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The upper surface of blokc C is horizont...

The upper surface of blokc C is horizontal and its right part is inclined to the horizontal at angle `37^(@)`. The mass of blocks A and B are `m_(1) = 1.4kg and m_(2)=5.5kg`, respectively. Neglect friction and mass of the pulley. Calculate acceleration a with which block C should be moved to the right so that A and B can remain stationary relative to it.

Text Solution

Verified by Experts

The correct Answer is:
`20//11 ms^(-2)`.
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)`.
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