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We have boxes and we are picking up a wood and put in between our hand and the box. We start pushing wood with 20N force. This force Is transferred from wood to the box and box return 20N to wood (Newton’s third law). So we have 20N from left to right and 20N from right to left that have an effect on box1. So why the wood (box1) start moving? (You can use any force instead 20N.) (Please check image).

Arastoo haghighi
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 21 '21 at 18:00
  • The 3d law statement "for every action there is always an equal and opposite reaction" can indeed sound confusing, and sounds more like a clever wordplay. Did Newton the great physicist/mathematician actually wrote that exact sentence as his definition? –  Oct 21 '21 at 18:23
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    Does this answer your question? Given Newton's third law, why are things capable of moving? – garyp Oct 21 '21 at 18:30
  • Actually there are some difference in our question. – Arastoo haghighi Oct 23 '21 at 08:27
  • I asked this question again at https://physics.stackexchange.com/questions/673162/newton-s-third-law-understanding – Arastoo haghighi Oct 23 '21 at 08:34
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    Please don’t re-ask a closed question. Instead, [edit] to clarify why the answers at the duplicate, or the answers here so far, are inadequate. – rob Oct 23 '21 at 10:52
  • Newton's Third Law does imply that the other forces (block 2's exertion on block 1 and vice versa) are also $20\text{ N}$. The Law only implies that the force block 1 exerts on you is $20\text{ N}$. The other magnitudes are natures' decisions, and with different masses you will find different values. – gmz Oct 23 '21 at 12:34

2 Answers2

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You can't say that the contact force has a magnitude of 20 N.

To find the contact force, firstly find the acceleration of the system, which will be $\frac{net\: external\: force}{net\: mass}$. After this, you can draw the individual f.b.ds to find the contact force. This I leave to you.

The major error was directly stating that the first block exerts 20N on the second block

gmz
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imposter
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  • Hello. Your say is true for time that we accept that system is start moving, but in this case we don't have acceleration because I think there is no pure force. – Arastoo haghighi Oct 23 '21 at 08:21
  • @Arastoohaghighi what do you mean by pure force . The user hasn't mentioned presence of friction which can resist motion and 20N is the external force due to which acceleration has to be present – imposter Oct 23 '21 at 15:57
  • Can u add the actual diagram or statement – imposter Oct 23 '21 at 15:59
  • I mean at start the system (especially block1) not move And after that the net force is zero for block1 so it shouldn't start moving. But you say that 20N isn't transferred completely, for example imagine16N is transferred. So what happen for that 4N that remain? – Arastoo haghighi Oct 23 '21 at 16:58
  • That is used for providing acceleration – imposter Oct 24 '21 at 06:13
  • Thanks for your great answering – Arastoo haghighi Oct 24 '21 at 09:42
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Your free body diagram already has the answer. There are two 20N vectors pointing right and one pointing left. $\sum F = ma$. If you follow the chain of forces back through your hand, arm, body, etc, you'll keep finding one extra instance of 20N right until you get to the planet you're standing on, which balances out with one extra instance of 20N left.

g s
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  • Actually, there isn't two force and it is one force that Is transferred. In this case I want to check out the force that has an effect on object number 1 – Arastoo haghighi Oct 23 '21 at 17:05