Question

Two blocks of masses $M$ and $m$ are held on a smooth wedge. The wedge is placed on smooth plane and is free to move on plane. Now both blocks are released, and it is found that block $m$ remains stationary w.r.t the wedge, the ratio of $M/m$ is $6k$. Find the value $k$.

Answer 1

$A$: acceleration of the wedge w.r.t. the ground. (in leftward direction)
a: acceleration of $M$ w.r.t. the wedge.
Equation of mass $m$
$N\cos {60}^o=mA$ ...$\left(1\right)$
$N\sin {60}^o=mg$ ...$\left(2\right)$
From equation $\left(2\right)/\left(1\right)$
$A=g\cot {60}^o=g/\sqrt{3}$ ...$\left(3\right)$
Solving $\left(1\right)$ and $\left(2\right)$
$N=2mg/\sqrt{3}$ ...$\left(4\right)$
Equation of the wedge: $N_1\sin {30}^o\_N\sin {30}^o=5mA$ (w.r.t. wedge)
$\frac{N_1}{2}\_\frac{mg}{\sqrt{3}}=\frac{5mg}{\sqrt{3}}$ $\Rightarrow N_1=\frac{12mg}{\sqrt{3}}$
Equation of mass $M$: $N_1+MA\cos {60}^o=Mg\cos {30}^o$, perpendicular to the wedge.
$\frac{12mg}{\sqrt{3}}+\frac{mg}{2\sqrt{3}}=\frac{\sqrt{3}Mg}{2}$ $\Rightarrow M=12m$
$M:m=12:1$