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Newton's Laws of Motion

Three blocks ...

Question

Three blocks of mass $m_{1}, m_{2}$ and $m_{3}$ are connected as shown in the figure. All the surfaces are frictionless and the string and the pulleys are light. Find the acceleration of $m_{1} .$

a_{0} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}

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a_{0} = \frac{2 g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}

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a_{0} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{2}{m_{2}} + \frac{1}{m_{3}})}

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a_{0} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{2}{m_{3}})}

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Solution

The correct option is A $a_{0} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}$
Suppose the acceleration of $m_{1}$ is $a_{0}$ towards right.
The acceleration of pulley B will also be $a_{0}$ downwards because the string connecting $m_{1}$ and B is constant in length.
Also the string connecting $m_{2}$ and $m_{3}$ has a constant length.
This implies that the decrease in the separation between $m_{2}$ and and B equals the increase in the separation between $m_{3}$ and B. So, the upward acceleration of $m_{3}$ with respect to B.
Let this acceleration Be $a_{r} .$
The acceleration of $m_{2}$ with respect to the ground $= a_{0} - a_{r}$ (downward) and the acceleration of $m_{3}$ with respect to the ground $a_{0} + a_{r}$ (downward).
Let the tension be $T_{1}$ in the upper string and $T_{3}$ in the lower string.
Consider the motion of the pulley B.
The forces on this light pulley are
a) $T_{1}$ upwards by the upper string and
b) $2 T_{2}$ downwards by the lower string.
As the mass of the pulley is negligible, $2 T_{2} - T_{1} = 0$ $⟹ T_{2} = \frac{T_{1}}{2}$ ... (i)
Motion of $m_{1}$ : In the horizontal direction, the equation is
$T_{1} = m_{1} a_{0}$ ...(ii)
Motion of $m_{2}$ : $m_{2} g - T_{1} / 2 = m_{2} (a_{0} - a_{r})$ ...(iii)
Motion of $m_{3}$ : $m_{3} g - T_{1} / 2 = m_{3} (a_{0} + a_{r})$ ...(iv)
Solving these four equations, we get $a_{0} = \frac{g}{1 + \frac{m_{1}}{4} (\frac{1}{m_{2}} + \frac{1}{m_{3}})}$

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