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I was studying friction and couldn't grasp the concept of normal force. Can someone kindly explain it?

Edit: why is the value of the normal force equal to $mg$ even though it is acting opposite to Earth's gravity? Shouldn't it be $−mg$?

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    $\begingroup$ What about the description your textbook (or other source) gives you could you not understand about normal force? $\endgroup$ – Kyle Kanos Feb 11 '17 at 22:06
  • $\begingroup$ Actually everything is messy in my textbook.Normal force is not defined at all.I came across this in an example problem solved in the same book. $\endgroup$ – AmoghaTr Feb 12 '17 at 11:42
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    $\begingroup$ Imagine you and your friend holding a metal rod (ノಠ益ಠ)ノ━━━ノ(°—°ノ). When you try to push it in any direction, your friend pushes it in opposite direction, so the rod never ever moves. Now imagime you holding the rod which is is immured in a wall ▓▓▓▓▓━━━ノ(°—°ノ). When you try to push it in any direction, the normal force (which is there because of Newton's third law) pushes it in opposite direction, so the rod again never ever moves. You can outsmart your friend, but you can't outsmart concrete wall's normal force. $\endgroup$ – Sanya_Zol Feb 12 '17 at 13:27
  • $\begingroup$ @Sanya_Zol A̶S̶C̶I̶I̶ Unicode art in explanations of physics... this is a good idea ツ $\endgroup$ – Ruslan Feb 12 '17 at 16:14
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Normal force is a "holding back" force.

Place an apple on a table, and the table holds it up. This "holding up" force is always perpendicular, which is why it is called the normal force ("normal to" means "perpendicular to").

Any stiff surface will hold back against something that pushes on it, like this weight of an apple, and that "hold back" force is given the name normal force.

why is the value of normal force equal to mg even though it is acting opposite to earth's gravity? Should it be -mg?

Forget about that right away! This is not true. It might be in some cases, but it is not a general rule. Never say that normal force must be $mg$. Never say that it must be anything - it depends on the situation every time.

  • The apple on the table has a weight $w=mg$. The table must hold back with exactly $mg$ as well to hold up the apple. This comes from Newton's 1st law (positive direction upwards): $$\sum F=0\Leftrightarrow n+(-mg) =0\Leftrightarrow n=mg$$

  • Now push down on top of the apple. The table must now hold back against the apple weight $w=mg$ and the pushing force $F_{push}$: $$\sum F=0\Leftrightarrow n+(-mg) +(-F_{push}) =0\Leftrightarrow n=mg+F_{push}$$

  • Now push on a wall. Nothing puts weight on the wall, but you give a horizontal pushing force. The wall's normal force appears to hold back, this time horizontally (positive direction outwards from the wall):

$$\sum F=0\Leftrightarrow n+(-F_{push}) =0\Leftrightarrow n=F_{push}$$

A new value of $n$ in a new situation. Doesn't have anything to do with weight.

I hope this also cleared out the signs. You are right that the weight $w=mg$ and normal force $n$ in the apple example will have opposite signs, and they also do when you put them into the first law: the weight $mg$ pulls down (negative) while $n$ holds back upwards (positive). But they are on the same side of the equal sign, so rearranging changes a sign.

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    $\begingroup$ I wonder if it might help to also talk about an example of a normal force not related to gravity. E.g., pushing on a wall or something like that. It seems to be one of the sources of confusion about the normal force is that it's so often presented in the context of gravity that it can seem like they are always linked. $\endgroup$ – Todd Wilcox Feb 11 '17 at 17:05
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    $\begingroup$ @ToddWilcox Good point. I added a third example. $\endgroup$ – Steeven Feb 11 '17 at 17:20
  • $\begingroup$ That's a summation symbol, not a subtraction symbol; the first example should be n+mg=0 ⇔ n=-mg $\endgroup$ – ShadSterling Feb 11 '17 at 19:31
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    $\begingroup$ "Any surface will hold back against something that pushes on it (unless it breaks)," I would recommend a small correction: "unless it deforms". Put a heavy object on a spring and it will (slowly) contract. Of course SigmaF is not 0, but there's still some normal. $\endgroup$ – user1803551 Feb 11 '17 at 21:20
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    $\begingroup$ @user1803551 Perhaps, more cautiously, state that that rule holds in a static scenario. Even with a spring, once it reaches equilibrium these rules hold. Your "unless it deforms" seems more in reference to dynamic scenarios, which are certainly more complicated! $\endgroup$ – Cort Ammon Feb 12 '17 at 2:30
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An object at rest stays at rest unless an external force acts upon it. If a force is exerted on the object, then the object will accelerate (via F=ma). If I put an object on a table, then one expects gravity to "pull down" the object. But the object on the table will appear to be at rest. This means there is no no net external force acting on the object, which means the table is "pushing up" with an equal and opposite force acting on the object (since mg - mg = 0). Pulling down can be -mg while pushing up is +mg.

