11If you consider the FBD of the small block:
mgsinC-T=ma(where a is the acceleration of the small block and subsequently the wedge.)
or, T=mgsinC-ma
If we consider the FBD of the wedge:
T-TcosC+mgsinCcosC=Ma
By replacing the value of T we get:
mgsinC-ma-mgsinCcosC+macosC+mgsinCcosC=Ma
By solving we get:
a(acceleration)=mgsinCM+m-mcosC(Ans.)
Anik Chatterjee i dont think that acc. of wedge=acc. of small block..
i think it must be acc of block=(acc of wedge)sinCUpvote·0· Reply ·2013-05-29 01:56:38
Akash Anand Acc. of block and acc. of wedge will be different. Acc. of block will have an acceleration down the inclined along with in horizontal direction as well. 