Question

Find the mass $M$ of the hanging block as shown in the figure which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

Answer 1

Force acting on smaller $\left(m\right)$ and triangular block $\left(N\right)$

Factor on hanging block $\left(N\right)$

so, in order to prevent slipping if smaller block,

$ma\cos \theta =mg\sin \theta$

$\Rightarrow a=g\tan \theta ---\left(1\right)$

Downward force on the black hanging $M$ is $(Mg-T)$ and acceleration is $a$

$Mg-T=Ma\Rightarrow T=Mg-Ma---\left(2\right)$

Consider the forces on the triangular block along the acceleration

now, horizontal force $T$ is cousing the movement of both blocks ($M$ and $m$)

$T=(M+m)a---\left(3\right)$

Solving $2$ and $3$

$Mg-Ma=(m+M)a$

$M=\frac{(m+M)a}{g-a}$

$M=\frac{(m+M)g\tan \theta }{g-g\tan \theta }$

as $(a=g\tan \theta )$

canceling and dividing both $N$ and $D$ by $\tan$

$M=\frac{M^{\prime }+m}{\cot \theta -1}$

mass of hanging volume