# How do I find the normal force?

1. Feb 25, 2013

### Gamerex

1. The problem statement, all variables and given/known data

Let's say I'm holding a crate up against the ceiling with a force of 'P', and the crate is stationary.
What is the normal force?

2. Relevant equations

Newton's 1st law: ƩF=0

3. The attempt at a solution

I have three forces acting on the crate - The normal force, the gravitational force, and the applied force.

According to Newton's first law, Fg+Fn+Fa=0

Thus, Fn=-Fg-Fa

Since Fg pulls down Fg=-mg
Since Fa pushes up, Fa=P

Thus, Fn=mg-P

However, my book says Fn=P-mg. What am I doing wrong?

2. Feb 25, 2013

### Staff: Mentor

3. Feb 25, 2013

### Gamerex

Why is it I only apply this negative to the normal force? If a downward force is negative, then Fg should be negative too, and

-Fg-Fn+Fa=0

Fn=Fa-Fg=P+mg

Which still isn't the right answer :/

4. Feb 25, 2013

### Gamerex

I made a force table that I can use to analyze the component of each force and the net force that results.

Force - y-component
Fg:____-mg
Fa:____+P
Fn:____mg-P?
__________________
ma:____0

Adding the entries in this table, it seems obvious that the normal force must equal mg-P. Otherwise, the sum of this table wouldn't be zero.

5. Feb 25, 2013

### Staff: Mentor

OK. Here's my version of it, where the positive direction of force is in the upward direction:

Fa-mg-Fn=0

This gives Fn=P-mg

In these equations Fn stands only for the magnitude of the normal force. The direction of the normal force is downward.

6. Feb 25, 2013

### tms

It will be easier to set up the forces so that they are all positive, and you deal with direction explicitly with the signs. Since $F_n$ and $F_g$ are both down, and $F_a$ is up, you should write
$$-F_g - F_n + F_a = 0.$$
From which the book's answer comes. Your answer is correct, too, but the value of $F_n$ is then negative.

Edit: Oops, someone beat me to it.