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General FBD issue

  1. Nov 10, 2012 #1
    Ok so here is the issue I'm having:

    For problems with inclined planes I have always broken the Fg force down into its components:

    Fgy= mgcos(theta)
    Fgx = mgsin(theta)

    Usually the Fnety= Fgy+ Fn(there is no motion in the Y direction for lets say a car going around a banked curve) Fnety=0 so we cant just say -Fgy= Fn, so Fn = -m(-g)cos(theta) finally arriving at Fn= mgcos(theta)

    If however I were to break the normal force vector into components and not the Gravitational force I get: Fn= mg/Cos(theta)


    Fny= Fncos(theta)
    Fnx = Fnsin(theta)

    Fnety = Fny+ mg;
    0 = Fny+ mg;
    Fny= mg;
    Fncos(theta)= mg;

    Fn = mg/ cos(theta)

    Can someone clarify why, I get two different normal force vectors depending on whether I use the Fg force or the Fn force to orient the axis and break the vector into the components ?

    Thank you.
     
  2. jcsd
  3. Nov 10, 2012 #2

    Simon Bridge

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    Here, let me tidy up:
    If you put your axis so +y is normal to the slope, +x points down the slope, and the slope has angle ##\theta## to the horizontal (as the usual setup) then you will have, in the absence of any other, applied, force:
    * a normal force ##\vec{N}=N\hat{\jmath}##,
    * a weight (gravitational force) ##\vec{w}=mg\sin(\theta)\hat{\imath}-mg\cos(\theta)\hat{\jmath}##; and
    * a friction force ##\vec{f}=-\mu N \hat{\imath}##
    (note - I find it is useful to avoid subscripts where I can.)

    For a car going around a banked curve, there is a net unbalanced force acting radially inwards... so it is not correct to write that ##F_{net}{}_y=0##

    In this situation it may be more useful to adopt a different coordinate system. i.e.
    If you orient your axis so +y = ##-\vec{w}## (i.e. "upwards") and +x pointing at the center of the turn (i.e not into the ground): then the three forces resolve into:
    * a normal force ##\vec{N}=N\sin(\theta)\hat{\imath}+N\cos(\theta) \hat{\jmath}##,
    * a weight (gravitational force) ##\vec{w}=-mg\hat{\jmath}##; and
    * a friction force ##\vec{f}=-\mu N\cos(\theta)\hat{\imath}+\mu N\sin(\theta)\hat{\jmath}##

    Thus, ##F_{net}{}_y = N_y-mg+f_y=0 \Rightarrow N_y=mg-f_y##
    For the sake of an argument - say that ##f_y=0##, then we recover your:
    ##N_y=N\cos(\theta)=mg \Rightarrow N_y=mg/\cos(\theta)##

    So lets see what the issue is:
    Well, clearly, you shouldn't. The trouble is that you are comparing situations that are not equivalent. I suspect you are treating the normal force as an fixed applied force rather than a dynamical force.



    Lets take the case of a block accelerating down a frictionless incline ... in that case, the forces normal to the slope should cancel.

    We would expect to find that ##N=mg\cos(\theta)## using either coordinate system.

    Using gravity to set the coordinates - it is not correct to write ##N\cos(\theta)=mg## because some of the gravity force must be unbalanced to give the acceleration down the slope. In fact ##\vec{N}## will always work so that ##\vec{N}+\vec{w}## will point down the slope.

    Now take the case of a block sitting stationary on a slope with friction.
    In this case normal components will cancel and so will perpendicular components.
    We'd expect to find that ##N=mg\cos{\theta}## and ##\mu N=mg\sin(\theta)## no matter what coordinates we use.

    I think you should be able to do it from there.
     
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