# Action and reaction pair problem

If I sit on a chair, the first action and reaction pair force is my weight and the force acting on the earth by me. And the 2nd pair is the force acting on the chair by me and the opposite force.

According to newton third law, the action and reaction force must be equalI. 'm confused because how come the force acting on the chair by me is equal to normal force?

F=gmm/r2

• Your weight is the force you exert on the chair, therefore it can not pair with the gravitational pull you exert on the earth. Your weight and the normal force of the chair form a pair. – KvdLingen Mar 17 '14 at 6:55
• I dont think it's right since weight is the force adding on me by gravity – user127595 Mar 17 '14 at 10:08
• Also according to newtons third law, action reaction pair must exert on different body, therefore weight and normal force from chair is not action reaction pair since they are acting on the same body. – user127595 Mar 17 '14 at 10:12
• No. gravity is the force the Earth exerts on you. As a result you exert a force on the supporting surface, here the chair. This last force is called weight. – KvdLingen Mar 17 '14 at 15:20

Where does the 'pairing' in your first pair come into place? I.e. what is the force counteracting the earth's gravity pull on you (in your theory)?

The relevant force pair in your example is the attractive force between your body and the earth (gravitational pull) and the repulsive force between your body and the chair's surface (its lack of compressability).

You can either see the chair as part of the earth in this scenario OR you can use a force chain in which the repulsive force the earth has on the bottom of the chair transfers via the chair to your body.

It's getting more complicated by the fact that thereby the force the earth has on you is mostly translated into the deformation of your body.

The application of Newton's laws is very much about abstraction and simplification or in other words macroscopic effects, that are in fact the result of a LOT of microscopic effects. (Electromagnetic repulsion vs. attraction on atomic level vs. gravity making up the bulk but not all of the forces ar work here.)

• The relevant force pair your describing are not a force pair according to Newtons third law, as both are acting upon the body of the sitting person. – KvdLingen Mar 17 '14 at 6:52
• Ok, changing the wording, so you can see it is. – Sascha Rambeaud Mar 17 '14 at 9:31

I'm not exactly sure what you are confused about but in my experience of TAing intro physics there is a common confusion when students compare two forces that happen to be equal but are not third law pairs.

You have stated the correct third law pairs, there are two of them, the gravitation force exerted by the earth on you and on the earth by you, and the contact forces of the chair on your body and your body on the chair.

Now when you make a free body diagram for you, you only draw the forces that are acting on you, this means there is an upwards contact(normal) force due to the chair acting on you and then there is the downwards gravitational force by the earth acting on you.

Newton's 2nd law demands that your net force is zero since you are not accelerating and are in equilibrium. This is why you can say that the magnitude of the contact force by the chair is equal to the gravitational force by the earth acting on you.

Newton's Second Law states: $$\Sigma F = ma$$ Net force equals mass times acceleration. If the acceleration is zero, we have equilibrium. constant motion, or rest. They are both equivalent. If the case of the chair, you do not move. Net force equals zero. Analyzing a force body diagram for the situation, we have the downward force of gravity pulling on you and the chair, $(m_{chair} + m_{you})g$, which is picture in red. The force the floor exerts on the chair is picture in blue by $F_{N,F}$, and the force the chair exerts on you is pictured in green by \$ F_{N,C}. We have 2 force in the up direction, 1 in the down direction, and they all must add up to zero. The green and blue arrows are then normal forces created by whichever surfaces are interfacing.

I think I get the right idea. Since the gravity is dragging me down, the force acting on the chair is equal to mg, so the normal reaction force by the chair is also mg. However, weight and the normal reaction force from the chair ARE NOT action and reaction pair since they are acting on the same body.

• They do form an action reaction pair. The person-chair system does, and the chair-floor system does. You can model these both as one situation and reduce it down to one force pair however, by addition of the green and blue vectors in my answer. This is analogous with making those two arrows an arrow of one color. Then you just have gravity pulling with mg down, and the other arrow mg up. – Doryan Miller Mar 17 '14 at 10:27
• weight is pair with the force acting on the earth by the body, isnt it? Because weight is acting on me by the earth. And the opposit force should be the force acting on the earth by me. In person-floor case, normal reaction force does pair with weight, but in this case, it doesnt. Am I wrong? – user127595 Mar 17 '14 at 10:37
• When you remove the chair, yes. You exert your weight on the Earth, and the Earth in turn exerts it's normal force on you. – Doryan Miller Mar 17 '14 at 10:44
• So in this case, it is the same. I also exert my weight on the Earth. I am concentrated in the right pair of forces. I do agree that the magnitude of force acting on the chair is equal to normal reaction force from the chair, but I dont agree weight and reaction f by chair are a action reaction pair. – user127595 Mar 17 '14 at 10:48

Pure semantics, The body weight generates an equal and opposite reaction from the chair, on the other hand the chair and the weight of the person as a pair generate a force equal and in the opposite direction of the pair caused by the floor where the system rests. in that order of ideas there are forces as pairs reacting in the different surfaces with opposite and equal value.

To clarify why normal reaction is equal to weight you can alwas think graphically. Equilibrium equations are anlytical notation of vector forces. If you draw free body diagram of a body in equilibrium it's always much easier to understand problem and set equilibrium equations. Drawing free body diagram you draw force vectors or better say draw graphical representation of acting forces.

Here is free body diagram of your problem, normal reaction forces act in pairs. You can analyze each body individually or use solidification principle. In case where person's feet do not touch ground:

If you now set equilibrium equations you could calculate each normal reaction and in your case prove that normal reaction or as you said force with which chair acts on you is same as your weight (in case when your feet do not touch the ground).

Also you could use solidification and analyze human and chair as one body. And have only 2 normal reaction (if feet are not touching ground).

If you think in view of gravitational field it may seem to you that this normal force of chair is invented and you may be right but also you cannot dismiss that chair is between body and earth and gravitational fiel tends to bring body cleser by deforming this chair. The rigid matter of chair or better say connections between particles oppose this tendency in the same ammount as the weight of body. The weight of chair is very small in comparison to earth and it does not significaly affect mass of Earth.