Chapter 7, Problem 23A

Each of these boxes is pulled by the same force F to the left. All boxes have the same mass and slide on a friction-free surface.

Rank the following from greatest to least.

a. the acceleration of the boxes

b. the tension in the rope connected to the boxes on the right in B and in C

To determine

The rank of the acceleration of the boxes from greatest to least.

Answer to Problem 23A

The rank of the acceleration of the boxes from greatest to least is $a_{\text{A}}>a_{\text{B}}>a_{\text{C}}$ .

Explanation of Solution

Introduction:

The expression for Newton’s second law of acceleration as follows:

  $F=ma$

Here, $m$ is the mass and $a$ is the acceleration.

Sketch the Free Body diagram of the forces acting on the system as shown in Figure.

  

Refer to Figure.

The acceleration of box A as follows:

  $\begin{array}{l}F=ma\\ a_{\text{A}}=\frac{F}{m}\end{array}$

The acceleration of box B as follows:

  $\begin{array}{l}F=ma\\ a_{\text{B}}=\frac{F}{m+m}\\ a_{\text{B}}=\frac{F}{2m}\end{array}$

The acceleration of box C as follows:

  $\begin{array}{l}F=ma\\ a_{\text{C}}=\frac{F}{m+m+m}\\ a_{\text{C}}=\frac{F}{3m}\end{array}$

Conclusion:

Thus, the rank of the acceleration of the boxes from greatest to least is $a_{\text{A}}>a_{\text{B}}>a_{\text{C}}$ .

To determine

The rank of the tension in the rope connected to the boxes on the right in B and in C from greatest to least.

Answer to Problem 23A

The rank of the tension in the rope connected to the boxes on the right in B and in C from greatest to least from greatest to least is $T_{\text{B}}>T_{\text{C}}$ .

Explanation of Solution

Sketch the Free Body diagram of the tension in the rope connected to the boxes on the right in B and in C as shown in Figure.

  

Refer to Figure.

The tension in the rope connected to the boxes on the right in B as follows

  $\begin{array}{l}F=T\\ T_{\text{B}}=F\end{array}$

The tension in the rope connected to the boxes on the right in C as follows

  $\begin{array}{l}F=T\\ F=T_{\text{C}}+T_{\text{C}}\\ F=2T_{\text{C}}\\ T_{\text{C}}=\frac{F}{2}\end{array}$

Conclusion:

Thus, the rank of the tension in the rope connected to the boxes on the right in B and in C from greatest to least from greatest to least is $T_{\text{B}}>T_{\text{C}}$ .