Acceleration of a Block Sliding Down a Wedge

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Bling Fizikst
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Consider the wedge as '1' and block as '2' : i need to find ##a_{2/F}=a_{2/1}+a_{1/F}## .
The FBD of '1' : $$Mg\sin\theta - f=Ma_{1/F}$$ $$Mg\cos\theta=N$$
The FBD of '2' : $$mg - T =ma_{2/1}$$
Assuming kinetic friction : ##f=\mu Mg\cos\theta##. But how do i find ##T##?
 
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You are overlooking some forces. The string exerts a force on the pulley, while the wedge and block exert contact forces on each other.
You also have the constraint that the block does not penetrate the wedge.

But, unless I am missing something, the question is much simpler than it looks. Think about how the block moves relative to the wedge.
 
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Is this correct? Seems fairly complicated . How do i handle all these tension forces , moreover since it's an ideal pully , the net force acting on it would be zero .
 
Bling Fizikst said:
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Is this correct? Seems fairly complicated . How do i handle all these tension forces , moreover since it's an ideal pully , the net force acting on it would be zero .
The net force on the pulley is zero, but the string exerts a force on it, so that force is transferred to the wedge.
As I wrote, think first about the relative motions of the two bodies.