Minimum value of F so that 'm' falls freely is given by? And also find acceleration with which Wedge M moves and the contact force b/w m and M

Dear Student ,
Here in this find the diagram in the below :

As we can see from the diagram :
For block of mass m :mgsinθ+macosθ =ma' ....(1)N =masinθ-mgcosθ .....(2)For block M :Nsinθ = Ma ....(3)So, N =Masinθ......(4)Substituting 4 in 2 we have ,Masinθ=masinθ-mgcosθMa=masin2θ-mgsinθcosθF =masin2θ-mgsinθcosθF=masin2θ-mgsin2θ2=Minimum value of force(F) so that m falls freely.Now from this above equation ,msin2θ-Ma=mgsinθcosθa=mgsinθcosθmsin2θ-M =Acceleration of the wedge 
As the block m falls freely so the inclined plane of the wedge is smooth. So there is no contact force between m and M.
Regards

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Jardini
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