# Normal Force and Force of Gravity: When and why are they equal?

by Trooper100
Tags: force, gravity
 P: 15 1. The problem statement, all variables and given/known data My teacher explained to my class that Normal force equals Force of Gravity. Given a free body diagram on a flat surface, Normal force is directed up, and Force of Gravity is directed down. 2. Relevant equations So my teacher did the following: Fg = Force of gravity Ff = force of friction Fn = normal force u = co-efficient of friction _________ Ff = uFn Ff = uFg Ff = u*m*g (mass times gravity) How is this possible? normal force is equal to the force of gravity? but they oppose each other, don't they? Thank you so much for any help at all.
 P: 649 Welcome to Physics Forums. They are equal in magnitude but opposite in direction on a flat horizontal surface. So when it comes to finding how large the normal force is, on a horizontal surface it equals the weight of the object. The frictional force depends on the normal reaction.
 Mentor P: 15,065 Your teacher meant equal in magnitude, not direction. And your teacher was wrong. The normal force is nearly, but not exactly, equal in magnitude to the force of gravity. Suppose an object is at rest with respect to the Earth. Unless the object's location is the north or south pole, the object iundergoes uniform circular motion. The Earth, after all, rotates once per day (once per sidereal day, to be picky) and the object is at rest with respect to the rotating Earth. That means there must be some non-zero net force acting on the object, $$\vec F_{\text{net}} = \frac{m \vec r}{\omega^2}$$ where $$\vec r$$ is the vector from the object to the Earth's rotation axis and $$\omega$$ is the Earth's rotation rate, 2*pi/sidereal day. There are other smaller accelerations in play as well. The Earth orbits about the Sun, for example. There are also other forces involved as well. What is the normal force on a helium balloon resting on the ground that has lost just enough helium so that the buoyant force is exactly equal to its weight? (Answer: Zero.) The air has a buoyant force on you, too. It's just a tiny fraction of your weight rather than equal to your weight. What if you are standing still on a hillside? Now the normal force is not even close to equal to the force of gravity. The normal force is normal, and the normal to the surface is not pointing upward in this example. In this example, static friction provides some of the force needed to keep you in uniform circular motion about the Earth's rotation axis. The normal force and static friction are examples of constraint forces. Anthropomorphizing and mystifying things way too much, these constraint forces "know" exactly how much force to apply to keep an object from sinking into the Earth (normal force) or moving along the surface of the Earth (static friction). The magic disappears when you look at things from a quantum perspective. When you are standing still on the floor, the atoms at the very bottoms of your shoes are not quite touching the atoms at the very top of the floor. The electrons in those atoms repel each other. You are, in a sense, floating just above the floor. Suppose you pick up a heavy book. The force due to gravity is pulling down on you a bit more. Your shoes descend a tiny, tiny bit to narrow the tiny (very tiny!) gap between the bottoms of your shoes and the floor. This increases the electrical repulsion, compensating for the added weight of the book.
HW Helper
P: 6,206
Normal Force and Force of Gravity: When and why are they equal?

 Quote by Trooper100 How is this possible? normal force is equal to the force of gravity? but they oppose each other, don't they? Thank you so much for any help at all.
Well if a mass m is just resting on a surface, the weight mg acts downwards. If there were no normal reaction (R), then that mass should fall through the surface right? (As there is no force opposing the downward force)

But it does not fall through the surface, so there must be a force R or a normal reaction, acting to oppose the weight, this reaction is perpendicular to the surface. Since the weigh is also perpendicular to the surface, one can sum the forces and equate to zero for equilibrium.

∴ R-mg=0 ⇒ R=mg

Understand it better?
 P: 649 DH. Though your answer is correct, I doubt very much that its level of detail and complexity is appropriate in this case. The student's question concerns magnitude and direction of a force, and the meaning of "equal" in this context. The difficulty is easily explained. I'm certain from reading the question, that a far simpler answer would suffice. Indeed, going into such a level of detail will only confuse the issue at this stage. Let's take things one step at a time. Also, telling a student his teacher is wrong, without knowing the context of the lesson, is ill advised.
P: 15
 Quote by rock.freak667 Well if a mass m is just resting on a surface, the weight mg acts downwards. If there were no normal reaction (R), then that mass should fall through the surface right? (As there is no force opposing the downward force) But it does not fall through the surface, so there must be a force R or a normal reaction, acting to oppose the weight, this reaction is perpendicular to the surface. Since the weigh is also perpendicular to the surface, one can sum the forces and equate to zero for equilibrium. ∴ R-mg=0 ⇒ R=mg Understand it better?
Oh right. so this means mg and normal forces cancel each other, because if one force was greater, then there is a vertical displacement as well. (Eg the brick on the table could vertically move up or down)

I see I see.

Thank you to everyone as well. Thank you!

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