All surfaces are smooth in following figure. Find `F`, such that block remain stationary with respect to wedge.

Options

A. `F=(M+m)g tan theta`
B. `F=(M-m)g tan theta`
C. `F=(M+m)/g tan theta`
D. `F=(M+m)g sin theta`

Answer

A

Explanation

Acceleration of (block+ wedge)
`a=(F)/((M+m))`
Let us solve the problem by both the methods.
(i) From Inertia Frame of Reference (Ground) FBD of block w.r.t. ground (apply real forces) with respect to ground, block is moving with an acceleration a

`therefore sumF_(y)=0`
`implies N cos theta=mg`...........(i)
and `sumF_(x)=ma`
`implies N sin theta=ma`............(ii)
From Eqs. (i) and (ii)
`a= g tan theta`
`therefore F=(M+m)a `
`=(M+m)g tan theta`
(ii) From Non-inertial Frame of Reference (Wedge) FBD of block w.r.t. wedge (real forces + pseudo force)
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w.r.t. wedge, block is stationery
`therefore sumF_(y)=0 implies N cos theta=mg`............(iii)
`sum F_(x)=0 implies N sin theta= ma`...............(iv)
In this way the block's net acceleration becomes zero because all the forces acting on them balance each other for an observer on the wedge and not for an observer on the ground.
From Eqs. (iii) and (iv) , we will get the same result
i.e., `F=(M+m)g tan theta`.