Can I figure out the force exerted on an object if i know

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To calculate the force exerted by a block sliding down an inclined plane on a frictionless surface, the mass of both the block and the plane, as well as the angle of the incline, are essential variables. The discussion emphasizes using Newton's second law (F=ma) to determine the force required for the inclined plane's acceleration. The normal force acting on the inclined plane is derived from the x component of this force, which is influenced by the block's motion. Clarifications about the direction of forces indicate that while the ramp moves left, it does not slide down; thus, only the horizontal component of the normal force affects the ramp's movement. Overall, the problem can be solved without additional complexities, focusing solely on the normal force's x component.
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Homework Statement


a block is sliding down an inclined plane that is resting on a frictionless terrain. so as the block slides down the ramp, the ramp also moves. Calculate the force exerted by the block on the plane.

I have the mass of both the object and the plane as variable m and M, the angle of contact theta, and using Lagrangians (or is it Hamiltonians... I am not quite sure at this point of the difference. feel free to explain that to me for extra credit lol :) but using L = T-U I have found the equation for the acceleration of x and X.

Homework Equations


f=ma

The Attempt at a Solution


I am thinking that I just need to say F=ma. the force required to accelerate the the inclined plane to the left is its mass M times its acceleration. That will give me the x component of the Normal Force acting on the Inclined plane by the block. Then using trigonometry, I can solve for the Normal Force. and I believe that is what the problem is looking for.
I am not entirely sure though. a little voice in my head is telling me that that is too easy. that maybe the mass of the little block should also be a factor in calculating the required force. (the mass is present in the equation of acceleration btw, but I am not sure if it should be part of the m in the F=ma equation.).

What do you think? is my first impulse a very wrong oversimplification... or is it fine?
 
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To be sure, you can draw a free body diagram. The Normal Force acts perpendicular to the incline, but the ramp does not slip in the direction of the Normal Force.
 
Jazz said:
To be sure, you can draw a free body diagram. The Normal Force acts perpendicular to the incline, but the ramp does not slip in the direction of the Normal Force.
what do you mean? it does slip in the direction of the normal force doesn't it? if the block slides down to the right and the ramp slides to the left.
 
grandpa2390 said:
what do you mean? it does slip in the direction of the normal force doesn't it? if the block slides down to the right and the ramp slides to the left.

The ramp does not slide down to the left, but straight to the left instead (the Normal Force is not straight to the left). There must be a (component) net force toward that direction that is being supplied by the block.
 
grandpa2390 said:
what do you mean? it does slip in the direction of the normal force doesn't it? if the block slides down to the right and the ramp slides to the left.

well of course. hence why I said it would be the x component of the normal force. but is that all I need to do. I don't have to factor in the mass of the small block or do another lagrangian or anything?
 
As far as I understand the problem, nope. You don't need to do anything else.

As long as the box is on the incline, only the horizontal component of the normal force will make the ramp to move to the left (until the lower right edge of the box reaches level ground).
 
grandpa2390 said:
well of course. hence why I said it would be the x component of the normal force. but is that all I need to do. I don't have to factor in the mass of the small block or do another lagrangian or anything?
As a check, how about you post your answer as an algebraic expression?
 
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