How do I find the normal force?

[Gamerex]

Homework Statement

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?

Homework Equations

Newton's 1st law: ƩF=0

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?

 

[Chestermiller]

F_n is pushing downward, so its sign in your force balance equation should be negative.

 

[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 :/

 

[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.

 

[Chestermiller]

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

F_a-mg-F_n=0

This gives F_n=P-mg

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

 

[tms]

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

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.

 

[xenxander]

Homework Statement

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?

Homework Equations

Newton's 1st law: ƩF=0

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?

You are both the applied force and the normal force, because the object is directly above you and hence you are 90 degrees from the plane. Thus they are the same force.
If you were on an incline, the normal force can be derived via some basic trig.

Fn = Fw / ( sin(normal angle) + cos(normal angle)*tan(angle elevation)

Fw: Force of weight, in newtons, or pounds, etc.
Normal angle: elevation angle + 90 degrees

 

[kuruman]

Please note that this thread is more than 9 years old. Thank you for your contribution, but it is unlikely to help the original poster with his/her homework this late in the game.

 

[xenxander]

Please note that this thread is more than 9 years old. Thank you for your contribution, but it is unlikely to help the original poster with his/her homework this late in the game.

Yes. though if someone were searching for something similar and found this forum, as I had, then perhaps some of these replies, mine or others, will help future seekers.

 

[haruspex]

Yes. though if someone were searching for something similar and found this forum, as I had, then perhaps some of these replies, mine or others, will help future seekers.

Fair enough, but unfortunately you have missed what has confused the OP. It is simply a matter of sign convention. In writing Fg+Fn+Fa=0, Gamerex is taking all forces acting on the box as positive up; the book has taken each force as positive in its actual direction, making the normal force positive down.

 

[kuruman]

Summary for the benefit of future help seekers

Without reference to force direction or any equations, one can use a bit of logic to answer this question. Say the person holding the crate lifts it up at constant velocity until it just barely touches the ceiling. Up to that point, the applied force is equal to the crate's weight, $mg$. The crate cannot be pushed through the ceiling because the ceiling exerts a normal force on it. Any additional increase of the applied force past $mg$, is matched by the opposing normal force. Thus, the normal force is the applied force less the weight of the crate.

 

[haruspex]

Summary for the benefit of future help seekers

Without reference to force direction or any equations, one can use a bit of logic to answer this question. Say the person holding the crate lifts it up at constant velocity until it just barely touches the ceiling. Up to that point, the applied force is equal to the crate's weight, $mg$. The crate cannot be pushed through the ceiling because the ceiling exerts a normal force on it. Any additional increase of the applied force past $mg$, is matched by the opposing normal force. Thus, the normal force is the applied force less the weight of the crate.

See my comment in post #10.

 

[kuruman]

See my comment in post #10.

I saw it before posting #11 which was not meant as a reply or comment to #10 but as an independent summary for the casual visitor to the thread.

 

[xenxander]

Fair enough, but unfortunately you have missed what has confused the OP. It is simply a matter of sign convention. In writing Fg+Fn+Fa=0, Gamerex is taking all forces acting on the box as positive up; the book has taken each force as positive in its actual direction, making the normal force positive down.

Well Fg + Fn = zero when it's vertical. now the Fa has an opposing force, the force from the ceiling pushing back on the person. Do you want to call that Fnc? So that's another normal force. The person holding the object upright is supporting the object exactly with normal force until they apply another force (against the ceiling)
So the person is the normal force, their force applied, Fa if you will, is the force past the normal force. So that's a net applied force, (Fnet = Fa - Fn), is then countered with Fnc. So four forces will equal zero, not three.

 

[kuruman]

We are talking about the forces on the crate. There are 3 entities exerting forces on it:

1. The Earth exerts force mg down on the crate.
2. The person exerts force F_app up on the crate.
3. The ceiling exerts force F_n down on the crate.

Forces are exerted by entities and you get one force per entity. The difference F_app - F_n is not a separate force exerted by a fourth entity. It's just a partial sum of the terms that make up the net force.

 

[xenxander]

We are talking about the forces on the crate. There are 3 entities exerting forces on it:

1. The Earth exerts force mg down on the crate.
2. The person exerts force F_app up on the crate.
3. The ceiling exerts force F_n down on the crate.

Forces are exerted by entities and you get one force per entity. The difference F_app - F_n is not a separate force exerted by a fourth entity. It's just a partial sum of the terms that make up the net force.

I know, I was trying to state that there's a difference - a subtraction. You stated that as well.
I think I merely didn't state it clearly enough. Partial sums calculate up to the entirety of the upward force. It's vector algebra with forces.

 

[haruspex]

Well Fg + Fn = zero when it's vertical. now the Fa has an opposing force, the force from the ceiling pushing back on the person. Do you want to call that Fnc? So that's another normal force. The person holding the object upright is supporting the object exactly with normal force until they apply another force (against the ceiling)
So the person is the normal force, their force applied, Fa if you will, is the force past the normal force. So that's a net applied force, (Fnet = Fa - Fn), is then countered with Fnc. So four forces will equal zero, not three.

I have no idea what you are trying to say, or how it relates to my post.
There is no force from the ceiling pushing against the person for the simple reason that they are not in contact. The force from the ceiling is increased because the person is pushing up, but that's not the same thing. The force exerted by the ceiling is on the box.

In responding to threads created by students, it is important to try to understand and address what specific misunderstanding is blocking them. It may not be that helpful merely to show how to solve the given problem. That is the point of my post #10. Do you understand that post?