Chapter 4, Problem 5SP

Two blocks tied together by a horizontal string are being pulled across the table by a horizontal force of 46 N, as shown in the diagram below. The 3-kg block has a 6-N frictional force exerted on it by the table, and the 7-kg block has an 8-N frictional force acting on it.

1. a. What is the net force acting on the entire two-block system?
2. b. What is the acceleration of this system?
3. c. What force is exerted on the 3-kg block by the connecting string? (Consider only the forces acting on this block. Its acceleration is the same as that of the entire system.)
4. d. Find the net force acting on the 7-kg block and calculate its acceleration. How does this value compare to that found in part b?

To determine

The net force acting on the entire two-block system.

Answer to Problem 5SP

The net force acting on the entire two-block system is $32\text{N}$.

Explanation of Solution

Given info: The horizontal force acting on $7\text{kg}$ mass along the right direction is $46\text{N}$, that along the left direction is $8\text{N}$ and horizontal force acting on $3\text{kg}$ mass along left direction is $6\text{N}$.

Write the expression for the net horizontal force.

$F_{\text{net}}=F_{\text{right}}-F_{\text{left}}$

Here,

$F_{\text{net }}$ is the net force acting on the system

$F_{\text{right}}$ is the horizontal force acting in the right direction

$F_{\text{left}}$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6\text{N}$ and $8\text{N}$.

That is,

$\begin{array}{c}{F}_{\text{left}}=6\text{N+8}\text{N}\\ \text{=14}\text{N}\end{array}$

Substitute $\text{14}\text{N}$ for $F_{\text{left}}$ and $46\text{N}$ for $F_{\text{right}}$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=46\text{N}-\text{14}\text{N}\\ =32\text{N}\end{array}$

Conclusion:

Thus, the net force acting on the entire two-block system is $32\text{N}$.

To determine

The acceleration of the system.

Answer to Problem 5SP

The acceleration of the system is $3.2\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The masses of the blocks are $3\text{kg}$ and $7\text{kg}$.

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=\frac{F_{\text{net}}}{m}$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $3\text{kg}\text{+7}\text{kg}\text{=10}\text{kg}$.

Substitute $32\text{N}$ for $F_{\text{net}}$ and $10\text{kg}$ for $m$ in the above equation to get $a$.

$\begin{array}{c}a=\frac{32\text{N}}{10\text{kg}}\\ =3.2\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the acceleration of the system is $3.2\text{m}/{\text{s}}^{\text{2}}$.

To determine

The force acting on the $3\text{kg}$ mass by the connecting string.

Answer to Problem 5SP

The force acting on the $3\text{kg}$ mass by the connecting string is $15.6\text{N}$.

Explanation of Solution

Given info: The horizontal force acting on $3\text{kg}$ mass along the left direction is $6\text{N}$.

Let $m_1$ be the mass of $3\text{kg }$ block and $m_2$ be the mass of $7\text{kg}$ block.

Write the expression for the net force on $3\text{kg}$ mass.

$F_{\text{net1}}=m_{1}a$

Here,

$m_1$ is the mass of $3\text{kg}$ block

$F_{\text{net1}}$ is the net horizontal force on the mass $3\text{kg}$

Substitute $3\text{kg}$ for $m_1$ and $3.2\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to get $F_{\text{net1}}$.

${}_{\begin{array}{c}{F}_{\text{net1}}=\left(3\text{kg}\right)\left(3.2\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.6\text{N}\end{array}}$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $3\text{kg}$ mass.

$F_{\text{net1}}=F_{\text{tension}}-F_{\text{left}}$

Here,

$F_{\text{left}}$ is the horizontal force on the left direction of mass $3\text{kg}$

$F_{\text{tension}}$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $F_{\text{left}}$ is along the left direction.

Substitute $6\text{N}$ for $F_{\text{left}}$ and $9.4\text{N}$ for $F_{\text{net1}}$ in the above equation to get $F_{\text{tension}}$.

$\begin{array}{c}9.6\text{N}=F_{\text{tension}}-6\text{N}\\ F_{\text{tension}}=15.6\text{N}\end{array}$

Therefore, the force acting on the $3\text{kg}$ by the string is equal to $15.6\text{N}$.

Conclusion:

Thus, the force acting on the $3\text{kg}$ mass by the connecting string is $15.6\text{N}$.

To determine

The net force and acceleration of $7\text{kg}$ block.

Answer to Problem 5SP

The net force on the $7\text{kg}$ block is $22.4\text{N}$ and net acceleration of the block is $3.2\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The horizontal force acting on $7\text{kg}$ mass along the right direction is $46\text{N}$ and that along the left direction is $8\text{N}$

Write the expression for the net force on the $7\text{kg}$ mass.

$F_{\text{net2}}=F_{\text{right}}-\left(F_{\text{tension}}+F_{\text{left}}\right)$

Here,

$F_{\text{left}}$ is the horizontal force on the left direction of mass $7\text{kg}$

$F_{\text{right}}$ is the horizontal force on the right direction of mass $7\text{kg}$

$F_{\text{tension}}$ is the tension on the string

Substitute $46\text{N}$ for $F_{\text{right}}$ ,$15.6\text{N}$ for $F_{\text{tension}}$ and $8\text{N}$ for $F_{\text{left}}$ in the above equation to get $F_{\text{net2}}$.

$\begin{array}{c}{F}_{\text{net2}}=46\text{N}-\left(15.6\text{N}+8\text{N}\right)\\ =22.4\text{N}\end{array}$

Write the expression for the net acceleration on $7\text{kg}$ mass.

$a=\frac{F_{\text{net2}}}{m_2}$

Here,

$m_2$ is the $7\text{kg}$ mass

$a$ is the net acceleration

Substitute $7\text{kg}$ for $m_2$ and $22.4\text{N}$ for $F_{\text{net2}}$ in the above equation to get $F_{\text{net2}}$.

$\begin{array}{c}a=\frac{22.4\text{N}}{7\text{kg}}\\ =3.2\text{m}/{\text{s}}^{\text{2}}\end{array}$

This is same as the net acceleration obtained in part $b$.

Conclusion:

Thus, the net force on the $7\text{kg}$ block is $22.4\text{N}$ and net acceleration of the block is $3.2\text{m}/{\text{s}}^{\text{2}}$.