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A system is pushed by a force F as shown...

A system is pushed by a force `F` as shown in figure All surfaces are smooth except between `B` and `C` is `mu`. Minimum value fo `F` to prevent block `B` from down ward slipping is

A

`((3)/(2 mu))mg`

B

`((5)/(2 mu))mg`

C

`((5)/(2))mu mg`

D

`((3)/(2))mu mg`

Text Solution

Verified by Experts

The correct Answer is:
b

Horizontal acceleration of the system is
`a = (F)/(2m + m+ 2m) = (F)/(2m)`
Let `N` be the normal reaction between B and C free body diagram of `C` gives
ltBRgt `n = 2 MA = (2)/(5)f`
Now B will not slide downward if
`muN ge m_(B)g`
or `mu((2)/(5) f) ge mg` or `F ge (5)/(2mu) mg`
so `F_(min) = (5)/(2 mu)mg`
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