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The mass of wedge, shown in figure is M ...

The mass of wedge, shown in figure is M and that of the block is m. Neglecting friction at all the places and mass of the pulley. Calculate the acceleration of wedge. Thread is inextensible.

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Method 1: Let the acceleration of wedge be a rightward. Then acceleration of wedge be a rightward. Then acceleration of block relative to wedge will aslo be a, but down in incline. Hence the net acceleration of the block will be equal to the vector sum of these two as shown in fig.

`(vec(a)_(m))_(x)=(vec(a)_(mM))_(x)+(vec(a)_(m))_(x)`
`(vec(a)_(m))_(x)=a cos theta+a=-a(1+cos theta)`
`(vec(a)_(m))_(y)=(vec(a)_(mM))_(y)= a sin theta + 0=a sin theta`
Now considering FBDs.
For the wedge: `N_(2)=N_(1) cos theta + T sin theta+Mg`...(i)
`T+T cos theta-T sin theta= ma sin theta=Ma` (ii)
For the block:
`N_(1) sin theta-T cos theta = m(a+a cos theta)`...(iii)
`mg-N_(1) cos theta-T sin theta = ma sin theta`..(iv)
from (iii), `N_(1)=(T cos theta +m(a=a cos theta))/(sin theta)`
Putting the value of `N_(1)` in (iv)
`mg-(cos theta)/(sin theta)(T cos theta+m(a+a cos theta))-T sin theta=ma sin theta`
Simplify to get `T=mg sin theta-ma(1+cos theta)` (v)
From (ii) and (iii), `T=Ma+m(a+a cos theta)` (iv)
Solving (v) and (vi), we get `a=(mg sin theta)/(M+am(1+cos theta)`
Method 2: The acceleration of wedge and block will be same in a direction perpendicular to the incline of wedge which is a `sin theta`

Writing the equation of block for this direction, we get
`N_(1) -mg cos theta=ma sin theta`..(viii)
Now solving (ii), (iii) and (iv) we get the same answer as earlier.
Method 3: If we consider block and wedge as a system, then the only horizontal force acting on the system is tension in the string as shown in fig.
Writing the equation for the combined system in horizontal direction, we get
`T=Ma + m (a+a cos theta)` ...(viii)

Consider the motion of block parallel to sloping surface, and writing the equation of motion w.r.t. ground
`(vec(a)_(m)_(x'))=(vec(a)_(m,M))+(vec(a)_(M))_(x')=a_a cos theta`
`mg sin theta-T = m(a+a cos theta)`
`T=mg sin theta-m(a+a cos theta)`...(ix)
Solving (viii) and (ix), we get
`a=(mg sin theta)/(M+2m(1+cos theta)`
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