# Why is the force of gravity not equal to the normal force on an inclined plane?

When an object is at rest the force of gravity and normal force cancel out. However, when place on an inclined plane the downward force of gravity is not equal to the normal force that is perpendicular to the ramp. Why is that?

My second question is why is normal force perpendicular to the ramp?

• Here's a recent question about why normal forces act along the normal of the plane: math.stackexchange.com/q/3811091/168433 Sep 7, 2020 at 14:10
• It's called the normal force because it acts perpendicular to the surface. This is what the term normal means in a geometrical context. Oct 14, 2021 at 3:21

Because you are forgetting friction.

If there is no friction then the force of gravity on the box will be greater than just the normal force from the plane - hence the forces won't cancel out and the box will slide down along the inclined plane. If you add friction to the surface between the box and the plane then if the friction is great enough it will make a force upward and along the plane that adds to the normal force from the surface and the box will be at rest.

The normal force will always act perpendicular to the surface. You can imagine that it is a force preventing the box from going straight through the plane. If it wasn't acting perpendicular then the box would not slide downward along the plane but rather move in some peculiar (and magical) way.

Even without friction, the normal force is defined as being perpendicular to the surface. If the surface is tilted, the normal force cannot be equal and opposite to gravity.

The net force on a body is $$\mathbf F = m\mathbf a$$. If friction is small for example, an object can slide along the inclined plane with an acceleration. The net force has the same direction.

But another concept of force relates it to elastic deformation instead of acceleration: $$\mathbf F = k\mathbf x$$. A perfect cubic object, lying in the ground by its own weight, has the vertical dimension smaller than the other two due to that elastic deformation.

If it is lying on a wedge, there is a reduction in the dimension in the normal direction ($$\mathbf N = k\mathbf x$$), and an angular deformation. The later is a measure of the couple caused because the friction force is not applied in the center of mass.

The forces can be measured by strain gages attached to the cube. Small deformations produce electric signals proportional to them. It is the principle behind most load cells.

One more thing to note, is that normal force is force of constraint. We know, that the object cannot move under the surface and there must be some force that prevents this from happening. Such force must be perpendicular to the surface, because it only prohibits things to go under it and does not influence anything else. This is how normal force is defined.

But of course, the true contact force from the ramp might be more complicated than that. We call this additional complication friction.

Turn the question on its head. Ask yourself: Why would you expect the normal force to be equal to the gravitational force?

There is no law stating this. There is no reason to thing that this should be the case. In fact it often is not the case. The law we have that links together forces in a stationary case is Newton's 1st law, $$\sum F=0$$.

• Sure, when a book is lying still on a flat surface, then we do see the normal force $$n$$ equalling the gravitational force, also called the weight, $$w$$. Because, when plugged into Newton's 1st law we see: $$\sum F=0\Leftrightarrow n-w=0\Leftrightarrow n=w,$$ where the weight is negative because we think of an upwards-pointing axis.

• But what if you were pushing the book horizontally onto a wall, holding it fixed with your strong push $$F_\text{push}$$? The wall responds with a normal force $$n$$ the opposite way in order to not break (this is actually due to Newton's 3rd law). If we choose an axis-direction pointing away from the wall, then we see $$\sum F=0\Leftrightarrow n-F_\text{push}=0\Leftrightarrow n=F_\text{push}.$$ Now the weight of - the gravitational force on - the book is not even a part of the expression for the normal vector.

So, as we see, there is no reason to expect the normal force to have got anything to do with the gravitational force. It only is equal to it in certain special cases. That we happen to often use such special cases with objects on flat ground in textbooks is just because they are easy to work with.

In you case with a e.g. a box resting on a sloped incline, then you have the force of gravity $$w$$, the normal force from the surface $$n$$, and a friction force $$f$$ present. Applying Newton's 1st law, while choosing an axis pointing directly away from the slope, gives us: $$\sum F=0\Leftrightarrow n-w_x=0\Leftrightarrow n-w\cos(\theta)=0\Leftrightarrow n=w\cos(\theta),$$ where I let $$\theta$$ represent the slope angle (but the exactly relationship depends on your specific scenario). You could choose different axis-directions, which would result in other expressions for the normal force (although the result in the end will always be the same), but you cannot choose it so that the normal force is equal to the weight, simply because they do not point the same way.

When you ask why the normal force is always perpendicular to the surface, then the answer is that this is just the name we give to the perpendicular force. When you exert a force on a surface, then as per Newton's 3rd law of action/reaction the surface will respond with a counteracting reaction force in order to avoid breaking under the exerted force.

This reaction force can always be split into a parallel component and a perpendicular component (parallel and perpendicular to the surface, that is).

• The parallel component is then given the name friction force.
• The perpendicular component is given the name normal force.

So, the fact that the normal vector is always precisely perpendicular to the surface is just due to the naming convention.