Question

In the arrangement shown, neglect the mass of the ropes and pulley. What must be the value of $m$ to keep the system in equilibrium. There is no friction anywhere.
Given: For $M$, force acting down the plane is $Mg\sin {30}^{\circ }$.

• A. $M$
• B. $2M$
• C. $\frac{M}{2}$
• D. $\frac{M}{4}$

Answer 1

The correct option is C $\frac{M}{2}$

As the block $m$ is in equilibrium
$⟹mg=T^{\prime \prime }c$
For the bottommost pulley, we can say $T^{\prime }=2T^{\prime \prime }$ and $2T=T^{\prime }$
( as pulleys are assumed to be light)
$\therefore T=T^{\prime \prime }=mg$
For mass $M$ on the inclined plane, we can say :
$T=Mg\sin 30=Mg/2$
$mg=Mg/2$
$\therefore m=M/2$