# How does normal force work?

1. Normal force is the force that prevents objects from passing through eachother,which is the force of the repulsion from the charge.

2. The normal force will get as large as required to prevent objects from penetrating each other.

My question is about the scenario of a person inside an elevator:

The elevator has a mass of $$1000kg$$ and the person has a mass of $$10kg$$

At the first few seconds the variables are($$_e$$ is for "elevator" and $$_p$$ is for "person", i'm assuming that the acceleration due to gravity is $$-10m/s^2$$, "-" is for downward):

$$v_e$$ = $$0m/s$$

$$a_e$$ = $$0m/s^2$$

$$v_p$$ = $$0m/s$$

$$a_p$$ = $$0m/s^2$$

And the forces are:

The force of gravity on the elevator $$f_g(elevator)=m_e*-10/s^2$$

The force of gravity on the person $$f_g(person)=m_p*-10m/s^2$$

The force of the wire keeping the elevator in place(without considering the weight of the person becuase that's one of my questions) $$f_w = +f_g(elevator)$$

Now, there's a force of gravity applied on the person which is $$f_g=10kg*-10m/s^2=-100n$$

So the person is supposed to accelerate downward,but it can't go through the elevator becuase of the normal force which I said what I think it does at the start of the question

Here's what I think is happening:

If the normal force were to be applied on the elevator by the person's feet, then it would be greater than if it were to be applied on the person's feet by the elevator(becuase the mass of the person would require less force for the elevator to stop it,than the mass of the elevator would require for the person to get the elevator moving with her/him so she/he doesn't penetrate the elevator)

Therefore the normal force is applied on the person by the elevator (as small as it can be) for them to not penetrate eachother, $$f_n=f_g(person)$$

When there is a net force on the elevator which accelerates it upward,the normal force is applied on the person by the elevator to prevent them from penetrating eachother because that way it is less than if the normal force were applied on the elevator by the person(becuase the mass of the person would require less force for the elevator to get the person moving with it,than the mass of the elevator would require for the person to get the elevator to stop,so they don't penetrate).

And the normal force in that case is $$f_n=m_p*(a_g+a_e)$$ applied on the person by the elevator.

The main thing:

1. IlIs my interpretation of normal force correct??,or does the normal force have to be applied on the "moving" object??

1. I heard a lot that when the elevator starts decelerating(acclerating in the downward direction) the elevator would apply a normal force on the person which is as small as it can be to prevent her/him from penetrating the elevator,and because the elevator is decelerating,the force will be less than gravity(assuming that the person has the velocity of the elevator before it was decelerating)

But if the elevator is slowing down(the same goes if the velocity was negative), that means for sometime the person wouldn't be in contact with the elevator(because the person's velocity has to be the same as the elevator's for her/him to not penetrate the elevator,the elevator has to change its velocity first before the velocity of the person can change due to gravity's downward accleration)

So how can there be a normal force applied??

1. Does normal force come in pairs?? and if it does, in what way??

If not,what is the opposite and equal force to the normal force??

I tried to make my question as clear as possible.......(:

• please do not use so much empty space. Also, you do not need to use <\br> for new paragraph just press enter twice. The equations are also better to write in the form \$\$ equation \$\$, when you want it on new line and \$equation\$ in the line. There is also no point in writting \$l.h.s.\$ = \$r.h.s.\$, just write \$l.h.s=r.h.s\$. I would also not write $*-10m/s^2$, but rather $*(-10m/s^2)$ or perhaps even $\cdot(-10ms^{-2})$ Aug 20, 2020 at 11:00
• also try to use dot for ending the sentence and capital letter for starting the sentence. All in all, the question is really hard for me to read. Aug 20, 2020 at 11:03
• the normal force will get as large as required to prevent objects from penetrating each other Certainly not true. If it would be true - no bullet would penetrate the target. Aug 20, 2020 at 12:36
• @Umaxo thx for the informations Aug 21, 2020 at 6:49

Yes, normal forces come in pairs - the elevator exerts a normal force on the person and the person exerts a normal force on the elevator. These two normal forces are equal in magnitude and opposite in direction - this is Newton's Third Law.

The best and simplest approach to this type of problem is to consider each object separately, work out the forces on each object, and use Newton's Second Law $$F=ma$$ to relate the forces to the acceleration of the object. Then you can see if you have enough information to determine the values of any unknown forces or accelerations. It might help if you draw a diagram for each object showing the forces acting just on that object - these are called "free body" diagrams.

When the person and the elevator are stationary, we know there are two forces on the person:

1. Gravity, which produces a force of $$100$$ Newtons downwards (by the way, $$10$$ kg is a very small person, but that is the figure you gave for their mass).
2. The normal force from the floor of the lift - let's call this $$N$$ Newtons upwards.

The person has an acceleration of $$0$$, so Newton's Second Law tells us that the net force on the person must be $$0$$. So $$100-N=0$$, and so we know that $$N=100$$ Newtons.

