Basic block on wedge problem.

  • #1

Homework Statement



A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is placed on the wedge and a horizontal force F is applied to the wedge. What must the magnitude of F be if the block is to remain at a constant height above the tabletop?

Homework Equations



F=ma of course...
Ffriction = coefficient * normal

The Attempt at a Solution



I know the final answer is (M+m)g tan(theta). This is a practice homework so we were given the answer but no work going up to it
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
Hi coppersauce, welcome to PF.
When you apply force F to the system, the acceleration of the system is
a = F/(ma + Mb)..........(1). It is parallel to the ground.
Since block A is not attached to the block B, due the inertia, it will experience an equal and opposite acceleration. The block A will remain rest when component of the weight of the block A balances the component of the acceleration.
Equate these components and find a. Put this value in equation(1) to get F.
 
  • #3
Hi coppersauce, welcome to PF.
When you apply force F to the system, the acceleration of the system is
a = F/(ma + Mb)..........(1). It is parallel to the ground.
Since block A is not attached to the block B, due the inertia, it will experience an equal and opposite acceleration. The block A will remain rest when component of the weight of the block A balances the component of the acceleration.
Equate these components and find a. Put this value in equation(1) to get F.


Trying to figure it out =/ that's about as far as I originally made it. You know the force on the system has to be ma, so since mass is (m+M), the acceleration of the system is a = F/(m+M). Now I assume since the acceleration is moving forward, the block on top of the wedge would have to move backwards (the reason you stated) with some force, and then the pull of gravity sliding it down would have to cancel that out in order for it to remain at the same height. Now I took Physics too long ago to remember some of the basic things =/ as in does the block on top move backwards with the same force applied to the whole system (and thus has a y component of F(sin(theta))? this is where I'm getting confused) and I also forget the force at which gravity pulls down the block... would it be g(sin(theta)) or g*m*sin theta? Bleh I don't remember any of this from Freshman year of high school I'm embarrassed.

I have a general understanding of the problem I just forget equations/specifics...
 
  • #4
I was being a little silly on the gravity pull part :/ I was thinking in terms of acceleration not in terms of Force (which obv. would have m =/ ). I just haven't done this in so long things are coming back to me.
 
  • #5
rl.bhat
Homework Helper
4,433
8
Component g along the wedge is g*sinθ.
Since block and the wedge move together, the acceleration of the whole system is the acceleration of the block in the opposite direction. Its component along the wedge is a*cosθ and is in the opposite direction to g*sinθ.
For equilibrium of the block g*sinθ = a*cosθ.
Find a and equate it to F/(m+M) to find F.
 
  • #6
Sigh that's so easy... I wish I had a textbook ;( thank you
 

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