Two block A and B of mass M & 3M are connected through a light string. One end of the string is connected to the blcok B and its other end is connected to a fixed point S as shown in fig. Now a force F is applied to block B. find the acceleration of block A & B.

Explanation

Taking reference line through support S let `x_(A)` and `x_(B)` are the distances of blocks A and B, respectively, from S. The total length of the string, `l=4x_(A)+5x_(B)+l_(0)`. Where `l_(0)` is some part of string which is over pulley and somewhere else which remains constant.

Differentiating l w.r.t time, we get
`(dl)/(dt)=(d)/(dt)(4x_(A)5x_(B)+l_(0))`
or `0=4(dx_(a))/(dt)+5(dx_(B))/(dt)=(dx_(A))/(dt) = 5/4 (dx_(B))/(dt)`
`(dx_(A))/(dt)=-v_(A)` (-because `x_(A)` is decreasing with time)
`(dx_(B))/(dt)=-v_(B)` (+because `x_(B)` is decreasing with time)
or `v_(A)=5/4 v_(B), also a_(A)=5/4 a_(B)`
Method 2: charge in the length of segment I, `Delta l_(1,2)=0+(x_B)`

Segment II
`Delta l_(3.4)=(-x_(A))+(x_B)`
Segment III
`Delta l_(5,6)=(-x_(A))+(x_B)`
Segment IV
`Delta l_(7,8)=(-x_(A))+(x_B)`
Segment V
`Delta l_(9,10)=(-x_(A))+(x_B)`
Total charge in segment lengths should be zero. Therefore,
`Delta l =Delta l_(1,2)+Delta l_(3,4)+Delta l_(5,6)+Delta l_(7,8)+Delta l_(9,10)=0`
`=x_(B) +[-x_(A)+x_(B)] + [-x_(A)+x_(B)] `
`+[-x_(A)+x_(B)]+[-x_(A)+x_(B)]=0`
`=-4x_(A)+5x_(B)=0`
or `x_(A)=5/4 x_(B) implies v_(A)=5/4 v_(B) implies a_(A)=5/4 a_(B)`
Equation of motion

For A:
`4T=m_(A)a_(A)=Ma_(A)`...(i)
For B:
`F-5T=m_(B)a_(B)=3Ma_(B)` ..(ii)
`a_(B)=(16F)/(73)`
and `a_(A)=5/4 xx((16F)/(73))=(20F)/(73)`.