How do i calculate a normal force when gravity is not the only force involved?

[SamuraiDave]

How do i calculate a normal force when gravity is not the only force involved??

Dear everyone,
This is my first post on the physics forum and I have a problem which I think relates to calculating the normal force (perhaps I'm wrong).

I'm programming a game and I have a particle which has a mass (M) - a scalar, velocity (V) -a 2d vector...The game has gravity (G) - a constant scalar value.

Also in the game is a line - (y=mx+c) which represents the surface or floor which the particle can rest on.

For now, imagine that the line is perfectly horizontal and the particle is resting on the line.
Due to gravity, the normal force is making sure the particles velocity remains at 0.

But, imagine another force (other than gravity) acting on the particle at a 45 degree angle...i.e. pushing the particle down and right.

My main question is, in this case, how would the particle move? It is obvious that not all of the force will be added to velocity (otherwise the particle would go through the line), as the sum of this force + gravity will be greater than the normal force.

I assume in this case the particle should move to the right, perpendicular to the normal force? but how? and how much?

Sorry if this is very difficult to visualize. I hope somebody can help, as this problem has been going through my mind a lot.

Cheers,

David

 

[||spoon||]

from what I understand of your question you want to know what happens to a box, when it on a horizontal floor and a force acting on it's upper edge at a downward 45 degree angle (in this case towards the right). Is that correct?

If it is then the box will move to the right. If you decompose the vector into its components you will find that one is perpendicular to the floor, and the other is parallel to the floor. The one perpendicular to the floor does no work, the normal force due to the table will be the sum of the weight and the vertical component of the pushing force.

So the vertical forces cancel and you are left with a net force to the right (the parallel component of the pushing force).

Hope that makes sense and helps, am typing on iPod so is a bit weird.

-spoon

 

[SamuraiDave]

If you decompose the vector into its components you will find that one is perpendicular to the floor, and the other is parallel to the floor. The one perpendicular to the floor does no work, the normal force due to the table will be the sum of the weight and the vertical component of the pushing force.

Hmmm, I understand how due to the normal force the vertical component of the pushing force will be canceled (as will the weight)...does this mean that the remaining horizontal component of the pushing force is the new horizontal velocity of the box, and that the vertical component is now zero?

 

[||spoon||]

he horizontal component is NOT the horizontal velocity (if the vector we are tslking about is a pushing force). The horizontal component will cause an acceleration given by f=ma (Newtons second law). The velocity of the box will depend on how long this force is exerted (and wether there is friction involved in the calculation).

The result is quit intuitive if you think about, say, pushing on the corner of an ice cube on a bench. It would move horizontally of course.

 

[SamuraiDave]

OK, got it now. thanks very much for the great help! :)