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Without getting into the deep details of what it means for two objects to touch each other, touching involves interactions of electric fields which hold each object together and prevent them from passing through each other.

These interactions manifest themselves in what we call the normal force: When object A touches object B, A exerts a force on B which is away from A and perpendicular to the contact surface (or perpendicular to the tangent line through the contact point). B simultaneously exerts an equal and opposite force on A, directed away from B, perpendicular to the contact surface.

Normal forces occur when objects touch, and their magnitudes vary depending on how tightly or loosely the surfaces are pushed together. In this sense, the magnitude of the normal forces might be called "reactionary." But that is misleading, because one could say that the gravitational force magnitude reacts to a change in separation distance.

The normal force is a name for the result of a complicated electric field interaction. In classical mechanics, we calculate the magnitude based on what it needs to be when, combined with other forces, we get a calculational result that agrees with the observed acceleration.

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If an object is place on a horizontal solid surface and the object is at rest the object must have no net force acting on it.
The two equal in magnitude and opposite in direction (but not a Newton third law pair) are the weight of the object acting downwards and the upward force that the surface exerts on the object which is called the normal reaction (forces).
So the magnitude of the normal reaction is $mg$ but the sign which is assigned to the normal reaction depends on the direction which has been chosen to be positive.
So if the weight is positive then the normal reaction will be negative.

The expression which links the frictional force $F$ and the normal reaction $N$ via the coefficient of friction $\mu$ just links the magnitudes of the two forces, $F=\mu N$, and the signs for the two force are assigned as appropriate to the problem in hand.

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  • $\begingroup$ I'm glad that you pointed out that the normal force on the object and the weight of the object are not a Newton's 3rd law (N3L) pair, but why do you use the word "reaction?" I believe that is misleading to novices because they associate reaction with N3L. I believe a better word is "interaction" (if you even need some qualifying word). The normal force is a force due to the interaction of the material structure of the object with the material structure of the surface. $\endgroup$ – Bill N Feb 11 '17 at 18:06
  • $\begingroup$ Actually, "interaction" force would be redundant. All forces are related to interactions. $\endgroup$ – Bill N Feb 11 '17 at 18:12
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In your case you can observe that the object isn't going into the earth and is stationary wrt the earth's surface. The balancing force that acts perpendicularly to the point of contact between earth and the object and nullifies the weight of the body is the normal force.

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The normal force is an interesting one because it shows some of the freedom we have with building up our equations. You're used to forces identified by their source, such as "the force of the baseball bat on the ball," or forces named after a behavior which causes that force, such as "the force of friction" or "the force of gravity." "The normal force" is another category: forces grouped based on a mathematical property of that force.

Let's take a trip down to the microscopic world of atoms. At the atomic scale objects are almost entirely empty space. However, they feel solid because of things like electrostatic repulsion. As you bring two atoms close together, their electrons start to repel each other and this prevents one object from passing through another. You could technically say the only reason you don't fall through the floor is because the electrons in your feet push against the electrons in the floor!

Now this is all accurate, but not very useful. The equations governing the electrostatics of the trillions of atoms in your feet are complex, and you don't really need to calculate them all out. What you really want is a way to handwave away all of those complications and make your life easy!

