Question

A block of mass M is placed on the floor of an elevator. With what acceleration the elevator
should descend so that the block exerts a force 1/3Mg on the floor of the elevator?​

Answer 1

Answer:

$\frac{2}{3} g$

Explanation:

Consider the attachment for FBD,

Let the acceleration the elevator was going down with be a,

Since the mass will remain constant,

we have,

M(g-a) = $\frac{1}{3} Mg$

g- a = $\frac{1}{3} g$

a=g-$\frac{1}{3} g$

a=$\frac{2}{3} g$