# Why is normal force perpendicular?

I realize "normal" is just another word for perpendicular, but semantics aside, what causes normal force to be always perpendicular to the plane?

For example, on an inclined plane, why is it that the plane pushes the box outwards rather than directly against gravity? It seems to me that in order for there to be normal force, there must be a force trying to push the box directly into the plane, but gravity is going directly downwards, not into the plane.

• the other component is the tangential force (here labeled friction). If you add tangential and normal you will get the opposite of the weight. – ratchet freak Oct 11 '15 at 20:03
• I have a feeling that a complete explanation of why the intermolecular contact forces wind up pushing the block to the left (as it slides down the ramp) is gonna be just as nasty as a complete explanation of friction. – zwol Oct 12 '15 at 0:21
• For the bounty: some nice diagrams paired with simple explanations would be nice, please. – moonman239 May 13 at 5:35
• The normal force N here prevents the block from "passing" through the incline. There's a normal force from block pushing back on the incline - it's called the normal component of gravity - or component of gravity perpendicular to the incline but directed in the direction opposite to N with the same magnitude as N. – Cinaed Simson May 13 at 22:04
• Do you need further explanation of my answer? – Bob D May 15 at 0:12

Let's talk about what the normal force is.

First of all, it is a feature of solids, which is to say materials that resist penetration by other materials. When two solids are in contact they resist interpenetration; they resist occupying the same space.

Now, if something (be it gravity, your own hands, or simple motion) brings two solids toward one another and they are prevented from moving into the volume occupied by the other there must be a force involved. Where it comes from on the molecular level is complicated, but on the human level it is simply an expression of the resistance of solids to occupy the same space.

We define "the normal force" as that force which resists an attempt to cause two solids to occupy the same space. As such it points perpendicularly to the surface of contact; because motion along the surface of contact is not interpenetrating.

Of course there is a force related to motion along the surface of contact, too, but it goes by a different name---friction---and follow different rules.

Newton's third law: when two bodies interact, they apply equal but opposite forces on one another.

Whenever you go to apply this, you have to be really careful that you're consistent about which two bodies you're talking about.

The weight is the force that the Earth as a whole applies to the box. It has nothing to do with the ramp. By Newton's 3rd law, the box also pulls the Earth up a tiny bit, but we're ignoring that in this problem because we're not considering how the Earth moves, and besides a little box barely matters to the Earth's motion.

So the interaction we're concerned about in this problem is between the box and the ramp. The box pushes down and to the right on the ramp, and the ramp pushes up and to the left by an equal amount. That satisfies Newton's 3rd law.

We choose to decompose the weight vector into components along the ramp and perpendicular to the ramp in order to make it easier to solve this problem. The box is not accelerating through the ramp, although it may slide along the ramp. That tells us that the force components directly into and out of the ramp must be balanced. I.e. they must add to 0. Only the components along the ramp may or may not be 0, depending on whether or not the box is sliding.

• "The box pushes down and to the right on the ramp" That's the part I'm somewhat confused by, why does the box push to the right when the acting force of gravity is directly downwards? – manwithquestions Oct 11 '15 at 18:40
• Because the box is free to slide. If the box were pinned to the ramp (not free to slide) then it would push the ramp directly downwards. Imagine a vertical ramp (i.e. cliff). If the box is just barely grazing the cliff does it exert any force into the wall? – Jabavu Adams Oct 11 '15 at 18:46
• Let's look at this a different way. It's the way it is, because that's what matches the motion we observe. Imagine that the force box->ramp was straight down. Then the force ramp->box would have to be straight up, by Newton's 3rd law. So the net (total) force on the box would be either 0, straight up, or straight down. So it would either never move, jump upwards off the ramp, or somehow go through the ramp. None of this is what we observe. We have to prevent the box from going through the ramp, while allowing the possibility of sliding along the ramp. – Jabavu Adams Oct 11 '15 at 18:52
• @JabavuAdams Your first comment argument reads "If it was a cliff there's no rightward force, but it isn't, so there is." Your second demonstrates that there is a rightward force, but doesn't tell us why. -- I think you're better off focusing on the weight vector decomposition argument: in the frame of the ramp, the weight vector is the sum of a ramp-tangent vector (down and left), and a ramp-inward vector (down and right). It's this ramp-inward vector that the normal force opposes. (Generally, all vector decompositions are valid, although only some are useful.) – R.M. Oct 11 '15 at 19:36

If your ramp has friction holding the block in place, then you could talk about the total force the ramp applies to the block, and this total force would point directly opposite the force of gravity, and it would have the same magnitude, giving a net force of 0. It might be more useful to consider the case where there is no friction whatsoever. In that case, there can be no force resisting the sliding of the box down the ramp. However, if your normal force somehow pointed in a direction other than perpendicular to the ramp's surface, then there would be a component of the normal force reducing the block's acceleration down the ramp (assuming you make your normal force point up), but that's friction's job, right? If there's no friction, there is no force parallel to the ramp besides a component of the force of gravity pulling the block downward.

"Normal" is a mathematical synonym for "perpendicular"

Yes, I read the question too fast. The normal force is caused by very small compressions of the intermolecular bonds connecting the first layer of molecules to the second. If those bonds are perpendicular to the surface, it's clear that the force will be perpendicular. Generally the bonds don't all point normal to the surface. However, on average, the horizontal components cancel, leaving only the perpendicular force.

• This doesn't answer the "why" of it. The questioner is confusing the interaction of the box and ramp with the interaction of the box and Earth. – Jabavu Adams Oct 11 '15 at 18:25

I think it is obvious in the diagram that you can move the box along the plane without "disturbing" it (lift it a fraction of a mm so that they are not in contact anymore and move your box freely along the surface and you will not make any effect on the plane), this is the direction in which the friction would act, the sliding motion.

If we now turn to the other possibility, perpendicular motion, you can see that we do affect the plane. Namely, we are applying a force that compresses the material at that point. Since the atoms inside the material are "happy" at their current distance (solids keep their shape when put in a container because their atoms are "happily linked together") trying to push them together will cause a force, which will oppose the movement of the box which is going in the normal direction. It follows that the force should go in this same direction.

For example, on an inclined plane, why is it that the plane pushes the box outwards rather than directly against gravity?

The plane does push against the box directly upward against gravity, See the free body diagram below (force components not shown to scale).

In the diagram, I am showing the reaction of the plane directly upwards against gravity. Any such force can be resolved into perpendicular and parallel components to the plane. In this case the normal component is that perpendicular to the plane. But the parallel component (what I show as ??) can only be the friction force, since that is the only force the plane can exert parallel to its surface. That is, for a frictionless surface the parallel component of the force would have to be zero..

Hope this helps.

• Leaving aside philosophical questions there is a problem with the lower element of your figure. The arrows representing $\text{Weight} \sin \theta$ and $\text{Weight} \cos \theta$ need to be shorter than the one representing $\text{Weight}$ as the former certainly don't have larger magnitudes than the latter. – dmckee May 13 at 22:30
• Thanks. I agree, but I never meant them to be to scale – Bob D May 14 at 5:52
• I’ve edited my answer to indicate that. – Bob D May 14 at 6:01
• Hi Bob, I am considering awarding the bounty to you for your excellent explanation and diagram. I do have one question that I feel could use a little more elaboration: Why is the normal force mg*cos theta? – moonman239 May 14 at 23:09
• First of all, thank you for your consideration. The cosine of 90 is zero, the cosine of 0 is 1. When the angle of the incline goes to zero, the cosine goes to 1 and that means the normal force goes has to equal the weight of the box. $mg$. Hope that helps. – Bob D May 14 at 23:17

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