Chapter 4, Problem 68AP

An object of mass m₁ hangs from a string that passes over a very light fixed pulley P₁ as shown in Figure P4.68. The string connects to a second very light pulley P₂. A second string passes around this pulley with one end attached to a wall and the other to an object of mass m₂ on a frictionless, horizontal table, (a) If a₁ and a₂, are the accelerations of m₁ and m₂, respectively, what is the relation between these accelerations? Find expressions for (b) the tensions in the strings and (c) the accelerations a₁ and a₂? in terms of the masses m₁ and m₂. and g.

Figure P4.68

To determine

The relation between the accelerations of the two masses.

Answer to Problem 68AP

Introduction:

The acceleration produces on the masses or bodies due to the application of external forces on it. In the presence of external force the acceleration produced in the same direction as that of the force.

Explanation of Solution

Here the two masses are affected by tensions and those are the reasons for the acceleration of the two masses $m_{\text{1}}$ and $m_2$. The motion of mass ${\text{m}}_{\text{1}}$ is affected by two forces one is tension in the upward direction and another is its weight. The motion of mass $m_{\text{2}}$ is affected by two tensions in the direction of acceleration.

The tensions those are affected on the strings are same. Thus the acceleration on the mass $m_{\text{2}}$ is two times in magnitude as that of the acceleration on the mass $m_{\text{1}}$. This difference in acceleration is because of the arrangements of the two masses.

Conclusion:

Therefore, the acceleration on mass $m_{\text{2}}$ is $a_2$, whose magnitude is twice in magnitude as that of the mass $m_{\text{1}}$ whose acceleration  is $a_1$.

To determine

The magnitude of the tensions.

Answer to Problem 68AP

Solution:

The magnitude of tensions are $\frac{4m_1{m}_{2}g}{4m_2+m_1}$ and $\frac{2m_1{m}_{2}g}{4m_2+m_1}$.

Explanation of Solution

Given Info:

The acceleration of mass $m_1$ is $a_1$ and acceleration of mass $m_2$ is $a_2$. The free body diagrams are given by,

Write the expressions for tensions on two strings attached by two masses.

$m_1{a}_1=m_{1}g-T_1$ (I)

$m_2{a}_2=T_2$ (II)

Write (I) in terms of $T_1$.

$\begin{array}{c}{T}_1=m_{1}g-m_1{a}_1\\ =m_1\left(g-a_1\right)\end{array}$ (III)

Write the relation between the two accelerations.

$a_1=\frac{a_2}{2}$ (IV)

Rewrite the equation (III) using (IV).

$T_1=m_1\left(g-\frac{a_2}{2}\right)$ (V)

Rewrite (II) in terms of $a_2$.

$a_2=\frac{T_2}{m_2}$ (VI)

Write the expression for tension in pulley 2.

$T_1-2T_2=0$

Rewrite the above expression in terms of $T_1$.

$T_1=2T_2$ (VII)

Substitute (VI) and (VII) in (V) to calculate $T_2$.

$\begin{array}{c}2T_2=m_1\left(g-\frac{T_2}{2m_2}\right)\\ 2T_2=m_{1}g-\frac{T_2{m}_1}{2m_2}\\ 2T_2+\frac{T_2{m}_1}{2m_2}=m_{1}g\\ T_2\left(2+\frac{m_1}{2m_2}\right)=m_{1}g\end{array}$

$\begin{array}{c}{T}_2\left(\frac{4m_2+m_1}{2m_2}\right)=m_{1}g\\ T_2=\frac{2m_1{m}_{2}g}{4m_2+m_1}\end{array}$ (VIII)

Substitute (VIII) in (VII) to calculate $T_1$

$T_1={\frac{4m_1{m}_{2}g}{4m_2+m_1}}_2$

Conclusion:

Therefore, the magnitude of tensions are $\frac{4m_1{m}_{2}g}{4m_2+m_1}$ and $\frac{2m_1{m}_{2}g}{4m_2+m_1}$.

To determine

The magnitude of the acceleration.

Answer to Problem 68AP

Solution:

The magnitude of accelerations $a_1$ and $a_2$ are respectively $\frac{m_{1}g}{4m_2+m_1}$ and $\frac{2m_{1}g}{4m_2+m_1}$.

Explanation of Solution

Given Info:

Write the formula to calculate $a_2$.

$a_2=\frac{T_2}{m_2}$

Use $\frac{2m_1{m}_{2}g}{4m_2+m_1}$ for $T_2$ in the above equation to calculate $a_2$.

$\begin{array}{c}{a}_2=\frac{1}{m_2}\left(\frac{2m_1{m}_{2}g}{4m_2+m_1}\right)\\ =\frac{2m_{1}g}{4m_2+m_1}\end{array}$

Substitute the above expression in (IV) to calculate $a_1$.

$\begin{array}{c}{a}_1=\frac{1}{2}\left(\frac{2m_{1}g}{4m_2+m_1}\right)\\ =\frac{m_{1}g}{4m_2+m_1}\end{array}$

Conclusion:

Therefore, the magnitude of accelerations $a_1$ and $a_2$ are respectively $\frac{m_{1}g}{4m_2+m_1}$ and $\frac{2m_{1}g}{4m_2+m_1}$