Chapter 6, Problem 23A

Each block on the friction-free lab bench is connected by a string and pulled by a second falling block.

Rank each of the following from greatest to least.

a. the acceleration of the two-block systems

b. the tension in the strings

To determine

The rank of the acceleration of the two-block systems from greatest to least.

Answer to Problem 23A

The rank of the acceleration of the two-block systems from greatest to least is B, C, A.

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.

In case A, the mass of the block on the frictionless bench 3 kg is greater than that of the mass of thehanging block 1 kg. Hence, the acceleration of the system in A will be less.

In case B, the mass of the block on the friction less bench 1 kg is less than that of the mass of the hanging block 3 kg. Thus, the acceleration of the system in B will be the greatest.

In case C, the mass of the block on the friction less bench 3 kg is same as that of the mass of the hanging block 3 kg. Thus, the acceleration of the system in this case will be in between the A and B.

Hence, the rank of the acceleration of the two-block systems from greatest to least is B, C, A.

Conclusion:

Thus, the rank of the acceleration of the two-block systems from greatest to least is B, C, A.

To determine

The rank of the tension in the strings from greatest to least.

Answer to Problem 23A

Thus, the rank of the tension in the strings from greatest to least is C, A=B.

Explanation of Solution

Formula used:

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

  $F=ma$

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

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Sketch the Free Body diagram of the forces acting on the system as shown below:

  

For the two block system:

Apply the equilibrium of vertical force as follows:

  $\begin{array}{l}{\displaystyle \sum F_y}=m_{2}g-T_A\\ m_{2}a=m_{2}g-T_A\\ m_{2}a=m_{2}g-m_{1}a\\ a=\frac{m_{2}g}{m_1+m_2}\end{array}$

Find the tension in the string as follows:

  $\begin{array}{l}T=m_{1}a\\ T=m_1\left(\frac{m_{2}g}{m_1+m_2}\right)\\ T=\frac{m_1{m}_{2}g}{m_1+m_2}\end{array}$ …… (1)

For case A:

Find the tension in string using Equation (1) as follows:

  $\begin{array}{l}T=\frac{m_1{m}_{2}g}{m_1+m_2}\\ T=\frac{\left(3.0\text{ kg}\right)\left(1.0\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(3.0\text{ kg}\right)+\left(1.0\text{ kg}\right)}\\ T=7.35\text{m}/{\text{s}}^{\text{2}}\end{array}$

For case B:

Find the tension in string using Equation (1) as follows:

  $\begin{array}{l}T=\frac{m_1{m}_{2}g}{m_1+m_2}\\ T=\frac{\left(1.0\text{ kg}\right)\left(3.0\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(1.0\text{ kg}\right)+\left(3.0\text{ kg}\right)}\\ T=7.35\text{m}/{\text{s}}^{\text{2}}\end{array}$

For case C:

Find the tension in string using Equation (1) as follows:

  $\begin{array}{l}T=\frac{m_1{m}_{2}g}{m_1+m_2}\\ T=\frac{\left(3.0\text{ kg}\right)\left(3.0\text{ kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\left(3.0\text{ kg}\right)+\left(3.0\text{ kg}\right)}\\ T=14.7\text{m}/{\text{s}}^{\text{2}}\end{array}$

Hence, the rank of the tension in the strings from greatest to least is C, A=B.

Conclusion:

Thus, the rank of the tension in the strings from greatest to least is C, A=B.