- #1
Kugan
- 3
- 0
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 let's say a car going around a banked curve) Fnety=0 so we can't 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.
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 let's say a car going around a banked curve) Fnety=0 so we can't 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.