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Problem on a moving wedge and acceleration

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    For the friction we have:T<=fs*N,where N is the normal reaction force.
    For the block:Wx=mgsinφ,Wy=mgcosφ


    3. 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!
     
  2. jcsd
  3. Nov 10, 2013 #2

    haruspex

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    There will be a range of accelerations for which the block will stay in place on the wedge. They may all be in the direction the wedge points, but 0 may be in the range, meaning there's a range of accelerations in each direction.
    ... which could be negative.
    You are asked for an acceleration, not a force. You must therefore assume the acceleration is achieved in whatever way necessary.
    Again, you're finding an acceleration, not a force. It follows that the masses of the wedge and block are irrelevant.
     
  4. Nov 10, 2013 #3
    Thank you very much for your answers!Actually,our professor asked us to find the minimum acceleration needed for the problem and the acceleration to be horizontal.Also,he didn't seem to care for any torques,but the correct isn't that we should?And my answer you think is correct(for amin?)
     
  5. Nov 10, 2013 #4

    haruspex

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    Your answer is correct except that I think you should discuss the case where the expression goes negative. What do you think 'minimum' means in this case? I would think it should be interpreted as minimum magnitude.
    No. The question asks for an acceleration. How that acceleration is achieved is immaterial. If it asked for a force, and where that force should be applied, then you might worry about torque.
     
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