Question

Why is the force of gravity larger than the normal force if the object is resting on a sloped surface?

Answer 1

The normal force is defined as the force exerted by a surface onto an object in the direction PERPENDICULAR to the surface. The normal force prevents the object from going through the surface.

First consider a stationary object on a flat surface. You know the force of gravity acting on that object is mg (in the downward direction), yet that object does not accelerate through the surface (the net force is 0). Therefore, there must be another force equal but opposite to mg - in this case, it is the normal force (acting in the upward direction).

Now consider your sloped surface. If you place an object on the sloped surface, the object doesn't crash through the surface or fly away from it - in other words, the net force PERPENDICULAR to the surface is 0. The normal force is acting in this dimension, "away" from the surface, but it must be balanced by something. This something is the component of mg perpendicular to the surface, acting "toward" it. However, there is still an additional component of mg acting PARALLEL to the surface (which will cause the object to slide if it outweighs the force of friction up the slope). If you add the component vectors of mg parallel and perpendicular to the surface, you will get the complete mg vector, which goes straight down.

In this case, the normal force only balances PART of the force of gravity, and so if you were to compare the magnitude of the (full) force of gravity to the normal force, you'll find that it's larger. Hope this helps!

Also this is a physics question :P