Question

A block of mass $M$ is connected with a particle of mass $m$ by a light inextensible string as shown in the figure. Assuming all contacting surfaces as smooth, find the acceleration of the wedge after releasing the system.

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Accepted Answer

From the constraint relationship shown in the figure,

$y + 4 x = l$

$\Rightarrow \frac{d y}{d t} + \frac{4 d x}{d t} = 0$

$\Rightarrow \frac{d^{2} y}{d t^{2}} + \frac{4 d^{2} x}{d t^{2}} = 0$

$\Rightarrow a_{2} = 4 a_{1}$ (numerically) $. . . (1)$

where $a_{1}$ is the acceleration of $M$ and $a_{2}$ is the acceleration of $m$ with respect to $M .$

Equation for $M$

$Σ F_{x} = M a_{1} \Rightarrow 4 T - N' = M a_{1}$

For $m, Σ F_{y} = m a_{2} \Rightarrow m g - T = m a_{2} \dots (3)$

$Σ F_{x} = m a_{1} \Rightarrow N' = m a_{1} \dots (4)$

as $m$ is always in contact with $M .$ Hence, the acceleration of $m$ in horizontal direction should be same as the acceleration of $M .$

$(2)$ $+$ $(4)$ $\Rightarrow a_{1} = \frac{4 T}{M + m} \dots (5)$

Putting $a_{2} = 4 a_{1}$ from $(1)$ and $T = \frac{(M + m) a_{1}}{4}$

From $(5)$ in $(3)$ , we obtain,

$m g - \frac{(M + m) a_{1}}{4} = m (4 a_{1})$

$\Rightarrow a_{1} = \frac{4 m g}{M + 17 m}$

Approach 2:

If we take block $(M)$ and particle $(m)$ as system, then the system has acceleration $a_{1}$ in horizontal direction. The external force on the system in horizontal direction is $4 T .$ Now consider the system in horizontal direction.

We can write equation of motion for system

$4 T = (M + m) a_{1} \dots (4)$

***Now consider the motion of the particle $(m)$ in vertical direction.

Equation of motion of $m : m g - T = m a_{2} \dots (5)$

or $m g - T = m (4 a_{1}) \dots (6)$

[As $a_{2} = 4 a_{1}] \dots (7)$

Now on solving equations $(6)$ and $(7)$ , we get, $a_{1} = \frac{4 m g}{(M + 17 m)}$

B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 59: ILLUSTRATION 6.54

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Newton's Laws of Motion (Without Friction), Exercise 59: ILLUSTRATION 6.54

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