Chapter 5, Problem 118P

(a)

To determine

The acceleration of center of mass of two block cylinder string system.

Explanation of Solution

Given:

The mass of cylinder is $m_c$ .

Formula used:

Write expression for total force on the system.

  $F_{cm}=F_1+F_2+F_c$

Substitute $M{\overrightarrow{a}}_{cm}$ for $F_{cm}$ , $m_1{a}_1$ for $F_1$ , $m_2{a}_2$ for $F_2$ and $m_c{a}_c$ for $F_c$ .

  $M{\overrightarrow{a}}_{cm}=m_1{a}_1+m_2{a}_2+m_c{a}_c$

Substitute $m_1+m_2+m_3$ for $M$ , $a$ for $a_1$ , $-a$ for $a_2$ and $0$ for $a_3$ in above expression.

  ${\overrightarrow{a}}_{cm}=a\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)$  .......(1)

Calculation:

Draw free body diagram of blocks of Atwood’s machine.

  

Write expression for net force in block of mass $m_2$ .

  ${\displaystyle \sum F_2=0}$

Substitute $T-m_{2}g-m_2{a}_2$ for ${{\displaystyle \sum F}}_2$ in above expression.

  $T-m_{2}g-m_2{a}_2=0$  .......(2)

Write expression for net force in block of mass $m_1$ .

  ${\displaystyle \sum F_1}=0$

Substitute $m_{1}g-T-m_1{a}_1$ for ${\displaystyle \sum F_1}$ in above expression.

  $m_{1}g-T-m_1{a}_1=0$

Rearrange above expression for $T$ .

  $T=m_1{a}_1-m_{1}g$

Substitute $m_1{a}_1-m_{1}g$ for $T$ in equation (2).

  $m_1{a}_1-m_{1}g-m_{2}g-m_2{a}_2=0$

Substitute $a$ for $a_1$ and $-a$ for $a_2$ in above expression and solve for $a$ .

  $a=\frac{m_1-m_2}{m_1+m_2}g$

Substitute $\frac{m_1-m_2}{m_1+m_2}g$ for $a$ in equation (1).

  $\begin{array}{l}{\overrightarrow{a}}_{cm}=\left(\frac{m_1-m_2}{m_1+m_2}g\right)\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)\\ {\overrightarrow{a}}_{cm}=\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g\end{array}$

Conclusion:

Thus, the acceleration of COM is $\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g$ .

(b)

To determine

The force exerted by the support.

Explanation of Solution

Given:

The mass of cylinder is $m_c$ .

Formula used:

Write expression for total force on the system.

  $F_{cm}=F_1+F_2+F_c$

Substitute $M{\overrightarrow{a}}_{cm}$ for $F_{cm}$ , $m_1{a}_1$ for $F_1$ , $m_2{a}_2$ for $F_2$ and $m_c{a}_c$ for $F_c$ .

  $M{\overrightarrow{a}}_{cm}=m_1{a}_1+m_2{a}_2+m_c{a}_c$

Substitute $m_1+m_2+m_3$ for $M$ , $a$ for $a_1$ , $-a$ for $a_2$ and $0$ for $a_3$ in above expression.

  ${\overrightarrow{a}}_{cm}=a\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)$  .......(1)

Write expression for force exerted by the support.

  $F=Mg-M{a}_{cm}$  .......(2)

Calculation:

Draw free body diagram of blocks of Atwood’s machine.

  

Write expression for net force in block of mass $m_2$ .

  ${\displaystyle \sum F_2=0}$

Substitute $T-m_{2}g-m_2{a}_2$ for ${{\displaystyle \sum F}}_2$ in above expression.

  $T-m_{2}g-m_2{a}_2=0$  .......(3)

Write expression for net force in block of mass $m_1$ .

  ${\displaystyle \sum F_1}=0$

Substitute $m_{1}g-T-m_1{a}_1$ for ${\displaystyle \sum F_1}$ in above expression.

  $m_{1}g-T-m_1{a}_1=0$

Rearrange above expression for $T$ .

  $T=m_1{a}_1-m_{1}g$

Substitute $m_1{a}_1-m_{1}g$ for $T$ in equation (3).

  $m_1{a}_1-m_{1}g-m_{2}g-m_2{a}_2=0$

Substitute $a$ for $a_1$ and $-a$ for $a_2$ in above expression and solve for $a$ .

