Chapter 5, Problem 39AP

Two blocks of masses m₁ and m₂, are placed on a table in contact with each other as discussed in Example 5.7 and shown in Figure 5.13a. The coefficient of kinetic friction between the block of mass m₁ and the table is μ₁, and that between the block of mass m₂ and the table is μ₂. A horizontal force of magnitude F is applied to the block of mass m₁. We wish to find P, the magnitude of the contact force between the blocks. (a) Draw diagrams showing the forces for each block. (b) What is the net force on the system of two blocks? (c) What is the net force acting on m₁? (d) What is the net force acting on m₂? (e) Write Newton’s second law in the x direction for each block. (f) Solve the two equations in two unknowns for the acceleration a of the blocks in terms of the masses, the applied force F, the coefficients of friction, and g. (g) Find the magnitude P of the contact force between the blocks in terms of the same quantities.

To determine

To draw: The free body diagram of each block to show the forces.

Introduction: The free body diagram of an object represents the direction and magnitude of forces acting on the body.

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

The free body diagram of the book is given below.

Figure (1)

The sum of all vertical forces is zero because the block moves on a horizontal surface. So the vertical acceleration $a_{\text{y}}=0$.

The Newton’s law second law is given as,

$\begin{array}{c}{\displaystyle \sum F_{\text{1y}}}=m_1{a}_{\text{y}}\\ -m_{1}g+n_1=0\\ n_1=m_{1}g\end{array}$

• $F_{\text{1y}}$ is the vertical force of block $1$.
• $n_1$ is the normal force of block $1$.
• $g$ is the acceleration due to gravity.

The Newton’s second law is,

$\begin{array}{c}{\displaystyle \sum F_{\text{2y}}}=m_2{a}_{\text{y}}\\ -m_{2}g+n_2=0\\ n_2=m_{2}g\end{array}$

• $F_{\text{2y}}$ is the vertical force of block $2$.
• $n_2$ is the normal force of block $2$.

The equation for kinetic friction for block $1$ is,

$\begin{array}{c}{f}_1={\mu }_1{n}_1\\ ={\mu }_1{m}_{1}g\end{array}$

The equation for kinetic friction for block $2$ is,

$\begin{array}{c}{f}_2={\mu }_2{n}_2\\ ={\mu }_2{m}_{2}g\end{array}$

Conclusion:

Therefore, the free body diagram of each block to show the forces is given in figure I.

To determine

The net force on the system of two blocks.

Answer to Problem 39AP

The net force on the system of two blocks is $F$.

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

Assume the friction force is too small.

From the figure I, The net force on the system of two blocks is equal to the magnitude of the force $F$.

Conclusion:

Therefore, the net force on the system of two blocks is $F$.

To determine

The net force acting on $m_1$.

Answer to Problem 39AP

The net force acting on $m_1$ is $F-P$.

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

Each block exerts an equal and opposite force on the other, so those forces have the same magnitude.

$\begin{array}{c}{P}_{12}=P_{21}\\ =P\end{array}$

From the figure I, the net force net force acting on $m_1$ is equal to the $F-P$.

Conclusion:

Therefore, the net force acting on $m_1$ is $F-P$.

To determine

The net force acting on $m_2$.

Answer to Problem 39AP

The net force acting on $m_2$ is $P$.

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

From the above figure the magnitude of the forces is $P$. So the net force net force acting on $m_2$ is equal to the $P$.

Conclusion:

Therefore, the net force acting on $m_2$ is $P$.

To determine

To write: The Newton’s second law in the $x$ direction for each block.

Answer to Problem 39AP

The Newton’s second law in the $x$ direction for $m_1$ is $F-P=m_{1}a$ and the Newton’s second law in the $x$ direction for $m_2$ is $P=m_{2}a$.

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

The block has on a horizontal acceleration $a_{\text{x}}=a$.

The Newton’s second law for block $1$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{1x}}}=m_1{a}_{\text{x}}\\ F-P=m_{1}a\end{array}$

The Newton’s second law for block $2$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{2x}}}=m_2{a}_{\text{x}}\\ P=m_{2}a\end{array}$

Conclusion:

Therefore, the Newton’s second law in the $x$ direction for $m_1$ is $F-P=m_{1}a$ and the Newton’s second law in the $x$ direction for $m_2$ is $P=m_{2}a$.

To determine

The solution of the two equations for the acceleration of the blocks.

Answer to Problem 39AP

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

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

Assume the friction force.

The Newton’s second law is for block $1$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{1x}}}=m_1{a}_{\text{x}}\\ F-P-f_1=m_{1}a\\ F-P-{\mu }_1{m}_{1}g=m_{1}a\end{array}$ (I)

The Newton’s second law is for block $2$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{2x}}}=m_2{a}_{\text{x}}\\ P-f_2=m_{2}a\\ P-{\mu }_2{m}_{2}g=m_{2}a\end{array}$ (II)

Add the equation (I) and equation (II) and solve for $a$.

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

Conclusion:

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

To determine

The magnitude $P$ of the contact force between the blocks.

Answer to Problem 39AP

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

Explanation of Solution

Given information: The mass of block $1$ is $m_1$, the mass of block $2$ is $m_2$, the coefficient of kinetic friction between $m_1$ and the table is ${\mu }_1$, the coefficient of kinetic friction between $m_2$ and the table is ${\mu }_2$ and the magnitude of horizontal force is $F$.

Recall the equation (II).

$\begin{array}{c}P-{\mu }_2{m}_{2}g=m_{2}a\\ P={\mu }_2{m}_{2}g+m_{2}a\end{array}$

Substitute $\left(\frac{F-{\mu }_1{m}_{1}g-{\mu }_2{m}_{2}g}{m_1+m_2}\right)$ for $a$ in above expression.

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

Conclusion:

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