Chapter 4, Problem 147P

(a)

To determine

The minimum horizontal force required to make both the blocks slide together along the floor.

Answer to Problem 147P

The minimum horizontal force required to make both the blocks slide together along the floor is $15.1\text{N}$.

Explanation of Solution

Draw the free body diagram of the system.

The net force in the vertical direction is zero. In the horizontal direction, the force must be equal to the maximum static friction between the bottom block and the floor.

Write the horizontal force equation.

$\begin{array}{c}{F}_x=F_{boy}-f_{s,\max }\\ =F_{boy}-{\mu }_{\text{s}}N\\ =F_{boy}-{\mu }_{\text{s}}\left(m_1+m_2\right)g\\ =0\\ F_{boy}={\mu }_{\text{s}}\left(m_1+m_2\right)g\end{array}$ (I)

Here, $F_x$ is the net force in the horizontal direction, $F_{boy}$ is the force applied by the boy on the block, ${\mu }_{\text{s}}$ is the coefficient of static friction, $m_1$ is the mass of block 1, $m_2$ is the mass of block 2 and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $0.220$ for ${\mu }_{\text{s}}$, $5.00\text{kg}$ for $m_1$, $2.00\text{kg}$ for $m_2$ and $9.80{\text{m/s}}^2$ for $g$ in equation (I) to find $F_{boy}$.

$\begin{array}{c}{F}_{boy}=\left(0.220\right)\left(5.00\text{kg}+2.00\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =15.1\text{N}\end{array}$

Thus, the minimum horizontal force required to make both the blocks slide together along the floor is $15.1\text{N}$.

(b)

To determine

The maximum force with which the boy can push the block without the top block slides of the lower block.

Answer to Problem 147P

The maximum force with which the boy can push the block without the top block slides of the lower block is $34.3\text{N}$.

Explanation of Solution

For the net system, find the expression for acceleration by finding the net force in the x direction.

$\begin{array}{c}{F}_{boy}-{\mu }_{\text{k}}\left(m_1+m_2\right)g=\left(m_1+m_2\right)a\\ a=\frac{F_{boy}}{m_1+m_2}-{\mu }_{\text{k}}g\end{array}$ (II)

Here, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Draw the free body diagram of the top block alone.

Use equation (II) to find the expression for force applied by the boy.

$\begin{array}{c}{F}_{boy}-{\mu }_{\text{s}}{m}_{1}g=m_{1}a\\ =m_1\left(\frac{F_{boy}}{m_1+m_2}-{\mu }_{\text{k}}g\right)\\ F_{boy}=\frac{\left({\mu }_{\text{s}}-{\mu }_{\text{k}}\right)m_{1}g\left(m_1+m_2\right)}{m_2}\end{array}$ (III)

Here, ${\mu }_{\text{s}}$ is the static friction between the blocks.

Conclusion:

Substitute $0.400$ for ${\mu }_{\text{s}}$, $0.200$ for ${\mu }_{\text{k}}$, $5.00\text{kg}$ for $m_1$, $2.00\text{kg}$ for $m_2$ and $9.80{\text{m/s}}^2$ for $g$ in equation (III) to find $F_{boy}$.

$\begin{array}{c}{F}_{boy}=\frac{\left(0.400-0.200\right)\left(5.00\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(5.00\text{kg}+2.00\text{kg}\right)}{2.00\text{kg}}\\ =34.3\text{N}\end{array}$

Thus, the maximum force with which the boy can push the block without the top block slides of the lower block is $34.3\text{N}$.