Question

In the arrangement shown in the figure, the rod of mass $m$ held by two smooth walls, remains always perpendicular to the surface of the wedge of mass $M$. Assuming all the surfaces are friction-less, find the acceleration of the rod and that of the wedge.

Answer 1

angle of inclination is '$\alpha$'.

let acceleration of wedge be $a_1$.

and acceleration of rod be $a_2$.

Now,

consider F.B.D of rod,

$mgcos\left(\alpha \right)-N_2=m{a}_2$

therefore,

$N_2=mgcos\left(\alpha \right)-m{a}_2$ eqn 1

consider F.B.D of wedge,

$N_1=Mg+N_{2}cos\left(\alpha \right)$ eqn 2

$N_{2}sin\left(\alpha \right)=M{a}_1$ eqn 3

Using constraint equation,

we've,

$a_{2}sin\left(\alpha \right)=a_1$ eqn 4

from eqn3 and eqn4 ,

we've, $N_2=\frac{M{a}_1}{sin\left(\alpha \right)}$

also , $a_2=\frac{a_1}{sin\left(\alpha \right)}$

therefore, $N_2=M{a}_2$ eqn 5

putting eqn5 in eqn1,

$M{a}_2=mgcos\left(\alpha \right)-m{a}_2$

ANS:

$a_2=\frac{mgcos\left(V\right)}{(m+M)}$

also,

$a_1=a_{2}sin\left(\alpha \right)=\frac{mgcos\left(V\right)}{(m+M)}sin\left(\alpha \right)$