Applied Force to a Sliding Frictionless Ramp

AI Thread Summary
A block on a frictionless ramp requires a specific horizontal force to prevent it from sliding. The correct force is calculated as (m+M)*g*tan(θ). The discussion highlights the importance of analyzing the forces acting on both the block and the ramp system. It confirms that the block and ramp share the same horizontal acceleration, allowing for the application of Newton's laws. Properly determining the normal force is essential for solving the problem accurately.
Rook2012
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Homework Statement



Hello all, kinda angry at myself for posting this but my brain has hit a brick wall.
This is an easy problem too... anyway.

A frictioness ramp of mass M and incline θ sits on a frictionless surface. A block of mass m sits on the ramp. What horizontal force F must be applied to the ramp to ensure the block does not move?

Homework Equations



F = ma
Newton's Third Law

The Attempt at a Solution



I know the answer is (m+M)*g*tan(θ). But I'm having trouble making myself believe it.

I started with a force diagram of the block, the only 2 forces acting on it are gravity directly downward and the normal force (N) at an angle of θ above the horizontal.

So the net force on the block is F_{block} = N sin(θ) i + (mg-(N cos(θ)) j = ma

Then I looked at the ramp+block system and determined
F_{system} = (M+m)a_{system}

Here for some reason the gears stop turning. I went on to say that the y components on the block sum to zero, so the Normal force is \frac{mg}{cosθ}.

From here I substitued in for N in the x direction and then equated "a" in both equations but I know that's wrong. Got me the right answer, but its the wrong way of doing it. Any help would be appreciated.
 
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Come to think of it, equating the acceleration of the system and the block is alright isn't it? They are both moving at the same acceleration which is all in the x direction.

If I let ma = Nsin(theta) because the forces in the vertical direction cancel, and then solve for "a" as I did... maybe that it the right way to go about it. Let me know if I'm delusional, peace.
 
Rook2012 said:
Come to think of it, equating the acceleration of the system and the block is alright isn't it? They are both moving at the same acceleration which is all in the x direction.
yes, correct
If I let ma = Nsin(theta) because the forces in the vertical direction cancel, and then solve for "a" as I did... maybe that it the right way to go about it. Let me know if I'm delusional, peace.
yes, just be sure to determine N by looking at sum of forces in the vertical y direction = 0.
 
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