Hi! I have what is probably a very elementary question, but one that has puzzled me for a long time.
In numerous physics texts I see a diagram of a crate on a friction-free ramp, with vectors indicating the forces on the crate. One vector points straight down, representing the downward force of gravity. The other vector, representing the force imposed on the crate by the ramp (the “normal force”)points away from the ramp at a 90 degree angle. It is said that this force is a third law force, created in reaction to the downward force on the ramp imposed by the crate.
Here’s my question. The third law says that two forces related by that law are opposite in direction. Since the force ON the ramp imposed by the crate’s weight is directly downward, why isn’t the force imposed BY the ramp on the crate directly upward instead off at an angle to the vertical (such that it is perpendicular to the ramp)?
If the ramp exerted an equal and opposite vertical force on the crate, those 2 would be the only forces and they would cancel to zero. How then to explain that the crate accelerates down the ramp?
The downward force of gravity on the crate is imposed by the Earth's gravitational field. That force can be resolved into perpendicular components. The choice of direction of the components to be considered can aid in analyzing why the crate accelerates down the ramp. Choosing a component of its weight that points down the ramp explains the acceleration. This component does not exert a force on the ramp since the ramp is friction-free. The other component of the weight must be perpendicular to the other component and is therefore perpendicular to the ramp. It is that "other component" that Newton's 3rd requires to have an "equal and opposite reaction" force. Since the other component is perpendicular to the ramp, the reaction force is also perpendicular to the ramp, but in the opposite direction.
Consider a case in which the ramp is not friction-free and the tilt is not enough to make the crate slide. The friction between the ramp and crate would be the equal and opposite the component of the weight pointing down the incline. The Newton's 3rd action/reaction pair would be the ramp's friction pushing up the slope and the crate exerting a force on the ramp (essentially trying to break off the bumps that cause the friction).
I hope this helps,