Two blocks of mass M and 3M (in kg) are connected by an inextensible light thread which passes over a light frictionless pulley. The whole system is placed over a fixed horizontal table as shown.The coefficient of friction between the blocks and table surface is `mu`. The pulley is pulled by a string sttached to its centre and accelerated to left with acceleration `a((m)/(s^2))`. Assume that gravity acts with constant acceleration g`((m)/(s^2))` downwards through the plane of table. Then:
(a) Find the forizontal accelration of both the blocks (Assume that both the blocks are moving.)
(b) What is the maximum acceleration a, for which the block of mass 3 M will remain stationary?

Explanation

(a) Let the acceleration of blocks of mass M and 3 M are `a_1` and `a_2` respectively. From constraint relation
`a_(pulley)=(a_1+a_2)/(2)=a` .(i)
F.B.D. of block of mass M
`T-muMg=Ma_1` .(ii)
F.B.D. of block of mass 3 M
`T=(-3mu)/(Mg)=3Ma_2` .(iii)
From (ii) and (iii):
`-muMg=3muMg=Ma_1-3Ma_2`
`implies2mug=a_1-3a_2`
From (i) and (iv)
`2mug=(2a-a_2)-3a_a`
`impliesa_2=(a-mug)/(2)`
From (i) and (v)
`a_1=2a-a_2=2a-((a-mug)/(2))=(3a)/(2)+(mug)/(2)=(3a+mug)/(2)`
(b) Since `a=(a_1+a_2)/(2)`
Acceleration a will be maximum when `a_1` is maximum (since`a_2=0`)
Acceleration `a_1` is maximum when tension in the string is maximum for block of mass 3 M to be stationary
`Tle3muMg`
`T_(max)=3muMg`
For block of mass M
`T_(max)-muMg=Ma_1`
`implies3muMg-muMg=Ma_1`
`implies2mug=a_1`
So maximum acceleration
`a=(a_1+a_2)/(2)=(2mug+0)/(2)=mug`