A block of mass m is at rest relative to the stationary wedge of mass M. The coefficient of friction between block and wedge is `mu`. The wedge is now pulled horizontally with acceleration a as shown in figure. Then the minimum magnitude of a for the friction between block and wedge to be zero is:

Answer

`20//11 ms^(-2)`.

Explanation

Since block A and B remain stationary relative to block C, it means motion of blocks A and B is identical with the block C or acceleration of all the three blocks is same as a horizontally (rightwards). Now considering FBD w.r.t wedge C

For block B,
`N_(2)=m_(2)g or, N_(2)=55 "newton"`
`T=m_(2)a` ...(i)
For block A, `N_(1) cos 37^(@)+T sin 37^(@)=m_(1)g` ..(ii)
`N_(1) sin 37^(@)-T cos 37^(@)=m_(1)a` ..(iii)
Solving above equations
`T=10N`
`N_(1)=20N`
and `a=20//11 ms^(-2)`.