# Question with Normal Force

1. Jun 26, 2007

### Gear300

1. The problem statement, all variables and given/known data
I was doing a problem with normal force on an inclined plane and found the normal force by finding the y-component of the gravity vector. Then I also found that if I find the components of the normal force, I can also get another answer.
-How are the 2 different equations available for the normal force the same

2. Relevant equations
Fg = mg
Fgy = -mgcos(theta)

Fny = Fncos(theta)

3. The attempt at a solution
Fgy = -mgcos(theta)
Fn - mgcos(theta) = 0 'movement along plane; not vertically
Fn = mgcos(theta)

Fny = Fncos(theta)
Fncos(theta) - mg = 0 'movement along plane; not vertically
Fn = mg/cos(theta)

I got 2 equations for Fn; Fn = mgcos(theta), Fn = mg/cos(theta)...I'm not understanding how the 2 are the same; can anyone help

Last edited: Jun 26, 2007
2. Jun 26, 2007

### andrevdh

The object is accelerating, $$a$$, along the direction of the incline.

This means that in a direction perpendicular to the incline it is not accelerating as you correctly assumed in the derivation of your first formula.

One can decompose the acceleration along the incline into x- and y- components though. This means that in the x and y direction one cannot assume that the acceleration of the object is zero. So your second formula is invalid.

3. Jun 26, 2007

### Gear300

I see; thanks. But, in that case, the second equation would work if lets say the object was moving perpendicular to the plane (such as a car on a banked road), right?

4. Jun 26, 2007

### andrevdh

Yes, because the resultant acceleration is horizontal and there is then no vertical accelleration component.

5. Jun 26, 2007

### Gear300

Alright; thanks