Chapter 9, Problem 74P

To determine

To Calculate: The acceleration of each block and the tensions in the segments of string between each block and the pulley.

Answer to Problem 74P

The acceleration of each block is $1.56{\text{ m/s}}^2$ .

The tension acting on the block of mass $m_1$ is $16.04\text{ N}$ .

The tension acting on the block of mass $m_2$ is $16.5\text{ N}$ .

The tension on the pulley is $0.47\text{ N}$ .

Explanation of Solution

Given Information:

Mass of block resting on the ledge, $m_1=4\text{kg}$

Mass of the block hanging from the pulley, $m_2=2\text{kg}$

Coefficient of kinetic friction between ledge and block, $\text{=0}\text{.25}$

Mass of the pulley, ${\text{m}}_P=0.60\text{kg}$

Radius of the disk of the pulley, $r_p=8\text{cm}$

Formula used:

By Newton’s second law of motion, the net force acting on a system is:

  $F=ma$

Where, m is the mass and a is the acceleration.

Calculation:

Free body diagram can be drawn as follows:

  

By using the Newton’s second law of motion, horizontal component of force on block $1$ is

  $\begin{array}{l}{F}_1=T_1-f_1\\ m_{1}a=T_1-{\mu }_K{F}_{nl}\mathrm{......}\left(1\right)\end{array}$

Normal force on block $1$ is

  $F_{nl}=m_{1}g$

Substitute this value in the equation $\left(1\right)$ :

  $\begin{array}{l}{m}_{1}a=T_1-{\mu }_K{m}_{1}g\\ T_1=m_{1}a+{\mu }_K{m}_{1}g\mathrm{.......}\left(2\right)\end{array}$

Write the equation of motion for the pulley of mass $m_p$ and use the Newton’s second law of motion to find torque acting on pulley.

  $\begin{array}{l}{\displaystyle \sum {\tau }_{\text{p}}=I_{\text{p}}}\alpha \\ \left(T_2-T_1\right)r=I_{\text{p}}\alpha \mathrm{........}\left(3\right)\end{array}$

Moment of inertia $\left(I_{\text{p}}\right)$ of the pulley (disk) is

  $I_{\text{p}}=\frac{1}{2}{m}_{\text{p}}{r}^2$

Angular acceleration $\left(\alpha \right)$ of the pulley is

  $\alpha =\frac{a}{r}$

Substitute the equations of $I_{\text{p}}$ and $\alpha$ in the equation $\left(3\right)$

  $\begin{array}{l}\left(T_2-T_1\right)r=\left(\frac{1}{2}{m}_{\text{p}}{r}^2\right)\left(\frac{a}{r}\right)\\ \left(T_2-T_1\right)r=\frac{1}{2}{m}_{\text{p}}ar\\ \left(T_2-T_1\right)=\frac{1}{2}{m}_{\text{p}}a\mathrm{......}\left(4\right)\end{array}$

Write the equation of motion for the hanging mass $m_2\left(=2.0\text{ kg}\right)$ by using the Newton’s second law of motion.

  $\begin{array}{l}{F}_2=m_{2}g-T_2\\ m_{2}a=m_{2}g-T_2\\ T_2=m_{2}g-m_{2}a\mathrm{.......}\left(5\right)\end{array}$

Substitute the equations $\left(2\right)$ and $\left(5\right)$ :

  $\begin{array}{l}{T}_2-T_1=\left(m_{2}g-m_{2}a\right)-\left(m_{1}a+{\mu }_K{m}_{1}g\right)\\ =m_{2}g-{\mu }_K{m}_{1}g-m_{1}a-m_{2}a\mathrm{.......}\left(6\right)\\ =g\left(m_2-{\mu }_K{m}_1\right)-a\left(m_1+m_2\right)\end{array}$