The beautiful secret to forces is that they sum together. If I have 10 forces, I can sum them all together into a single force (using vector addition, of course) and the result is the same. I can have 10 trillion small electrostatic forces, or I can just sum them all together into one or two big easy to understand forces. And that's what the normal force is.

In reality, when I bring my finger close to one of these keys to type, the atoms in my fingers get close to the atoms in the keys getting closer until the electrostatic forces balance out the force I'm using from my hand. If I push harder, the atoms in my fingers actually get closer to the atoms of the keys, causing the electrostatic forces to go up. However, they don't get to move much. Maybe they move 0.00000000001 meters closer. That's almost not moving at all. In fact, unless you're dealing with exotically powerful forces, such as those we find in particle accelerators, we might as well just say "it's not moving at all."

So we have these trillions of tiny electrostatic forces, all of which move those fractions of a nanometer to stay in equilibrium with whatever force is exerted on them. Since we don't really care about that movement, we ignore it, quickly lump all of those forces together, and call them the "normal force." This is the sum of all of that electrostatic complexity pushing outward from the surface.

This points out quickly why the normal force gets its name. The repulsive forces are always perpendicular to the surface (aka "normal" to the surface). They can't be in any other direction. Even if you have a block on a ramp being pulled down by gravity, the normal force is still perpendicular to the ramp's surface not directly opposing gravity.

We also see the same sort of behavior with friction. Friction is really the sum of millions of little interactions, like bumpy surfaces sliding past each other. We lump all of those together and call it "friction."

So why do we choose to lump them into "normal forces" and "friction forces?" Why not just lump them all into one giant "surface force?" Well, you can. Forces are additive, so you can do this all you like. However, what we have found is that it is convenient to split them into two parts because its easier to calculate that way. Normal forces are always normal to the surface, and they always are exactly strong enough to prevent one solid object from passing through another. Friction forces, on the other hand, are more complicated, with their coefficients of friction and different directions, and static vs. dynamic friction. They act differently enough that we find it convenient to keep them separate.

Remember, despite what it may seem, every choice made in physics is designed to help you predict what is going to happen in an experimental setup. We choose our variables to be convenient. We choose our coordinate systems to be convenient. It won't always feel convenient... sometimes it will feel outright asinine. But there's always a reason. Don't be afraid to ask your teacher why we do it one way or another. They may not be able to tell you everything, but much of why we do things the way we do can be explained!

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A normal force is a force involving a surface and in the direction of the normal vector of that surface.

The normal vector of a surface is perpendicular to the surface; for a flat surface (e.g. the top of a table) the normal vector is the same everywhere, but for a curved surface (e.g. the outside of a sphere) the normal vector can be different for every point on the surface.

On a flat horizontal surface, the normal vector points straight up, and if there is a normal force is it in the same direction. For an object resting on that surface, the normal force is the weight of the object but pushing straight up; $mg$ is the magnitude of the normal force, if you consider one dimension with up as negative then $-mg$ is the vector. (Of course, if it's at rest, the magnitude has to be $mg$.)

There's nothing special about the surface being horizontal; in problems about friction on a ramp, the normal force no longer points straight up, it points as far from vertical as the ramp is from horizontal. In this case, the normal force alone will not keep an object at rest, but the normal force plus friction might. (In principle there's also nothing special about the surface being flat, but in practice it's easier to understand interactions with flat surfaces.)

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I will explain very simply.

  1. Consider a mass $m$ kept on a surface.
  2. According to sign convention, the acceleration due to gravity $g$ in upward direction is positive and in downward direction is negative.
  3. We now know that the force of attraction by the object of mass $m$ is $F=-mg$ downward (which will be negative).
  4. Now for equilibrium of the object to stay at rest the sum of all forces should be zero.
  5. Therefore: $$\sum F_y=0\\-mg+\text{Reaction}=0\\∴ \quad \text{Reaction}=mg$$ This is in accordance with Newton's third law.

Now the question arises why the reaction force is not more than $mg$?

If this will happen then the earth is providing extra energy to the object, which in turn will provide lift off to the object, which is absurd.

This thing can be observed when something falls on a spring, as it provides a larger reaction force than the weight of the object, which causes the object to jump.

I think that I have now simplified your doubt.

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