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Question: A block is kept on a smooth inclined plane of angle of inclination 30<sup>0</sup> that moves with a ...

A block is kept on a smooth inclined plane of angle of inclination 300 that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let F1 be the contact force between the block and the plane. Now the inclined plane stops and let F2 be the contact force between the two in this case. Then F1/F2 is –

A

1

B

4/3

C

2

D

3/2

Answer

4/3

Explanation

Solution

When block does not slide w.r.t. incline plane,

F1 = normal reaction = mg cos  + ma sin 

 a = g tan 

 F1 = mg cos  + m(g tan ) sin 

= mg sec  = mg sec 300 = 2mg3\frac{2mg}{\sqrt{3}}

When incline plane is stationary,

F2 = mg cos  = mg cos 30ŗ = 3mg2\frac{\sqrt{3}mg}{2}

F1F2\frac{F_{1}}{F_{2}} = 43\frac{4}{3}