  $a=\frac{m_1-m_2}{m_1+m_2}g$

Substitute $\frac{m_1-m_2}{m_1+m_2}g$ for $a$ in equation (1).

  $\begin{array}{l}{\overrightarrow{a}}_{cm}=\left(\frac{m_1-m_2}{m_1+m_2}g\right)\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)\\ {\overrightarrow{a}}_{cm}=\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g\end{array}$

Substitute $\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g$ for $a_{cm}$ .

  $F=Mg-M\left(\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g\right)$

Substitute $m_1+m_2+m_3$ for $M$ in above expression and solve.

  $F=\left(\frac{4m_1{m}_2}{m_1+m_2}+m_c\right)g$

Conclusion:

Thus, force exerted by the support is $\left(\frac{4m_1{m}_2}{m_1+m_2}+m_c\right)g$ .

(c)

To determine

Tension $T$ in the string; to show that $F=m_{c}g+2T$ .

Explanation of Solution

Given:

Formula used:

Write expression for total force on the system.

  $F_{cm}=F_1+F_2+F_c$

Substitute $M{\overrightarrow{a}}_{cm}$ for $F_{cm}$ , $m_1{a}_1$ for $F_1$ , $m_2{a}_2$ for $F_2$ and $m_c{a}_c$ for $F_c$ .

  $M{\overrightarrow{a}}_{cm}=m_1{a}_1+m_2{a}_2+m_c{a}_c$

Substitute $m_1+m_2+m_3$ for $M$ , $a$ for $a_1$ , $-a$ for $a_2$ and $0$ for $a_3$ in above expression.

  ${\overrightarrow{a}}_{cm}=a\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)$  .......(1)

Write expression for force exerted by the support.

  $F=Mg-M{a}_{cm}$  .......(2)

Draw free body diagram of blocks of Atwood’s machine.

  

Write expression for net force in block of mass $m_2$ .

  ${\displaystyle \sum F_2=0}$

Substitute $T-m_{2}g-m_2{a}_2$ for ${{\displaystyle \sum F}}_2$ in above expression.

  $T-m_{2}g-m_2{a}_2=0$  .......(3)

Write expression for net force in block of mass $m_1$ .

  ${\displaystyle \sum F_1}=0$

Substitute $m_{1}g-T-m_1{a}_1$ for ${\displaystyle \sum F_1}$ in above expression.

  $m_{1}g-T-m_1{a}_1=0$

Rearrange above expression for $T$ .

  $T=m_1{a}_1-m_{1}g$  .......(4)

Substitute $m_1{a}_1-m_{1}g$ for $T$ in equation (3).

  $m_1{a}_1-m_{1}g-m_{2}g-m_2{a}_2=0$

Substitute $a$ for $a_1$ and $-a$ for $a_2$ in above expression and solve for $a$ .

  $a=\frac{m_1-m_2}{m_1+m_2}g$

Substitute $\frac{m_1-m_2}{m_1+m_2}g$ for $a_1$ in equation (4) and solve for $T$ .

  $T=\frac{2m_1{m}_{2}g}{m_1+m_2}$

Calculation:

Substitute $\frac{m_1-m_2}{m_1+m_2}g$ for $a$ in equation (1).

  $\begin{array}{l}{\overrightarrow{a}}_{cm}=\left(\frac{m_1-m_2}{m_1+m_2}g\right)\left(\frac{m_1-m_2}{m_1+m_2+m_3}\right)\\ {\overrightarrow{a}}_{cm}=\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g\end{array}$

Substitute $\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g$ for $a_{cm}$ .

  $F=Mg-M\left(\frac{{\left(m_1-m_2\right)}^2}{\left(m_1+m_2\right)\left(m_1+m_2+m_3\right)}g\right)$

Substitute $m_1+m_2+m_3$ for $M$ in above expression and solve.

  $F=\frac{4m_1{m}_2}{m_1+m_2}g+m_{c}g$

Substitute $T$ for $\frac{2m_1{m}_{2}g}{m_1+m_2}$ in above expression.

  $F=2T+m_{c}g$

Conclusion:

Thus, the tension in the string is $\frac{2m_1{m}_{2}g}{m_1+m_2}$ .