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chester20080
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Homework Statement
We have a wedge and at the top of it a block(consider it dimensionless)and the whole system initially is at rest.Given the gravitational acceleration g, the angle φ of the inclined plane,the mass m of the block,the mass M of the wedge,the coefficient of friction fs between the block and the ledge and that there is no friction between the wedge and the floor,find the acceleration needed for the block not to move with regards as the wedge.
Homework Equations
For the friction we have:T<=fs*N,where N is the normal reaction force.
For the block:Wx=mgsinφ,Wy=mgcosφ
The Attempt at a Solution
I considered an observer at rest somewhere out of the system and I thought that in respect to him the system will have a horizontal acceleration when we finally give the system the acceleration we need.I analyzed this acceleration vector to two vertical components,one vertical to the ledge and the other parallel to it.Then I made a free body diagram,drew the forces on the block (N,T,Wb) and analyzed the weight to two vertical components.Of course I considered X,Y axis.Then I wrote ΣFx=max and ΣFy=may (with vectors) and finally I got:ax=(mgsinφ-T)/m,ay=(N-mgcosφ)/m and given that ax=acosφ,ay=asinφ and T<=fs*N and solving for a:a>=(g(sinφ-fs*cosφ))/(fs*sinφ+cosφ) so the acceleration needed is aminimum=(g(sinφ-fs*cosφ))/(fs*sinφ+cosφ).
Furthermore,I want to ask if we must consider if the wedge could rotate due to the torque of the force we have to exert on it for the acceleration.In addition to that I'm confused why I didn't get M to the solution.I considered that as a system,whatever acceleration the block gets,the same the wedge gets and the same the whole system gets,although that bothers me for its physical meaning...What if we consider the wedge weightless?(M=0?)
I am not sure if any of those is correct,so please enlighten me!