Newton's third law.. please explain here

According to Newton's third law, the force acting on object is equal and opposite to the other object that is applying the force. If I push a table with 5N force then table is doing the same. But the table and I have different masses, so how could the force be same? Isn't $F=ma$?

• Possible duplicate of physics.stackexchange.com/questions/141913 & links therein.
– user36790
Nov 24 '14 at 11:16
• Not a moderator but will advise you, before posting any quo, google it. We are here for making this site the best collection of all the intuitive questions. We are not meant for answering the same quo again and again. But since you are new, to err is human !
– user36790
Nov 24 '14 at 11:19
• I googled it and searched here also but I didn't find any satisfactory answer. Now I think I am on the right site Nov 24 '14 at 11:23
• We have a whole bunch of questions asking "given N3 law, why can anything move?" but I don't see one asking how the force can be the same. So I think (but not sure) this may be addressing a different aspect of the law than most of those other questions. If that's the case, it might not be a duplicate. Nov 24 '14 at 17:57

Suppose you and the table are floating in space. If you push the table it will go in one direction and you will go in the other direction. Your and the table's acceleration will be different so you will end up travelling at different speeds. This is obvious from conservation of momentum. The momentum in the centre of mass frame is initially zero, so after the push your velocity $v$ and the table's velocity $V$ are related by:

$$MV = mv$$

where $M$ is the mass of the table and $m$ is your mass. Your velocity will therefore be:

$$v = \frac{M}{m} V$$

If you walk up to a table in your living room and push it then other forces come into play. These include the force on the table legs exerted by the living room floor and the force on your feet exerted by the living room floor. You would need to take these into account to analyse the situation properly.

• MV = mv $\rightarrow P = p$ the formula says that momenta are equal and not forces (F = m * a). The law is thought (action) to refer to forces but shouldn't it read "to any (action) momentum corresponds an equal momentum (reaction)" ?
– user59485
Nov 24 '14 at 11:16
• @Alba: change of momentum = force $\times$ time. The time you push on the table must be the same as the time the table pushes on you, so therefore the momenta being the same means the forces are the same and vice versa. Nov 24 '14 at 11:19
• Okk got your answer, table and I both will have equal forces exerting on each other.. and we will be moving in different directions.... Whether the speed will be uniform or accelerated?? Nov 24 '14 at 11:21
• @SuyashPrakash: as long as you are pushing on the table you and the table will both be accelerated. Once you stop pushing you and the table will move at a constant velocity. If you push with a constant force then your (and the table's) acceleration will be constant. If the force of your push varies then so will the acceleration. Nov 24 '14 at 13:16
• Constant acceleration means if I have pushed the table and the acceleration I received is 2 m per second square... So I will be accelerating at the speed of 2m/s every second??.. It means after duration of 60 seconds, my velocity would be 120 m/s and increasing with time to the infinity.. is it? Nov 24 '14 at 13:50

Your mass or your acceleration doesn't tell you anything about the forces acting on your body. F=ma simply states that if we add (vector addition) all the force acting on your body then the effect of the net force will be given by your mass times acceleration.
There may be a lot of forces acting on your body but if the net sum of these is zero, the body simply stays the way it was before (doesn't move if it were in rest).
The force exerted by the table in not the only force acting on you while you are pushing it. There is also force due to friction acting on you in opposite direction.
The net effect of these forces combined will give you your mass times acceleration.

This question has to be a duplicate, but just in case it isn't ...

You are confusing the force Newton's third law with the force in Newton's second law. Newton's third law addresses the individual forces on two different objects subject to a common interaction. Newton's second law addresses the net force on one object.

The net force on the book are a result of the gravitational interaction between the book and the Earth as a whole and the normal interaction between the book and the tabletop. Ignoring planet rotation, these cancel one another. The net force on the book is zero. Note that just because these two forces cancel one another does not mean that these are third law action-reaction forces.

The net force on the table is the result of the gravitational interaction between the table and the Earth as a whole, the normal interaction between the legs of the table and the floor, and the normal interaction between the tabletop and whatever is piled atop the table. Ignoring planet rotation, these collectively cancel one another. The net force on the table also is zero.

• @Alba - I haven't the faintest idea what you mean by that. In retribution, you deselected a very good answer I gave in favor of an alternative that is in actuality an incorrect answer? That makes no sense. Nov 24 '14 at 13:42
• @Alba - People regularly confuse the force in Newton's second law with the force in Newtons' third law. They are very different things. Two of Newton's corollaries to his three laws deduce that (in modern nomenclature) forces add vectorially. Unfortunately, Newton's proof of those corollaries is circular. In modern physics education, it's assumed that forces are vectors; they add vectorially because that's what vectors do. A few physics instructors teach that the vectorial nature of forces is in essence Newton's fourth law. This, IMHO, is the right path. Nov 24 '14 at 13:54
• @Alba - Flag me then. I don't care. Then are plenty of popular answers on this site that are completely wrong. Science is not a popularity contest. Nov 24 '14 at 14:49
• @DavidHammen I have been pinged to respond to a comment you made by an anonymous party. When you reference that Alba deselected an answer of yours in favour of an incorrect answer, are you referring to the question about the sky being green? If so, it might be more productive to address your issues to my post so that we might work out the problems together rather than berate someone that doesn't know which is correct or not for accepting the most upvoted answer. If that wasn't the case, then I retract my statement and I'll delete this comment
– Jim
Nov 24 '14 at 15:24
• @Alba I have had no reply from David and I'd rather not rush to conclusions. Besides, I assume even in the event that his comment was aimed at my post, he would be professional enough not to hold a grudge. My participation here is solely in the interest of determining what, if anything, he think is wrong with my post. Anything beyond that is outside of my business
– Jim
Nov 26 '14 at 16:05

Here there are two objects, One is the Person pushing, other is the Table. Both have applied force which is equal to (say) 5N, F=5N.

• Mass of the Person (say) m=20kg
• Mass of the Table (say) m'=10kg
• Acceleration of the Person= a m/s^2**
• Acceleration of the Table = a' m/s^2
• Force of the Person and Table= 5N ---- {According to Newton's Third law}
• Applying Newton's Third law to the Person --- F=ma implies 5=20a

Applying Newton's Third law to the Table --- F=ma' implies 5=10a'

• You see, acceleration is variable i.e. a is not equal to a' and as the Person has more momentum than of the Table, obviously the Table has to move but Force remaining Equal and Opposite.
• -1. This answer is fundamentally wrong. Newton's third law pertains to individual interactions between objects. The individual force exerted on the table by the object resting on the table is equal but opposite to the individual force exerted by that object on the table. There's no F=ma here because F=ma pertains to net force rather than individual force. Nov 24 '14 at 11:19