In the arrangement shown in figure pulleys are mass of block A,b and C is `m_(1) = 5 kg m_(2) = 4 kg` and `m_(3) = 2.5 kg` respectively .Co-efficient acceleration of friction for both the plane is `mu = 0.50` Calculate acceleration of which block when system is released from rest

Answer

`a_(1) = 4ms^(-2)` b. `a_(2) = 0` c. `a_(3) = 2ms^(-2)`

Explanation

When the system is released , block C tries to move vertically downward due to its own weight in the process blocks A and B are pulled by tension in the thread Hence block A tries to move down the plane and block B leftward.
Let the magnitude of acceleration of A and B be a and b respectively along the direction dicassed above . Then acceleration of block C will be `((a + b)/(2))` downward considering free body diagram

For force on block A `N_(1) = m_(2)g cos theta = 40 N`
`m_(2)g sin theta + T - mu N_(1) = m_(1) a or T - 10 = 5a`...(i)
For force on block C
`m_(2)g- 2T = m_(3)((a + b)/(2)) rArr 25 - 2T = 1.25 (a + b)`....(ii)
For force on blockB
`N_(2) = m_(2)g = 40 N `
`T - mu N_(2) = m_(2)b or (T - 20) = 4b`...(iii)
From above three equations
`T = (460)/(41) N, a = (174)/(41)ms^(-2)` and `b = (90)/(41) ms^(-2)`
Negative value of b indicans that block b does not slide along the assumed direction which is leftward from geomery it is clear that B camon move rightward . Therefore the block B remain stationary B is due to the fact tension T in thread is less than the limitiung friction `muN_(2)`
Hence `b = 0`
Considering free body diagram again

for force on block `A N_(1) = m_(2) g cos theta = 40` newton
`T + m_(2)g sin theta - muN_(1) = m_(1)a`
or `(T + 10) = 5a`....(iv)
For force on blocks `C m_(2)g - 2T = m_(3)((a)/(2))`
`(25 - 2T) = 1.25a`...(vi)
From Eqs (v) and (vi)
`T = 10 N and a= 4 ms^(-2)`
Acceleration of block `A a_(1) = 4ms^(-2)`
Acceleration of block `B a_(2) = 0`
Acceleration of block `CB a_(3) = 2ms^(-2)`