Chapter 5, Problem 53P

(a)

To determine

Draw free body diagram of each block.

Answer to Problem 53P

The free body diagram of mass $m_1$ is shown in Figure 1.

The free body diagram of mass $m_2$ is shown in Figure 2.

Explanation of Solution

The free body diagram is the graphical illustration used to visualize the movements and forces applied on a body.

Let $n$ be the normal force, $f$ is the frictional force, $mg$ is the force due to gravitation, $P$ is the magnitude of contact force between the blocks, and $F$ is the applied horizontal force.

The free body diagram of mass $m_1$ is shown in Figure 1.

The free body diagram of mass $m_2$ is shown in Figure 2.

Conclusion:

Therefore, the free body diagram of mass $m_1$ is shown in Figure 1, and the free body diagram of mass $m_1$ is shown in Figure 2.

(b)

To determine

The net force on the system of two blocks.

Answer to Problem 53P

The net force on the system of two blocks is $\underset{\_}{F-f_1-f_2}$.

Explanation of Solution

From the Figure 1, write the expression for net force action on the mass $m_1$ in the horizontal direction.

    ${\displaystyle \sum F_x}=F-f_1-P_{21}$        (I)

Here, $f_1$ is the frictional force acting the mass $m_1$, and $P_{12}$ is the contact force on mass $m_1$ due to $m_2$.

From the Figure 2, write the expression for net force action on the mass $m_2$ in the horizontal direction.

    ${\displaystyle \sum F_x}=P_{12}-f_2$        (II)

Here, $f_2$ is the frictional force acting the mass $m_2$, and $P_{21}$ is the contact force on mass $m_2$ due to $m_1$.

So the net force on the system is the sum of net force on the each block.

    $\begin{array}{c}{\displaystyle \sum F_x}=F-f_1-P_{21}+P_{12}-f_2\\ =F-f_1+f_2\end{array}$

Since contact force acting on each mass is equal.

The net force on the system of block is equal to the magnitude of force $F$ minus the frictional force on each block. Each block have the same acceleration.

Conclusion:

Therefore, the net force on the system of two blocks is $\underset{\_}{F-f_1-f_2}$.

(c)

To determine

The net force on the mass $m_1$.

Answer to Problem 53P

The net force on the mass $m_1$ is $\underset{\_}{F-f_1-P_{21}}$.

Explanation of Solution

From the Figure 1, write the expression for net force action on the mass $m_1$ in the horizontal direction.

    ${\displaystyle \sum F_x}=F-f_1-P_{21}$

The net force on the mass $m_1$ is equal to the force $F$ minus the frictional force on $m_1$ and the force $P_{21}$.

Conclusion:

Therefore, the net force on the mass $m_1$ is $\underset{\_}{F-f_1-P_{21}}$.

(d)

To determine

The net force acting on the $m_2$.

Answer to Problem 53P

The net force acting on the $m_2$ is $\underset{\_}{P_{12}-f_2}$.

Explanation of Solution

From the Figure 2, write the expression for net force action on the mass $m_2$ in the horizontal direction.

    ${\displaystyle \sum F_x}=P_{12}-f_2$

The net force acting on the $m_2$ is equal to the force $P_{12}$ minus the frictional force action on the $m_2$.

Conclusion:

Therefore, the net force acting on the $m_2$ is $\underset{\_}{P_{12}-f_2}$.

(e)

To determine

Newton’s second law in the $x$ direction of each block.

Answer to Problem 53P

Newton’s second law in the $x$ direction of block $1$ is $\underset{\_}{F-P-{\mu }_1{m}_{1}g=m_{1}a}$, and for the block $m_2$ is $\underset{\_}{P-{\mu }_2{m}_{2}g=m_{2}a}$.

Explanation of Solution

The blocks are pushed to the right, the acceleration on each block is same and the block exerts equal and opposite forces on each other, so these forces have the same magnitude.

Write the expression for Newton’s second law in the $x$ direction.

    ${\displaystyle \sum F_x}=m_{1}a$        (III)

Here, $a$ is the acceleration.

Write the expression for frictional force.

    $f=\mu mg$        (IV)

Here, $\mu$ is the coefficient of friction.

Use equation (III) and (IV) in (I).

    $F-{\mu }_1{m}_{1}g-P=m_{1}a$        (V)

Use equation (III) and (IV) in (II).

    $P-{\mu }_2{m}_{2}g=m_{2}a$        (VI)

Conclusion:

Therefore, Newton’s second law in the $x$ direction of block $1$ is $\underset{\_}{F-P-{\mu }_1{m}_{1}g=m_{1}a}$, and for the block $m_2$ is $\underset{\_}{P-{\mu }_2{m}_{2}g=m_{2}a}$.

(f)

To determine

Acceleration of the blocks.

Answer to Problem 53P

Acceleration of the blocks is $\underset{\_}{a=\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}}$.

Explanation of Solution

Add equation (V), and (VI), and solve for $a$.

    $\begin{array}{c}F-{\mu }_1{m}_{1}g-P+P-{\mu }_2{m}_{2}g=m_{1}a+m_{2}a\\ F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g=a\left(m_1+m_2\right)\\ a=\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}\end{array}$        (VII)

Conclusion:

Therefore, the acceleration of the blocks is $\underset{\_}{a=\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}}$.

(g)

To determine

The magnitude of contact force $P$.

Answer to Problem 53P

The magnitude of contact force $P$ is $\underset{\_}{\left(\frac{m_2}{m_1+m_2}\right)\left[F+\left({\mu }_2-{\mu }_1\right)m_{1}g\right]}$.

Explanation of Solution

Solve equation (IV) for $P$.

    $P={\mu }_2{m}_{2}g+m_{2}a$        (VIII)

Use equation (VII) in (VIII).

    $\begin{array}{c}P={\mu }_2{m}_{2}g+m_2\left(\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}\right)\\ =m_2\left[{\mu }_{2}g+\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}\right]\\ =m_2\left[\frac{\left(m_1+m_2\right){\mu }_{2}g+F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{\left(m_1+m_2\right)}\right]\end{array}$        (IX)

Simplify the equation (IX).

    $\begin{array}{c}P=\frac{m_2}{\left(m_1+m_2\right)}\left[m_1{\mu }_{2}g+m_2{\mu }_{2}g+F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g\right]\\ =\frac{m_2}{\left(m_1+m_2\right)}\left[m_1{\mu }_{2}g+F-{\mu }_1{m}_{1}g\right]\\ =\frac{m_2}{\left(m_1+m_2\right)}\left[F+\left({\mu }_2-{\mu }_1\right)m_{1}g\right]\end{array}$

Conclusion:

Therefore, the magnitude of contact force $P$ is $\underset{\_}{\left(\frac{m_2}{m_1+m_2}\right)\left[F+\left({\mu }_2-{\mu }_1\right)m_{1}g\right]}$.