Chapter 5, Problem 60P

(a)

To determine

The maximum force which is applied so that there is no relative motion between block and bracket.

Explanation of Solution

Given:

Mass of the block is $10.0\text{kg}$ .

Mass of the bracket is $5.0\text{kg}$ .

Value of the static friction is $0.40$ .

Value of the kinetic friction is $0.30$ .

Formula used:

Draw free body diagram of block and bracket.

  

Write expression for net force in x -direction on block.

  ${\displaystyle \sum F_x}=0$

Substitute $f_s-F-m_1{a}_1$ for ${\displaystyle \sum F_x}$ in above expression.

  $f_s-F-m_1{a}_1=0$

Here, $f_s$ is the force due to static friction, $m_1$ is the mass of block and $a_1$ is the acceleration of the block.

Substitute ${\mu }_s{m}_{1}g$ for $f_s$ in above expression and solve for $a_1$ .

  ${\mu }_s{m}_{1}g-F-m_1{a}_1=0$

Here, ${\mu }_s$ is the coefficient static friction and $g$ is the acceleration due to gravity.

Rearrange above expression for $F$ .

  $F={\mu }_s{m}_{1}g-m_1{a}_1$.......(1)

Write expression for net force in x -direction on block.

  ${\displaystyle \sum F_x}=0$

Substitute $2F-{\mu }_s{m}_{1}g-m_2{a}_2$ for ${\displaystyle \sum F_x}$ in above expression.

  $2F-{\mu }_s{m}_{1}g-m_2{a}_2=0$

Here, $m_2$ is the mass of bracket and $a_2$ is the acceleration of bracket.

Substitute $a_1$ for $a_2$ and ${\mu }_s{m}_{1}g-m_1{a}_1$ for $F$ in above expression and solve for $a_1$ .

  $a_1=\frac{{\mu }_s{m}_{1}g}{m_2+2m_1}$.......(2)

Calculation:

Substitute $0.40$ for ${\mu }_s$ , $10\text{kg}$ for $m_1$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $5.0\text{kg}$ for $m_2$ in equation (2).

  $\begin{array}{c}{a}_1=\frac{\left(0.40\right)\left(10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{5\text{kg}+2\left(10\text{kg}\right)}\\ =1.56\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $0.40$ for ${\mu }_s$ , $10\text{kg}$ for $m_1$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.56\text{m}/{\text{s}}^{\text{2}}$ for $a_1$ .

  $\begin{array}{c}F=\left(0.40\right)\left(10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)-\left(10\text{kg}\right)\left(1.56\text{m}/{\text{s}}^{\text{2}}\right)\\ =23.64\text{N}\end{array}$

Conclusion:

Thus, the maximum force is $23.64\text{N}$ .

(a)

To determine

The maximum force which is applied so that there is no relative motion between block and bracket.

Explanation of Solution

Given:

Mass of the block is $10.0\text{kg}$ .

Mass of the bracket is $5.0\text{kg}$ .

Value of the static friction is $0.40$ .

Value of the kinetic friction is $0.30$ .

Formula used:

Draw free body diagram of block and bracket.

  

Write expression for net force in x -direction on block.

  ${\displaystyle \sum F_x}=0$

Substitute $f_s-F-m_1{a}_1$ for ${\displaystyle \sum F_x}$ in above expression.

  $f_s-F-m_1{a}_1=0$

Substitute ${\mu }_s{m}_{1}g$ for $f_s$ in above expression and solve for $a_1$ .

  ${\mu }_s{m}_{1}g-F-m_1{a}_1=0$

Rearrange above expression for $F$ .

  $F={\mu }_s{m}_{1}g-m_1{a}_1$

Write expression for net force in x -direction on block.

  ${\displaystyle \sum F_x}=0$

Substitute $2F-{\mu }_s{m}_{1}g-m_2{a}_2$ for ${\displaystyle \sum F_x}$ in above expression.

  $2F-{\mu }_s{m}_{1}g-m_2{a}_2=0$

Substitute $a_1$ for $a_2$ and ${\mu }_s{m}_{1}g-m_1{a}_1$ for $F$ in above expression and solve for $a_1$ .

  $a_1=\frac{{\mu }_s{m}_{1}g}{m_2+2m_1}$.......(1)

Calculation:

Substitute $0.40$ for ${\mu }_s$ , $10\text{kg}$ for $m_1$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $5.0\text{kg}$ for $m_2$ in equation (1).

  $\begin{array}{c}{a}_1=\frac{\left(0.40\right)\left(10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{5\text{kg}+2\left(10\text{kg}\right)}\\ =1.56\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

The acceleration of the block is equal to the acceleration of the bracket. Thus, the acceleration of the bracket is $1.56\text{m}/{\text{s}}^{\text{2}}$ .