Turning now to the elevator, there are three forces on the elevator:

1. Gravity, which produces a force of $$10000$$ Newtons downwards.
2. The normal force from the person, which is a force of $$N$$ Newtons downwards. We know that $$N$$ here has the same value as the normal force acting on the person, because Newton's Third Law tells us that if the lift exerts a force on the person then the person exerts an equal and opposite force on the list.
3. The tension in the wire, which we will call $$T$$ Newtons upwards.

The elevator also has an acceleration of $$0$$, so we know that the net force on it must be $$0$$, so $$T = 10000 + N$$. But we know from our analysis of the person that $$N=100$$ Newtons. Therefore $$T=10100$$ Newtons. This makes intuitive sense, because the wire must support the weight of the elevator and the person.

Exactly the same analysis is true if the elevator is moving at a constant velocity (because its acceleration and the person's acceleration are still zero). However, if the elevator is accelerating upwards at an acceleration of $$a$$ metres per second squared, then the force equation for the person becomes:

$$N - 100 = 10a \\ \Rightarrow N=100+10a$$

In other words, the normal force $$N$$ increases (this is why you feel heavier in an elevator that is accelerating upwards - what you feel is the increased normal force on your feet).

And for the elevator we have

$$T - 10000 - N = 1000a \\ \Rightarrow T = 10000 + N + 1000a = 10100 + 1010a$$

In other words the tension in the wire increases because it must now support the weights of the elevator and the person and provide enough additional force to accelerate them both upwards at an acceleration of $$a$$. Notice that it does not matter whether the velocity of the elevator is zero, upwards or downwards - it is only the acceleration that matters.

Similarly, if the elevator is accelerating downwards, the normal forces and the tension in the wire will be reduced - but note that normal forces and tensions in wires cannot become negative. If we want to accelerate the elevator and the person downwards with an acceleration greater than $$10$$ m/s^2 then we would have to replace the wire with a stiff rod so that $$T$$ can act downwards, and we would have to give the person some means of gripping onto the floor so that $$N$$ can act downwards too.

Both the person's feet and the elevator's surface apply equal and opposite repulsive forces on each other, doesn't matter which one is trying to penetrate into the other.

The force of repulsion between charges grows to be very large as the separation between them becomes small. So when you try to rest your shoes on the elevator surface, the charges on both the surfaces (the shoe and the elevator) get very close to each other. As you stand still, the charges on each surface is constantly repelling the charges on the other surface with an equal and opposite force.

Say, you jump from an elevator surface and land on your feet afterwards. While you're in the air, gravity brings you down. When your shoes get very close to the elevator surface as you come down, the repulsive forces from the charges on the elevator's surface oppose your motion. You start decelerating until you are brought to rest. The fact that you are at rest now automatically means that the elevator's repulsive forces on you have adjusted to be equal and opposite to your weight.

While the the charges on the elevator's surface were trying to slow you down via repulsion, the charges on your shoes were also repelling the elevator in the downward direction with an equal force (Newton's third law). This will try to make the elevator accelerate downwards. Right now, two forces will be trying to accelerate the elevator downwards, the repulsion from your shoes and the elevator's weight. Luckily, the elevator is hanging from a rope. So it will try to penetrate through the rope surface first before being able to fall down. Repulsive forces will arise between the rope surface and the elevator surface in contact with each other. If the rope is strong enough, the repulsion force from it on the elevator will be able to counter both the force from your shoes on the elevator and the elevator's weight. In that case, the elevator won't accelerate at all.

While the repulsive forces from the rope try to balance the downward forces on the elevator, the charges on the part of the elevator in contact with the rope also apply an equal and opposite repulsive force on the rope, trying to accelerate it downwards. The rope will get tightened. Sometimes, this force will be enough to more than counter the attractive forces which hold the rope particles together (these are also forces arising from charges which constitute the rope). If this happens, the rope particles get separated and the rope breaks.

1. Normal force obeys Netwon's 3rd law, and it is applied in equal and opposite terms between the bodies involved. However, forces acting on the ground are often disregarded. But for all other pairs of bodies, the normal force needs to be accounted for on both bodies.

2. Always draw free body diagram showing each body separately and all the forces acting on them.

3. Normal forces are whatever they need to be in order for the bodies to obey all kinematic constraints. If two bodies must share a velocity component because of contact (like in the elevator example) then the normal force is found from this constraint.

4. In general the acceleration vector of two rotating contacting bodies does not match, and to bring the velocity constraint into acceleration form you have to treat the contact point as not moving. See this answer on this exact issue that I posted.

In your case since none of the bodies rotate, point #4 is moot. Using the equations of motion for each body in the free body diagram, as well as the constraint equation.

\begin{aligned} N - m_p g & = m_p a_p \\ T-N - m_e g & = m_e a_e \\ a_p &= a_e \end{aligned}

where $$T$$ is the cable tension of the elevator, and $$N$$ the normal force. See how $$+N$$ is acting on the person and $$-N$$ on the elevator? That is three equations with three unknowns, the two accelerations and the normal force.