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

In summary, the conversation discusses whether the normal force on a mass on an inclined plane depends on the chosen coordinate system. It is concluded that the normal force remains the same regardless of the coordinate system used. The conversation also mentions a similar scenario of a rotating rod about the edge of a table and clarifies the misconception about the vertical forces adding up to zero.
  • #1
cherian
3
0
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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  • #2
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 [tex] N = mg\cos\theta [/tex] where [tex] \theta [/tex] is the angle of inclination . In the second case remember [tex] \cos0 = 1 [/tex] because the plane is inclined at 0 degrees.
Sorry if I have misunderstood your question.
 
  • #3
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)
 
  • #4
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.
 
  • #5
I apologize, there was lack of thinking before I posted it, My bad
 
  • #6
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.)
 

1. What is normal force on an inclined plane?

The normal force on an inclined plane is the perpendicular force exerted by the surface of the inclined plane on an object placed on it. It is equal in magnitude and opposite in direction to the force of gravity acting on the object.

2. How do you calculate normal force on an inclined plane?

To calculate normal force on an inclined plane, you can use the equation FN = mg cosθ, where FN is the normal force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of inclination.

3. Does the normal force change on an inclined plane?

Yes, the normal force changes on an inclined plane based on the angle of inclination. As the angle increases, the normal force decreases and vice versa.

4. What is the relationship between normal force and friction on an inclined plane?

The normal force and friction on an inclined plane have an inverse relationship. As the angle of inclination increases, the normal force decreases and the friction force increases.

5. Can normal force be greater than the force of gravity on an inclined plane?

Yes, the normal force can be greater than the force of gravity on an inclined plane if the angle of inclination is less than 90 degrees. However, if the angle is greater than 90 degrees, the normal force will be less than the force of gravity.

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