Equation the equations $\left(4\right)$ and $\left(6\right)$ we get

  $\begin{array}{l}g\left(m_2-{\mu }_K{m}_1\right)-a\left(m_1+m_2\right)=\frac{1}{2}{m}_{\text{p}}a\\ g\left(m_2-{\mu }_K{m}_1\right)=a\left(m_1+m_2\right)+\frac{1}{2}{m}_{\text{p}}a\\ a\left(m_1+m_2+\frac{1}{2}{m}_{\text{p}}\right)=g\left(m_2-{\mu }_K{m}_1\right)\mathrm{......}\left(7\right)\\ a=\frac{\left(m_2-{\mu }_K{m}_1\right)}{\left(m_1+m_2+\frac{1}{2}{m}_{\text{p}}\right)}g\end{array}$

Substitute $4\text{ kg}$ for mass of block resting on a horizontal ledge $\left({\text{m}}_1\right)\text{,}\text{ 2 kg}$ mass of hanging block, $\left(m_2\right),0.25$ coefficient of kinetic friction between ledge and block, $\left({\mu }_k\right),0.60\text{ kg}$ mass of pulley $\left(m_p\right)$ and $8.0\text{ cm}$ radius of pulley, $\left(r\right)$ in the equation $\left(7\right)$ solve for acceleration

The acceleration of each block is

  $\begin{array}{l}a=\frac{\left[\left(2\text{ kg}\right)-\left(0.25\right)\left(4.0\text{ kg}\right)\right]}{\left(4.0\text{ kg}\right)+\left(2.0\text{kg}\right)+\frac{1}{2}\left(0.60\text{ kg}\right)}\left(9.8{\text{ m/s}}^2\right)\\ =1.56{\text{ m/s}}^2\end{array}$

Tension $\left(T_1\right)$ acting on the block of mass $m_1$ is

  $T_1=m_{1}a+{\mu }_K{m}_{1}g$

Substitute $4\text{ kg}$ mass of block resting on a horizontal ledge $\left({\text{m}}_1\right)\text{,}1.56{\text{m/s}}^2$ is the acceleration $\left(a\right),0.25$ for coefficient of friction, $\left({\mu }_k\right),$ and mass of pulley $\left(m_p\right)$ and $9.8{\text{m/s}}^2$ for acceleration to gravity $\left(g\right)$ in the above equation solve for $T_1$

  $\begin{array}{l}{T}_1=\left(4.0\text{ kg}\right)\left(1.56{\text{ m/s}}^2\right)+\left(0.25\right)\left(4.0\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\\ =16.04\text{ N}\end{array}$

Hence the tension acting on the block of mass $m_1$ is $16.04\text{N}$

Tension $\left(T_2\right)$ acting on the hanging block of mass $m_2$ is $T_2=m_{2}g-m_{2}a$

Substitute $\text{2 kg}$ mass of hanging block $\left({\text{m}}_2\right)\text{,}1.56{\text{m/s}}^2$ is the acceleration $\left(a\right),$ and $9.8{\text{m/s}}^2$ for acceleration due to gravity $\left(g\right),$ in the above equation solve for $T_2$

  $\begin{array}{l}{T}_2=\left(2.0\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)-\left(2.0\text{ kg}\right)\left(1.56{\text{ m/s}}^2\right)\\ =16.5\text{ N}\end{array}$

Hence the tension acting on the block of mass $m_2$ is $16.5\text{ N}$ .

Tension acting on the pulley of mass $m_{\text{p}}$ is

  $\begin{array}{l}\left(T_2-T_1\right)=\frac{1}{2}{m}_{\text{p}}a\\ =\frac{1}{2}\left(0.60\text{ kg}\right)\left(1.56{\text{ m/s}}^2\right)\\ =0.47\text{ N}\end{array}$

Hence, the tension on the pulley is $0.47\text{ N}$ .

Conclusion:

The acceleration of each block is $1.56{\text{ m/s}}^2$ .

The tension acting on the block of mass $m_1$ is $16.04\text{ N}$ .

The tension acting on the block of mass $m_2$ is $16.5\text{ N}$ .

The tension on the pulley is $0.47\text{ N}$ .