Chapter 4, Problem 32P

Two blocks of masses m₁ and m₂ (m₁ > m₂) are placed on a frictionless table in contact with each other. A horizontal force of magnitude F is applied to the block of mass m₁ in Figure P4.62. (a) If P is the magnitude of the contact force between the blocks, draw the free-body diagrams for each block. (b) What is the net force on the system consisting of both blocks? (c) What is the net force acting on m₁? (d) What is the net force acting on m₂? (e) Write the x-component of Newton’s second law for each block. (f) Solve the resulting system of two equations and two unknowns, expressing the acceleration a and contact force P in terms of the masses and force. (g) How would the answers change if the force had been applied to m₂ instead? (Hint: use symmetry; don’t calculate!) Is the contact force larger, smaller, or the same in this case? Why?

Figure P4.62

To determine

The free body diagram of each block.

Answer to Problem 32P

The normal forces $N_1$ and $N_2$ balance the weights of the blocks.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

The free body diagram of $m_1$ is,

The free body diagram of $m_2$ is,

Conclusion:

The normal forces $N_1$ and $N_2$ balance the weights of the blocks.

To determine

The net force on the system.

Answer to Problem 32P

The net force on the system is F.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

The free body diagram of $m_1$ is,

The free body diagram of $m_2$ is,

The net force on mass $m_1$ is,

$F_1=F-P$

The negative sign shows that the contact force is opposite to the horizontal force.

The net force on mass $m_2$ is,

$F_2=P$

The net force on the system is,

$F_{net}=F_1+F_2$

Substitute $F-P$ for $F_1$ and P for $F_2$ in the above expression to get $F_{net}$.

$\begin{array}{c}{F}_{net}=\left(F-P\right)+\left(P\right)\\ =F\end{array}$

Conclusion:

The net force on the system is F.

To determine

The net force on mass $m_1$

Answer to Problem 32P

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

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

The free body diagram of $m_1$ is,

From the diagram, the net force on mass $m_1$ is,

$F_1=F-P$

The negative sign shows that the contact force is opposite to the horizontal force.

Conclusion:

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

To determine

The net force on mass $m_2$.

Answer to Problem 32P

The net force on mass $m_2$ is P.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

The free body diagram of $m_2$ is,

From the diagram, the net force on mass $m_2$ is,

$F_2=P$

Conclusion:

The net force on mass $m_2$ is P.

To determine

The x component of Newton’s second law for $m_1$ and $m_2$.

Answer to Problem 32P

The x component of Newton’s second law for $m_1$ is $F-P=m_{1}a$.

The x component of Newton’s second law for $m_2$ is $P=m_{2}a$.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

The free body diagram of $m_1$ is,

The free body diagram of $m_2$ is,

The net force on mass $m_1$ is,

$F_1=F-P$ (I)

The negative sign shows that the contact force is opposite to the horizontal force.

The net force on mass $m_2$ is,

$F_2=P$ (II)

From Newton’s second law, the force on $m_1$ and $m_2$ is,

$\begin{array}{l}{F}_1=m_{1}a\\ F_2=m_{2}a\end{array}$ (III)

• $m_1$ and $m_2$ are the masses of the blocks.
• a is the acceleration.

From Equations (I), (II) and (III),

$F-P=m_{1}a$

$P=m_{2}a$

Conclusion:

The x component of Newton’s second law for $m_1$ is $F-P=m_{1}a$.

The x component of Newton’s second law for $m_2$ is $P=m_{2}a$.

To determine

The acceleration and contact force.

Answer to Problem 32P

The acceleration is $\frac{F}{m_1+m_2}$.

The contact force is $F\left(\frac{m_2}{m_1+m_2}\right)$.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

The net force on mass $m_1$ is,

$m_{1}a=F-P$ (IV)

The net force on mass $m_2$ is,

$m_{2}a=P$ (V)

Add Equations (IV) and (V).

$a\left(m_1+m_2\right)=F$

On Re-arranging,

$a=\frac{F}{m_1+m_2}$

The acceleration is $\frac{F}{m_1+m_2}$.

The net force on mass $m_2$ is,

$m_{2}a=P$

Substitute $\frac{F}{m_1+m_2}$ for a in the above expression to get P.

$P=F\left(\frac{m_2}{m_1+m_2}\right)$

The contact force is $F\left(\frac{m_2}{m_1+m_2}\right)$.

Conclusion:

The acceleration is $\frac{F}{m_1+m_2}$.

The contact force is $F\left(\frac{m_2}{m_1+m_2}\right)$.

To determine

The contact force when the horizontal force is applied on mass $m_2$.

Answer to Problem 32P

The contact force when the horizontal force is applied on mass $m_2$ is $F\left(\frac{m_1}{m_1+m_2}\right)$.

Explanation of Solution

Given Info: Masses of the blocks are $m_1$ and $m_2$. Contact force is P. The magnitude of horizontal force is F.

Explanation:

From (f), the contact force when the horizontal force is applied on mass $m_1$ is $F\left(\frac{m_2}{m_1+m_2}\right)$.

By Symmetry, the contact force when the horizontal force is applied on mass $m_2$ is obtained by interchanging $m_1$ and $m_2$ in the above expression.

$P=F\left(\frac{m_1}{m_1+m_2}\right)$

The contact force is now larger than the case of (f) because $m_1>m_2$.

Conclusion:

The contact force is $F\left(\frac{m_1}{m_1+m_2}\right)$.