Acceleration of a Block Sliding Down a Wedge

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To find the acceleration of the block sliding down the wedge, the equations of motion for both the wedge and the block must be analyzed, considering forces such as gravity, normal force, and friction. The tension in the string needs to be determined, as it affects the forces acting on the block and the wedge. The interaction between the block and the wedge is crucial, particularly since the block cannot penetrate the wedge, which introduces constraints on their relative motion. Despite the complexity of the system, understanding the relative movements simplifies the problem significantly. Analyzing these forces and constraints will lead to the correct expression for the accelerations involved.
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.
 
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