Question

A rod of mass m is supported on a wedge of mass M. Shown in $Fig.6.159$. Find the accelerations of rod a and wedge A in the arrangement. The friction between all contact surface is negligible.

Answer 1

The rod is constrained to move in the vertical direction (with the help of the guides) and the wedge will move along the surface in the horizontal direction. Initially, the system is held at rest.

Constraint relation : Approach 1

Let the acceleration of m w.r.t. ground be a vertically downwards and acceleration of M w.r.t. ground be A horizontally towards right.

The motion of the system is constrained by the fact that the 'bottom face of the rod must always be in contact with the inclined plane." If in time t, X is the displacement of the wedge and x is the displacement of the rod, then the constraint demands that

From Fig, we have $\frac{x}{X}=\tan \alpha$

$\therefore$ $x=X\tan \alpha$........(i)

Here $\alpha$ remains constant.

Differentiating (i) w.r.t. t twice, we get

$\left(\frac{d^{2}x}{{dt}^2}\right)=\left(\frac{d^{2}X}{{dt}^2}\right)\tan \alpha$ [$\tan \alpha$=constant]

Hence, $a=A\tan \alpha$

Approach 2 :

The fact that the rod (or a particle on the wedge) and the wedge must not lose contact is usually called wedge constraints. For this, the component of the acceleration of the rod perpendicular to the wedge plane= component of acceleration of the wedge perpendicular to wedge plane.

$a\cos \alpha =A\sin \alpha \Rightarrow a=A\tan \alpha$

The force acting on the rod are:

$\bullet$ The weight mg, vertically downwards.

$\bullet$ The normal force N, normal to the bottom surface of the rod

$\bullet$ The force ($F_{guide}$) exerted by the guide to nullify the horizontal component of N as for the rod $a_{horizontal}=0$.

Motion in vertical direction

$mg-N\cos \alpha =ma$......(ii)

and the forces acting on the wedge are

$\bullet$ The weight, Mg

$\bullet$ reaction of N acting on the rod

$\bullet$ $N_1$, normal force by the surface

The force equations are

$N_1-Mg-N\cos \alpha =0$.......(iii)

and $N\sin \alpha =MA$......(v)

Solving the above equations for a and A, we get

$a=\frac{mg\tan \alpha }{m\tan \alpha +M\cos \alpha }$ and $A=\frac{mg}{m\tan \alpha +M\cos \alpha }$