Inclined plane normal force varies by coordinate system?

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Discussion Overview

The discussion revolves around the calculation of the normal force acting on a block resting on an inclined plane, exploring how the choice of coordinate system affects the resulting values. Participants examine the implications of using different axes—one aligned with the incline and the other aligned with gravity—on the determination of normal and frictional forces.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that using a coordinate system aligned with the incline yields a different normal force compared to a system aligned with gravity.
  • Another participant points out that in the second coordinate system, the normal force has both x and y components, complicating the analysis.
  • A later reply suggests that the y-component of the normal force can be used to solve for the normal force, but acknowledges an error in their initial vector reflection calculations.
  • Another participant emphasizes the need to consider all forces as vectors and highlights the importance of breaking forces into components, especially when friction is involved.
  • It is mentioned that using the ramp's coordinate system simplifies the problem by requiring only the weight to be broken into components, while the ground system necessitates breaking multiple forces into components.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the normal force based on the chosen coordinate system. There is no consensus on which method is superior or whether the normal force can be equated across systems.

Contextual Notes

Participants acknowledge that the normal force is a vector quantity and that the presence of friction complicates the analysis. There are unresolved aspects regarding the correct application of vector components and the implications of different coordinate systems.

x86
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Suppose we have a block on an inclined plane.

If we choose the x-y axis to be parallel and perpendicular to the inclined plane, then we have

Fy = N - mgcos30 = 0

But if we choose our trivial x-y coordinate system, where y is parallel to the force of gravity, then we get:

Fy = Ncos30 - mg = 0

Of course, we get two different normal forces (and thus different frictional forces), for the exact same type of problem.

What is going on?
 
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In the second coordinate system the normal force is no longer given by Fy. There is also an x-component of the normal force in that system.
 
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DaleSpam said:
In the second coordinate system the normal force is no longer given by Fy. There is also an x-component of the normal force in that system.

Right, but can't we take the y component of the normal force (in coordinate system where x is the ground, not the ramp) (Ncos30) and use this to solve for N? I.e. N = mg/cos30

Should this not solve for the normal force?

EDIT: The problem was my vector reflection math was wrong. Also, as it turns out Fy != 0, rather F(normal direction) = 0

Then you get the same normal for both coordinate systems
 
Last edited:
I think that you are confusing yourself with your notation. The normal force is a vector, not a scalar. So you would not have N=mg/cos30, you would have Ny=mg. However, even that is not sufficient if there is friction as well as the normal force.

You need to break all three forces into components, so in vectors you have N+F+mg=0, and in components you would have Ny+Fy+mgy=0 and Nx+Fx+mgx=0 [where ##\mathbf{g}=(g_x,g_y)=(0,-|\mathbf{g}|)##, and ##\mathbf{N}=(N_x,N_y)=(-|\mathbf{N}|\sin 30,|\mathbf{N}|\cos 30)## and ##\mathbf{F}=(F_x,F_y)=(-|\mathbf{F}|\cos 30,|\mathbf{F}|\sin 30)##].

One of the reasons to use the ramp coordinate system is that you only need to break one vector (the weight) into components. Using the ground coordinate system you need to break both the normal and the friction force into components, and if there is any acceleration then you have to break that into components also.
 

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