Question

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

Answer 1

From the constraint relationship shown in Fig.

$y+4x=l$

$\Rightarrow$ $\frac{dy}{dt}+\frac{4dx}{dt}=0$

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

$\Rightarrow$ $a_1=4a_2$ (numerically).........(i)

Where $a_1$ is the acceleration of M and $a_2$ is the acceleration of m, with respect to M.

Equation for M

$\sum F_x=M{a}_1\Rightarrow 4T-N=M{a}_1$.....(ii)

For m, $\sum F_y=M{a}_2\Rightarrow mg-T={ma}_2$....(iii)

$\sum F_x=M{a}_1\Rightarrow N^{\prime }={ma}_1$......(iv)

As m is always in contact with M.

(ii)+(iv)$\Rightarrow a_1=\frac{4T}{M+m}$.......(v)

Putting $a_2=\frac{a_1}{4}$ from (i) and $T=\frac{(M+m)a_1}{4}$

from (v) in (iii),we obtain

$mg-\frac{(M+m)a_1}{4}=m\frac{a_1}{4}$

$\Rightarrow$ $a_1=\frac{4mg}{M+2m}$