Question

A block is kept on a smooth inclined plane of the angle of inclination $30^\circ $ that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let $F_{1}$ be the contact force between the block and the plane. Now the inclined plane stops and let $F_{2}$ be the contact force between the two in this case. Then $\frac{F_{1}}{F_{2}}$ is

• A. $1$
• B. $\frac{4}{3}$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

When block does not slide with respect to the inclined plane,
$F_{1}=$ normal reaction $=mgcos \theta +masin⁡\theta $
$\because a=gtan \theta $
$\therefore F_{1}=mgcos \theta +m\left(g tan ⁡ \theta \right)sin⁡\theta $
$=mgsec \theta =mgsec⁡30^\circ =\frac{2 m g}{\sqrt{3}}$
When incline plane is stationary,
$F_{2}=mgcos \theta =mgcos⁡30^\circ =\frac{\sqrt{3}}{2}mg$
$\therefore \frac{F_{1}}{F_{2}}=\frac{4}{3}$