Normal Force on an Inclined Plane: Does it Depend on the Coordinate System?

AI Thread Summary
The normal force on an inclined plane does not depend on the coordinate system used; it is always perpendicular to the surface. For a mass on an inclined plane, the correct expression for the normal force is N = mg cos(θ), regardless of the coordinate system orientation. The confusion arises when incorrectly applying the equations for forces in different coordinate systems, leading to erroneous results. It is essential to ensure that the forces in the perpendicular direction add to zero, which confirms the correct calculation of the normal force. Understanding this principle helps clarify common mistakes made by students in physics.
cherian
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I am a new user so if I am violating any rules I apologize


If we have a mass on an inclined plane ,

Does the Normal force depend on Coordinate used. i.e coordinate system along rod( tilted one) and coordinate system in the regular way

I know the answer is " NO" but I get


N= MgCos(theta) for tilted one
and N= Mg/cos(theta) for straight one

Same is happening for a rod rotating about the edge of a table


Thanks in advance
 
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Hi cherian, welcome to PF.

I am not really sure what you are asking here, but no it doesn't matter. If the inclined plane is stationary the you are correct in saying that N = mg\cos\theta where \theta is the angle of inclination . In the second case remember \cos0 = 1 because the plane is inclined at 0 degrees.
Sorry if I have misunderstood your question.
 
Sorry I can't express it very well. The plane is inclined in both cases. The cases that I mention is the choice of coordinate system. We can select a coordinate system in which X is parallel to the the mass and Y is perpendicular to the mass ( the tilted one) or
We can choose X and Y in regular way ( how we draw in regular way in paper)
 
cherian said:
N= MgCos(theta) for tilted one
and N= Mg/cos(theta) for straight one
I assume you are trying to figure out the normal force (which is always perpendicular to the surface) using different coordinate systems.

Taking coordinates parallel and perpendicular is easiest. Forces in the y-direction (perpendicular to the surface) must add to zero: N - Mg cos(theta) = 0. (Your first result.)

Your second equation comes from setting the vertical forces equal to zero: N cos(theta) -Mg = 0. The problem is: The vertical forces do not add to zero! So that equation is bogus.
 
I apologize, there was lack of thinking before I posted it, My bad
 
No need to apologize for asking a question! :smile: The only thing that matters is whether you received a useful answer.

(You'd be surprised at how many students make that same error! Switching coordinate systems can get confusing.)
 
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