In fig. mass `m` is being pulled on the incline of a wedge of mass M. All the surfaces are smooth. Find the acceleration of the wedge.

Options

A. `a=(F(1-cos theta)mg 2sin theta cos theta)/(M+2m sin^(2)theta)`
B. `a=(F(1+cos theta)+mg sin theta cos theta)/(M+m sin^(2)theta)`
C. `a=(F(1-cos theta)-mg sin theta cos theta)/(M+m sin^(2)theta)`
D. `a=(F(1+cos theta)-mg sin theta cos theta)/(M-m sin^(2)theta)`

Answer

C

Explanation

Method -1: Let the acceleration of block relative to wedge be `vec(a)_(mM)=vec(a)` and acceleration of wedge on ground be `vec(a_(M))=vec(A)`. Free body diagram as in fig.

Equation of motion of M,
`F-N sin theta-F cos theta=MA` ...(i)
Equation of motion of m in y-direction,
`N-mg cos theta=ma_(y)=mA sin theta` ..(ii)
Substituting the value of N from (ii) in (i), we can get the value of A.
`A=(F(1-cos theta)-mg sin theta cos theta)/(M+m sin^(2)theta)`
Method 2: If we make the FBD of m w.r.t. M, pseudo force `mA` is to be applied because M is non-inertial. W.r.t. M, there is no acceleration of m in y-direction, so balancing the forces in y-direction.
`N=mg cos theta+mA sin theta`
which is same as (ii).

Method 3: Taking block and wedge as system.
External force acting in horizontal direction is F.
`Sigma F_(x)=m(a_(m))_(x)+M(a_(M))_(x)`

Let the acceleration of block w.r.t. wedge is a. or `|vec(a)_(m,M)|=a`
`(vec(a)_(m))_(x)=(vec(a)_(m,M))_(x)+(vec(a)_(M))_(x)`
`|(vec(a)_(m))_(x)|=a cos theta +A`
Hence,
`F=m(a cos theta+A)+MA`
Now analyzing the sloping surface and writing equation w.r.t ground,

`Sigma vec(F)_(x') = m(vec(a)_(m))_(x')`
`(vec(a)_(m))_(x')=(vec(a)_(m,M))_(x')+(vec(a)_(m))_(x')`
`=a+A cos theta`
`F+mg sin theta = m(a+A cos theta)` ...(ii)
From (i) and (ii) ,
`A=(F(1-cos theta)-mg sin theta cos theta)/(M+m sin^(2)theta)`.