Chapter 5, Problem 5.98CP

Initially, the system of objects shown in Figure P5.49 is held motionless. The pulley and all surfaces and wheels are frictionless. Let the force $\stackrel{\to }{\text{F}}$ be zero and assume that m₁ can move only vertically. At the instant after the system of objects is released, Find (a) the tension T in the string, (b) the acceleration of m₂, (c) the acceleration of M, and (d) the acceleration of m₁. (Note: The pulley accelerates along with the cart.)

Figure P5.49 Problems 49 and 53

To determine

The tension in the string.

Answer to Problem 5.98CP

The tension in the string is $m_{2}g\left(\frac{m_{1}M}{m_{2}M+m_1\left(m_2+M\right)}\right)$.

Explanation of Solution

Consider the free body diagram given below,

Figure I

Here, $a$ is the acceleration of hanging block having mass $m_1$, $A$ is the acceleration of large block having mass $M$ and $a-A$ is the acceleration of top block having mass $m_2$.

Write the expression for the equilibrium condition for hanging block

    $\begin{array}{c}{m}_{1}g-T=m_{1}a\\ T=m_1\left(g-a\right)\end{array}$                                                                         (I)

Here, $m_1$ is the mass of the hanging block, $a$ is the acceleration of the hanging block, $g$ is the acceleration due to gravity and $T$ is the tension of the cord.

Write the expression for the equilibrium condition for top block

    $\begin{array}{c}T=m_2\left(a-A\right)\\ a=\frac{T}{m_2}+A\end{array}$                                                                                   (II)

Here, $m_2$ is the mass of the top block and $A$ is the acceleration of the top block

Write the expression for the equilibrium condition for large block

    $\begin{array}{c}MA=T\\ A=\frac{T}{M}\end{array}$                                                                                           (III)

Here, $M$ is the acceleration of the large mass.

Substitute $\left(\frac{T}{m_2}+A\right)$ for $a$ and $\frac{T}{M}$ for $A$ in equation (I) to find $T$.

    $\begin{array}{c}T=m_1\left(g-\left(\frac{T}{m_2}+A\right)\right)\\ =m_1\left(g-\left(\frac{T}{m_2}+\frac{T}{M}\right)\right)\\ =m_{1}g-\left(\frac{m_{1}T}{m_2}+\frac{m_{1}T}{M}\right)\\ =m_{1}g-m_{1}T\left(\frac{M+m_2}{M{m}_2}\right)\end{array}$

Further, solve for $T$.

    $\begin{array}{c}T=m_{1}g-m_{1}T\left(\frac{M+m_2}{M{m}_2}\right)\\ T+m_{1}T\left(\frac{M+m_2}{M{m}_2}\right)=m_{1}g\\ T=m_{2}g\left(\frac{m_{1}M}{m_{2}M+m_1\left(m_2+M\right)}\right)\end{array}$

Conclusion:

Therefore, the tension in the string is $m_{2}g\left(\frac{m_{1}M}{m_{2}M+m_1\left(m_2+M\right)}\right)$.

To determine

The acceleration of $m_2$.

Answer to Problem 5.98CP

The acceleration of $m_2$ is $\frac{m_{1}g\left(M+m_2\right)}{M{m}_2+m_1\left(M+m_2\right)}$.

Explanation of Solution

The force applied on the block of mass $M$ is zero initially and the block of mass $m_2$ has acceleration in synchronization with the big block so the net acceleration on the block is $a$.

Substitute $\frac{T}{M}$ for $A$ in equation (II).

    $\begin{array}{c}a=\frac{T}{m_2}+\frac{T}{M}\\ =T\left(\frac{M+m_2}{M{m}_2}\right)\end{array}$

Substitute $m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)$ for $T$ in above equation to find $a$.

    $\begin{array}{c}a=\left(m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)\right)\left(\frac{M+m_2}{M{m}_2}\right)\\ =\frac{m_{1}g\left(M+m_2\right)}{M{m}_2+m_1\left(M+m_2\right)}\end{array}$

Conclusion:

Therefore, the acceleration of $m_2$ is $\frac{m_{1}g\left(M+m_2\right)}{M{m}_2+m_1\left(M+m_2\right)}$.

To determine

The acceleration of $M$.

Answer to Problem 5.98CP

The acceleration of $M$ is $\frac{m_1{m}_{2}g}{m_{2}M+m_1\left(m_2+M\right)}$.

Explanation of Solution

The acceleration of $M$ is $A$.

Substitute $m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)$ for $T$ in equation (II).

    $\begin{array}{c}A=\frac{m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)}{M}\\ =\frac{m_1{m}_{2}g}{m_{2}M+m_1\left(m_2+M\right)}\end{array}$

Conclusion:

Therefore, the acceleration of $M$ is $\frac{m_1{m}_{2}g}{m_{2}M+m_1\left(m_2+M\right)}$.

To determine

The acceleration of $m_1$.

Answer to Problem 5.98CP

The acceleration of $m_1$ is $\frac{M{m}_{1}g}{M{m}_2+m_1\left(M+m_2\right)}$.

Explanation of Solution

The block of mass $m_1$ moves in vertical direction only but the net acceleration is the difference between the acceleration of the big block of mass $M$ and the acceleration $a$ of $m_2$.

Write the formula to calculate the acceleration of $m_1$

    $a_{m_1}=a-A$                                                                                   (IV)

Here, $a_{{\text{m}}_1}$ is the acceleration of mass $m_{\text{1}}$.

Substitute $m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)$ for $T$ in equation (II).

    $\begin{array}{c}A=\frac{m_{1}g\left(\frac{M{m}_2}{M{m}_2+m_1\left(M+m_2\right)}\right)}{M}\\ =\frac{m_1{m}_{2}g}{m_{2}M+m_1\left(m_2+M\right)}\end{array}$

Substitute $\left(\frac{m_{1}g\left(M+m_2\right)}{M{m}_2+m_1\left(M+m_2\right)}\right)$ for $a$ and $\frac{m_1{m}_{2}g}{m_{2}M+m_1\left(m_2+M\right)}$ for $A$ in equation (4) to find the value of $a-A$.

    $\begin{array}{c}a-A=\left(\frac{m_{1}g\left(M+m_2\right)}{M{m}_2+m_1\left(M+m_2\right)}\right)-\left(\frac{m_1{m}_{2}g}{M{m}_2+m_1\left(M+m_2\right)}\right)\\ =\left(\frac{M{m}_{1}g}{M{m}_2+m_1\left(M+m_2\right)}\right)\end{array}$

Conclusion:

Therefore, the acceleration of $m_1$ is $\frac{M{m}_{1}g}{M{m}_2+m_1\left(M+m_2\right)}$.