnitude of frictional force is $19.62\text{N}$ .

(d)

To determine

To show that the box will not fall if the acceleration is greater or equal to $\frac{g}{{\mu }_s}$ .

Explanation of Solution

Given:

Mass of box is $2.0\text{kg}$ .

Coefficient of friction between the box and the wall is $0.60$ .

Introduction:

Draw free body diagram of the box.

  

Write expression for net force in x -direction.

  ${\displaystyle \sum F_x}=F_n$

Here, $F_n$ is the normal force on the box.

Write expression for net force in y -direction.

  ${\displaystyle \sum F_y}=0$

Substitute $f_{s,\max }-mg$ for ${\displaystyle \sum F_y}$ in above expression.

  $f_{s,\max }-mg=0$

Here, $f_{s,\max }$ is the maximum friction force by the wall on the box.

Here, $g$ is acceleration due to gravity.

Substitute ${\mu }_s{F}_n$ for $f_{s,\max }$ in above expression.

  ${\mu }_s{F}_n-mg=0$

Here, ${\mu }_s$ is the coefficient of friction between box and wall.

Substitute $m{a}_{\min }$ for $F_n$ in above expression.

  ${\mu }_s\left(m{a}_{\min }\right)-mg=0$

Solve above expression for $a_{\min }$ .

  $a_{\min }=\frac{g}{{\mu }_s}$

The minimum acceleration required for the box to not fall in the ground is $\frac{g}{{\mu }_s}$ .

If the acceleration is greater than $\frac{g}{{\mu }_s}$ then also box will not fall.

Conclusion:

Thus, the box will not fall if the acceleration is greater or equal to $\frac{g}{{\mu }_s}$ .