# General FBD issue

1. Nov 10, 2012

### Kugan

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. Nov 10, 2012

### Simon Bridge

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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