Question

A block of mass m is placed on the inclined surface of a wedge as shown in $Fig.6.151$. Calculate the acceleration of the wedge and the block when the block is released. Assume all surface are frictionless.

Answer 1

Constraint relation. Approach 1

We can observe that the wedge M can only move in horizontal direction towards left, and the block m can slide on inclined surface of M always in contact with the wedge.

$\bullet$ Let us defined our x and y axes parallel to the incline and perpendicular to incline, respectively.

$\bullet$ We can observe that the displacement of m and M in -x' direction will be same as the block never lose contact with the wedge.

$\bullet$ If the wedge moves in the horizontal direction by a distance x', during this time, the block will move x in x' direction.

$\bullet$ We can relate these displacements x and X as

$\frac{y}{x}=\sin \theta \Rightarrow y=x\sin \theta$......(i)

Hence, velocity relation can be written as :

$vy=V\sin \theta$........(ii)

and acceleration relation can be written as :

$v_x=A\sin \theta$.....(iii)

Here $v_y$ and $a_y$ are the velocity and acceleration of the block, respectively, in the direction perpendicular to inclined surface,

Approach 2 : We consider the motion of the block parallel to incline and perpendicular to inclined surface. Let the components of acceleration of block with respect to ground along these direction are $a_x$ and $a_y$, respectively.

Then we can write $a_y=A\sin \theta$

Sol for wedge:

$N\sin \theta =MA$

For block : considering the block in the direction perpendicular to sloping surface.

$mg\cos \theta -N-{ma}_y$

But $a_y=A\sin \theta$

Hence,

$mg\cos \theta -N=mA\sin \theta$

From (i|) and (ii), we get

$A=\frac{mg\sin \theta \cos \theta }{M+m{sin}^2\theta }$