Understanding the normal vector

[nobahar]

Hello!
I have a question regarding the normal force. From what I have read, it is a force in repsonse to compression of the atom/molecules in an object. So that the molecules are forced closed togeather by a force, and, due to Newton's third law, there is a 'reaction' force. I can understand this for forces applied perpendicular to the surface of the object: the normal force will then also be perpendicular and in the opposite direction.
Now (I guess that you can see where this is going), when an object lies on a tilted surface, the normal force is perpendicular to the surface. I have clearly misunderstood what the normal force is or I have interpreted the compression incorrectly, as to me this would contradict the third law. Since the compression is due to gravity, would not the atoms/molecules be compressed in the direction of the force due to gravity, and therefore exert an equal and opposite force? Instead, the normal force is described by the equation -1*m*a*cos(x). I have included a rather pretty picture for your enjoyment, and because my description is probably quite poor.

 

[G01]

Remember that the force vector for gravity can be broken into components.

Let's break it up into two components, one along the direction of the slope, and one perpendicular to the surface of the slope.

It is only this second component that pulls the object into the surface of the slope. (The parallel component pulls the object along the slope, not into it.) Therefore, the normal force is opposite to this perpendicular component of gravity, not the full gravity vector.

Also:

While the normal force certainly wouldn't exist without gravity pulling the object into the surface, it is not correct to say that it is the reaction force to the gravitational force.

The actual "action-reaction pair" will be 1. The force of gravity of the Earth pulling on the box and 2. The force of gravity from the box pulling on the earth. We usually don't think about the second, since it has a negligible effect on the earth, but nevertheless, this is the true "reaction force."

The "normal force from the slope on the box" has a reaction pair as well. It's the "normal force from the box pushing on the slope," Again, we usually don't think about this force because we assume the slope is fixed and can't move. However, this is the true action reaction pair to which the normal force belongs.

 

[nobahar]

Thanks GO1, that was very clear and precise. It amazes me how a textbook can't explain it like that.
I'm guessing that the normal forces are both a product of the compression of the atoms in the objects.