Find the mass M of the hanging block in the figure which will prevent the smaller block from slipping over  the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

Ur question is not so clear to me(the figure)... so changed some things (Let the triangular block be M, smaller block on the wedge be m)

Let the mass of hanging block be  m`

acceleration of system is a=(m`g)/(m+M+m`)

The mass m will not slide over M if

substituting,

cross substitute and simplify for m`

u get

cheers !! rate if u like

Mg-T=M(acc)___________[1]

T=(M'-m)(acc)____________[2]

Mg=(M'+M+m)(acc.)

(acc.)=g(tan@)

Mg=(M+M'+m)g(tan@)

M/tan@=(M+M'+m)

Mcot@-M=M'+m

M(COT@-1)

M=(M'+m)/(cot@-1)

actually diagram is clear click on it and open it with the appropriate application

u will get the clear diagram

there r no solved sums on the link dat u gav varun

sorry dost mera vista hai na toh download karne par , it asks for various virus tests and stuff. Anyway soln is right.

ya buddy answer is right juzz u exchanged the variables actually hanging block is M & block on the incline is of mass m

varun 33 ke pehle method mein ma.cos = mg.sin kaise aaya

FBD diagram thoda help karega.......