Problem on block sliding on a wedge

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    Block Sliding Wedge
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The discussion centers on the dynamics of a block of mass 'm' resting on the hypotenuse of a right triangular wedge of mass 'M'. The normal force acting on the block is defined as N=mg cos(α), where α is the lower angle of the wedge. The horizontal force exerted between the block and the wedge is calculated as F=N sin(α)=mg cos(α) sin(α). Consequently, the acceleration of the wedge is derived as a=F/M=m g cos(α) sin(α)/M. The query regarding the condition under which 'm' remains stationary is clarified to involve its position relative to the wedge.

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gauravkukreja
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Consider a block of mass 'm' kept on the hypotenuse of a right triangular wedge of mass 'M'. Calculate the accelaration of the wedge and the block.
Hence find the force that should be applied to 'M' so that 'm' does not move?
 
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is there no friction?
in this case the interaction force between the block and the wedge is the normal vincular reaction, that is equal to the normal compnent of the block weight that is:
[tex]N=mg\cos\alpha[/tex]
where alpha is the lower angle of the wedge. So the horizontal force between the wedge and the block is
[tex]F=Nsen\alpha=mg\cos\alpha\, sen\alpha[/tex]
Considering the wedge it receives a force equal to F so it moves with an acceleration equal to
[tex]a=\frac{F}{M}=\frac{m}{M}g\cos\alpha\, sen\alpha[/tex]
 
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For the second question I don´t understand if m must not move respect a fix coordinate or respect to the wedge...
 

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