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?
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...