Chapter 4, Problem 75P

(a)

To determine

The acceleration of the objects and the tension in the string in terms of $\theta$ , $m_1$ and $m_2$ .

Answer to Problem 75P

The expression for the tension in the string is $\frac{m_{2}g\left(m_1+m_1\sin \theta \right)}{m_1+m_2}$ and the acceleration of the block is $\frac{g\left(m_2-m_1\sin \theta \right)}{m_1+m_2}$ .

Explanation of Solution

Calculation:

The free body diagram for the blocks is shown in Figure 1

  

Figure 1

Consider the tension in the string is $T$ and the acceleration of the system is $a$ . Then consider that the block of mass $m_1$ is accelerating to the right and the block of mass $m_2$ is accelerating downwards.

The expression for the tension $T$ for block $m_1$ is given by,

  $T=m_{1}g\sin \theta$

The expression for the tension in terms of the acceleration for mass $m_1$ is given by,

  $\begin{array}{l}T=m_{1}a\\ m_{1}g\sin \theta =m_{1}a\end{array}$

The expression for the tension $T$ for block $m_2$ is given by,

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

The expression for the tension in terms of the acceleration for mass $m_2$ is given by,

  $\begin{array}{l}T=m_2\left(g+a\right)\\ m_2\left(g+a\right)=m_{2}a\end{array}$

The expression for the acceleration is evaluated as,

  $\begin{array}{l}{m}_2\left(g-a\right)-m_{1}g\sin \theta =m_{1}a\\ a=\frac{g\left(m_2-m_1\sin \theta \right)}{m_1+m_2}\end{array}$

The expression for the tension in terms of acceleration is given by,

  $\begin{array}{c}T=m_2\left(g-a\right)\\ =m_2\left(g-\left(\frac{g\left(m_2-m_1\sin \theta \right)}{m_1+m_2}\right)\right)\\ =\frac{m_{2}g\left(m_1+m_1\sin \theta \right)}{m_1+m_2}\end{array}$

Conclusion:

Therefore, the expression for the tension in the string is $\frac{m_{2}g\left(m_1+m_1\sin \theta \right)}{m_1+m_2}$ and the acceleration of the block is $\frac{g\left(m_2-m_1\sin \theta \right)}{m_1+m_2}$ .

(b)

To determine

The acceleration and the tension of the string.

Answer to Problem 75P

The tension of the string is $36.75\text{N}$ and the acceleration of the string is $2.45\text{m}/{\text{s}}^2$ .

Explanation of Solution

Given:

The angle of inclination is $\theta =30°$ .

The mass of first block is $m_1=5.0\text{kg}$ .

The mass of the second block is $m_2=5.0\text{kg}$ .

Calculation:

The acceleration of the string is calculated as,

  $\begin{array}{c}a=\frac{g\left(m_2-m_1\sin \theta \right)}{m_1+m_2}\\ =\frac{\left(9.81\text{m}/{\text{s}}^2\right)\left(5.00\text{kg}-\left(5.00\text{kg}\right)\sin 30°\right)}{5.00\text{kg}+5.00\text{kg}}\\ =2.45\text{m}/{\text{s}}^2\end{array}$

The tension in the string is calculated as,

  $\begin{array}{c}T=\frac{m_{2}g\left(m_1+m_1\sin \theta \right)}{m_1+m_2}\\ =\frac{\left(5.00\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(5.00\text{kg}+\left(5.00\text{kg}\right)\sin 30°\right)}{5.00\text{kg}+5.00\text{kg}}\\ =36.75\text{N}\end{array}$

Conclusion:

Therefore, the tension of the string is $36.75\text{N}$ and the acceleration of the string is $2.45\text{m}/{\text{s}}^2$ .