Chapter 4, Problem 76GP

(a)

To determine

The mass of the sand added to bucket.

Answer to Problem 76GP

The mass of sand added to bucket is $11.25\text{kg}$ .

Explanation of Solution

Given:

The mass of the block is, $m_b=28\text{kg}$ .

The mass of the empty bucket is, $m_e=1.35\text{kg}$ .

The coefficient of static friction between table and block is, ${\mu }_s=0.450$ .

The coefficient of kinetic friction between table and block is, ${\mu }_k=0.320$ .

Formula Used:

Draw the free body diagram of the system.

Here, the force on the rope toward the block is $F_1$ and $F_2$ is the force acting in the downward.

The expression to calculate the mass of sand added to bucket is,

  $\begin{array}{c}{F}_1=F_2\end{array}$

  $\begin{array}{c}\left(m+m_e\right)g={\mu }_s{m}_{b}g\end{array}$

  $\begin{array}{c}m+m_e={\mu }_s{m}_b\end{array}$

  $\begin{array}{c}m={\mu }_s{m}_b-m_e\end{array}$

Where,

  $g$ is the acceleration due to gravity.

  $m$ is the mass sand added.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}m=\left(0.450\right)\left(28\text{kg}\right)-\left(1.35\text{kg}\right)\end{array}$

  $\begin{array}{c}=11.25\text{kg}\end{array}$

Conclusion:

Thus, the mass of sand added to bucket is $11.25\text{kg}$ .

(b)

To determine

The acceleration of the system.

Answer to Problem 76GP

The acceleration of the system is $0.31\text{m}/{\text{s}}^2$ .

Explanation of Solution

Given:

The mass of the block is, $m_b=28\text{kg}$ .

The mass of the empty bucket is, $m_e=1.35\text{kg}$ .

The coefficient of static friction between table and block is, ${\mu }_s=0.450$ .

The coefficient of kinetic friction between table and block is, ${\mu }_k=0.320$ .

Formula Used:

The expression to calculate the acceleration of the system is,

  $\begin{array}{c}\left(m_b+m_e\right)a=m_{e}g-{\mu }_k{m}_{e}g\end{array}$

  $\begin{array}{c}a=\left(\frac{m_e-{\mu }_k{m}_e}{m_b+m_e}\right)g\end{array}$

Where,

  $a$ is the acceleration.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}a=\left(\frac{1.35\text{kg}-\left(0.320\right)\left(1.35\text{kg}\right)}{28\text{kg}+1.35\text{kg}}\right)\left(9.8\text{m}/{\text{s}}^2\right)\end{array}$

  $\begin{array}{c}=0.31\text{m}/{\text{s}}^2\end{array}$

Conclusion:

Thus, the acceleration of the system is $0.31\text{m}/{\text{s}}^2$ .