Chapter 4, Problem 60A

(a)

To determine

The force (both magnitude and direction) exerted by the $7\text{kg}$ block on the $6\text{kg}$ block.

Answer to Problem 60A

The force exerted by the $7\text{kg}$ block on the $6\text{kg}$ block is $59\text{N}\left(\text{in the upward direction}\right)$ .

Explanation of Solution

Given:

A $6\text{kg}$ block is placed over the block of mass $7\text{kg}$ , which is placed over a horizontal surface.

Formula used:

Newton’s Second Law: The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Calculation:

Consider the block $B$ of mass $m_B=7\text{kg}$ is placed on a horizontal table.

Block $A$ of mass $m_A=6\text{kg}$ rest on the top of the block $B$ .

Show the assembly of the blocks as shown below:

  

Find the force exerted by the $7\text{kg}$ block on the $6\text{kg}$ as follows:

  $\begin{array}{l}{F}_{\text{net}}=N-m_{B}g\\ F_{\text{net}}=\left(m_{A}g+m_{B}g\right)-m_{B}g\\ F_{\text{net}}=m_{A}g\\ F_{\text{net}}=6\text{kg}\times \text{9}\text{.81}{\text{m/s}}^2\\ F_{\text{net}}=59\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{net}}=59\text{N}\left(\text{upward direction}\right)\end{array}$

Conclusion:

Hence, the force exerted by the $7\text{kg}$ block on the $6\text{kg}$ block is $59\text{N}$

  $\left(\text{in the upward direction}\right)$ .

(b)

To determine

The force (both magnitude and direction) exerted by the $6\text{kg}$ block on the $7\text{kg}$ block.

Answer to Problem 60A

The force (both magnitude and direction) exerted by the $6\text{kg}$ block on the $7\text{kg}$ block is $59\text{N}\left(\text{downward direction}\right)$ .

Explanation of Solution

Given:

A $6\text{kg}$ block is placed over the block of mass $7\text{kg}$ , which is placed over a horizontal surface.

Formula used:

Newton’s Third Law: For every action in nature, there is an equal and opposite reaction.

Calculation:

Refer to the result of part (a).

The force exerted by the block of mass $7\text{kg}$ on the block of mass $6\text{kg}$ is $59\text{N}\left(\text{upward direction}\right)$ .

Apply the Newton’ Third Law of Motion:

The block of mass $6\text{kg}$ also applies equal and opposite force on the block of mass $7\text{kg}$ .

Hence, the force exerted by the block of mass $6\text{kg}$ on the block of mass $7\text{kg}$ is $59\text{N}\left(\text{downward direction}\right)$ .

Conclusion:

The force exerted by the $6\text{kg}$ block on the $7\text{kg}$ block is $59\text{N}$

  $59\text{N}\left(\text{downward direction}\right)$ .

Conclusion:

Hence, the force exerted by the $6\text{kg}$ block on the $7\text{kg}$ block is $59\text{N}\left(\text{downward direction}\right)$ .