Chapter 2, Problem 122P

(a)

To determine

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

Answer to Problem 122P

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

Explanation of Solution

Write the expression for normal force acting on system of two blocks.

    $N=\left(m_1+m_2\right)g$        (I)

Here, $N$ is the normal force, $m_1$ is the mass of block 1, and $m_2$ is the mass of block 2, and $g$ is the acceleration due to gravity.

The minimum force required to slide both the blocks together along the floor is equal to the maximum static frictional force.

Write the expression for maximum static frictional force.

    $f_{\text{stat}}={\mu }_{\text{s}}N$        (II)

Here, ${\mu }_{\text{s}}$ is the coefficient of static friction, and $f_{\text{stat}}$ is the maximum static frictional force.

Use equation (I) in equation (II).

    $f_{\text{stat}}={\mu }_{\text{s}}\left(m_1+m_2\right)g$        (III)

Conclusion:

Substitute $5\text{kg}$ for $m_1$, $2\text{kg}$ for $m_2$, $0.22$ for ${\mu }_{\text{s}}$, and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (III), to find $f_{\text{stat}}$.

    $\begin{array}{c}{f}_{\text{stat}}=\left(0.22\right)\left(5\text{kg}+2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =15.1\text{N}\end{array}$

Therefore, the minimum force required to slide both the blocks together along the floor is $\underset{\_}{15.1\text{N}}$.

(b)

To determine

The maximum force required to push the top block that starts to slide off the lower block.

Answer to Problem 122P

The maximum force required to push the top block that starts to slide off the lower block is $\underset{\_}{33.35\text{N}}$.

Explanation of Solution

Write the expression for frictional force acting between the blocks.

    $f={\mu }_{\text{s}}{m}_{1}g$        (IV)

Here, $f$ is the frictional force acting between the blocks, ${\mu }_{\text{s}}$ is the static frictional force, $m_1$ is the mass of the upper block, and $g$ is the acceleration due to gravity.

Write the expression for net force acting on the system.

    $F={\mu }_{\text{k}}\left(m_1+m_2\right)g$        (V)

Here, $F$ is the net force acting on the system, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Write the expression for maximum force required to push the top block that starts to slide off the lower block.

    $F_{\text{max}}=F+f$        (VI)

Here, $F_{\text{max}}$ is the maximum force required to push the top block that starts to slide off the lower block.

Use equation (IV) and (V) in equation (VI).

    $F_{\text{max}}={\mu }_{\text{k}}\left(m_1+m_2\right)g+{\mu }_{\text{s}}{m}_{1}g$        (VII)

Conclusion:

Substitute $0.200$ for ${\mu }_{\text{k}}$, $5\text{kg}$ for $m_1$, $2\text{kg}$ for $m_2$, $9.81\text{m}/{\text{s}}^2$ for $g$, $0.4$ for ${\mu }_{\text{s}}$ in equation (VII), to find $F_{\text{max}}$.

    $\begin{array}{c}{F}_{\text{max}}=\left(0.200\right)\left(5\text{kg}+2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)+\left(0.4\right)\left(5\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =33.35\text{N}\end{array}$

Therefore, the maximum force required to push the top block that starts to slide off the lower block is $\underset{\_}{33.35\text{N}}$.