Question

With what acceleration 'a' should the lift descend so that the block of mass M exerts a force $\frac{Mg}{4}$ on the floor of the box?

• A. 4g
• B. g
• C. $\frac{3g}{4}$
• D. 2g

Answer 1

The correct option is C $\frac{3g}{4}$

In lift's frame, the lift is moving down with acceleration a. this means the pseudo force on the block will be Ma in upward direction.

Free Body Diagram of block

If normal by block on box is $\frac{Mg}{4}$ then its reaction will also be $\frac{Mg}{4}$

$\Rightarrow Ma+N=Mg$

$Ma+\frac{Mg}{4}=Mg$

$Ma=\frac{3Mg}{4}$

$a=\frac{3g}{4}$

Ground Frame

The block is at rest with respect to the box which is accelerated with respect to the ground. Hence, the acceleration of the block with respect to the ground is 'a' downward. The forces on the block are

(i) Mg downward (by the earth) and

(ii) N upward (by the floor)

The equation of motion of the block is, therefore

$Mg-N=Ma$

If $N=\frac{Mg}{4}$, the above equation gives a = $\frac{3g}{4}$. The block and hence the box should descend with an acceleration $\frac{3g}{4}$.