 # Relationship between applied and normal force?

## Relationship between applied and normal force?

The other answers here are generally correct. But it might be useful to add to this, since it is a common misconception especially with beginning physics students.

As stated elsewhere, normal here, just means perpendicular, that is, it is a mathematical term for when two lines or surfaces are at right angles to each other. So a normal force is, by definition, a force that is perpendicular to two surfaces which are in contact. Let’s look at some examples.

If you place a book on a table, the book exerts a normal force downward on the table while the table exerts an equal normal force upward on …

There are two issues that are, perhaps, causing your confusion. The first is a mix up between the force of gravity on the book and the force of the book on the table (I’ll deal with this last). The second is a mix up about what is meant by an ‘action/reaction pair”.

I really dislike the usual statement of Newton’s 3rd Law for this reason. “For every action there is an equal and opposite reaction” makes it sound like there are two things being described. But there is only ONE force, which we can describe from either end. Really the law should be called “The law of interaction.”

Forces are the …

Not at all. In fact, the only time the normal force on an object is equal to the object’s weight is when it is at rest on a horizontal surface which itself is not accelerating vertically — and there are no other forces acting on the object.

The normal force between two objects in contact is just the force they each exert on each other perpendicularly (i.e. normal) to the surfaces that are in contact. When you press your hand against the wall, the wall exerts an equal and opposite force back against your hand — we would call that force the normal force. It has nothing to do with weight. When …

Weight is a debatable terminology. In regular sense, weight is what the weighing machine registers, when you stand on it. However, there are two sides to this argument- The real weight and apparant weight.

If you stand on an elevator, and measure your weight, it will be different from what you measure on the ground. This different weight given by the machine is the apparant weight of the object, and is equal to the normal force acting on the body.

But however, we can call weight as the force with which we are attracted to the local gravitational well. In that case, our weight always remains s…

Both books are individually attracted by the whole mass of earth. That gives you the force \$mg\$. In principle also the books attract each other via gravity, but that force is so small that you can safely neglect it.

Let’s start with the lower book. It sits on the table, gravity is pulling it downwards. The table then resists the compression by the book exerts a normal force onto the book. The gravitational pull goes downwards, the normal force goes upwards. Both forces cancel each other out exactly, the book is at rest.

Now add the upper book. The gravitational pull also be there for the second book. The lower book resists the compression by exerting a normal force onto the upper book. This keeps the upper book at rest.
The lower book is now pressed harder onto the table. To resist the compression, the table has to double its normal force now.

To sum up: The upper book has the following forces:

• Gravitational pull from the earth (down)
• Normal force from lower book (up)

The lower book has the following forces:

• Gravitational pull from the earth (down)
• Pressure from upper book (down)
• Normal force from table (up)

When you are in a elevator going up, on the acceleration phase you will have a higher than usual “normal force” downwards so you will feel your weight is more than usual.

Or, if you live in a rotating space station -the ones that looks like a wheel- while you walk on the outer section, centripetal force -normal force- is felt as gravity even when there is no gravity at all.

After all, gravity is a downwards acceleration -from your body point of view- so you will “convert” any up-down forces -the normal forces- as a weight, regardless of it…

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Understanding this question already has an accepted answer, I was confused by this as well and did some additional research that I will share.

The first is the definition of weight. I first understood the definition of weight from the textbook I was reading as the gravitational force that exerts on a body. From this definition, there is no difference between $${F_g}$$, the gravitational force, and $$vec w$$, weight force.

That said, after re-reading and searching all the internet sufficiently, the definition more accurately used for weight is the gravitational force that the earth exerts on the body or «If you are on another planet, your weight
is the gravitational force that planet exerts on you.» Young, H. and Freedman, R. (2016). University Physics with Modern Physics (14th ed). Pearson Education, Inc, pp. 114.

With that in mind, typically weight force is a particular type of the gravitational force that refers to the gravitational force on a body (e.g. on the moon, on Mars, on Earth). If you are referring within space such as the gravitational force of Earth on the International Space Station (ISS) or an astronaut on the ISS, they are in constant free-fall, and for sake of clarity, the force causing that free-fall is just called gravitational force since there is the sensation of weightlessness. It would be a misnomer to say how much does an astronaut weigh on the ISS and more accurate to say what is the gravitational force the earth (or the ISS) exerts on the astronaut.

The same can be said of celestial bodies, such as the moon. One would not say, how much does the moon weigh. One would say, what is the gravitational force the earth has on the moon (and so forth for the Earth to the Sun and Mars to the Sun).

When being asked for the weight force, it is generally implicitly understood, it’s the gravitational force of an object on the earth (or it will be specifically called out on what body and the acceleration due to gravity on that body, e.g. find the weight force of a person of 68 kg mass on the moon which has an acceleration due to gravity of $$1.620 dfrac {m}{s^2}$$). When asked for the gravitational force, the 2 object’s mass will be provided or the needed data to calculate their mass.

Here is a comparison at the formula/math definitions.

Weight Force:$$vec w = mvec g$$ where m is the mass of the object and $$vec g$$ is the acceleration vector due to gravity (e.g. has both magnitude and direction).

Gravitational Force between 2 objects:

$${F_g} = G * frac {m_1 m_2}{r^2}$$ where G is the gravitational constant $$frac {6.67384(80) * 10^{-11} m^3}{kg * s^2}$$ and $$m_1$$ is the mass of object 1 and $$m_2$$ is the mass object 2 with $$r$$ being the distance between the 2 objects.

If we were to calculate both values for 150 lb person at sea level, we would find that both equal $$667 N$$ (rounding to 3 significant figures). To simplify the calculation of gravitational force on the Earth, Newton derived the weight force relationship.

Hope this helps.

Let’s draw a quick diagram to make it clear what we’re talking about: The condition for constant velocity is that the applied force \$F\$ and frictional forces \$mu mg\$ are equal so:

\$\$ F = mu mg \$\$

As you thought, the normal force is not equal to the applied force — well, not unless the coefficient of friction \$mu\$ happens to be equal to one.

The equation relating the frictional force \$F_f\$ to the normal force \$F_n\$:

\$\$ F_f = mu F_n \$\$

is generally called Amonton’s law. However this is an effective law not a basic principle, and in practice applies only in limited conditions. Why Amonton’s law is a useful approximation is discussed in Why is the equation for friction so simple? and in more detail in the paper On the origin of Amonton’s friction law by Persson et al, J. Phys.: Condens. Matter 20 (2008) 395006.

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