1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inclined plane normal force varies by coordinate system?

  1. Mar 2, 2015 #1
    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?
  2. jcsd
  3. Mar 2, 2015 #2


    Staff: Mentor

    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.
  4. Mar 2, 2015 #3
    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: Mar 2, 2015
  5. Mar 2, 2015 #4


    Staff: Mentor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook