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Two blocks A and B are placed in contace...

Two blocks `A` and `B` are placed in contace on a horizontal surface. Faces of block `A` and `B`, which are in contact , are inclined at `30^(@)` with the horizontal, as shown. There is no friction between block `A` and any surface which is in contact with this whereas friction coefficient between block `B` and the surface is `(1)/(sqrt(3))`. A force `F` is applied in horizontal direction on block `A`. What is the minimum value of `F` at which the block `B` just start moving rightwards ?

A

`(40)/(sqrt(3))N`

B

`(80)/(sqrt(3))N`

C

`100N`

D

For any value of `F`, motion will not start

Text Solution

Verified by Experts

The correct Answer is:
D
`F=Nsin theta, N=(F)/(sin theta)`
` N sin theta = mu (N cos theta+mg)`
`F=mu(F cot theta + mg)`
`F(1-mucot theta) = mumg`
`F=(mu mg)/(1-mucot theta)`

On putting `mu=(1)/(sqrt(3))` and `theta =30^(@_)`
`mu=(1)/(sqrt(3)) theta=30^(@)`
`F = oo`
Therefore motion will not start for any value of `F`.
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