IIT JEE , AIEEE Forum - Physics , Mechanics

INDIAN_ARMY19890

hi buddy
the solution is like this
when you will make free body dyagrame(FBD) OF BLOCK 1 YOU WILL GET THIS eq
T-M1g=M1a ..............(1) here T is tension in the string joing M1 to the second pulley a is acceleration of the block M1 wrt ground (or an inertial frame )
in the same way making FBD of M2 you will get this eq
T"-M2g=M2(a"-a) here see we have let the acceleration of the system containg M1 nd M2 a" but the net accleration of block will be a"-a
similarly making FBD of M3 u will get this eq
M3g-T"=M3(a"+a) of course the the net acceleration of M3 will be a+a"
you will get 3rd eq by making FBD of second pulley
the equation will be
T=2T" (as mass of pulley is very small)

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