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If the coefficient of friction between `A` and `B` is `mu`, the maximum acceleration of the wedge `A` for which `B` will remain at rest with respect to the wedge is

A

`mug`

B

`g((1+mu)/(1-mu))`

C

`(g)/(mu)`

D

`g((1-mu)/(1+mu))`

Text Solution

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The correct Answer is:
B, C, D


FBD of block B w.r.t. wedge A, for maximum `a`:
Perpendicular to wedge:
`sumf_y^`=(mgcostheta+masintheta-N)=0`
and `sumf_x^`=mgsintheta+muN-macostheta=0` (for maximum a)
`impliesmgsintheta+mu(mgcostheta+masintheta)-mcostheta=0`
`implies a((gsintheta+mugcostheta))/(costheta-musintheta)`
For `theta=45^@`
`a=((tan45^@+mu)/(cot45^@-mu))`,`a=g((1+mu)/(1-mu))`
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