i have seen newton's third law as every action has equal and opposite reactions and they act on different bodies,

my concern is is it only when both the bodies are at rest as while standing on water i m giving a force mg to water and the water should give me back mg keeping me in equilibrium,where m i going wrong.

i understand that its the normal reaction of water that should balance my weight to keep myself in equilibrium but my concern is why the normal reaction is not equal to mg,i understand it very well that as my foot presses water,the water yields,but my query is why all these things happen.

  • 1
    $\begingroup$ If your foot is going through the water, why do you think you are giving a force mg to the water? $\endgroup$
    – Jiminion
    Jan 17, 2018 at 16:37
  • $\begingroup$ The issue you face here is the fact that your weight is more force than the surface tension of the water can oppose. The ability of objects to give back a normal force equal to your weight is contingent on if they are able to take the pressure. To reconcile this with the third law, consider that the system your talking about is not as simple as examples with solids. You have the water being compressed and moved and being sucked over your foot because of the negative pressure there. It's a more complicated problem. $\endgroup$
    – Ben
    Jan 17, 2018 at 16:48
  • $\begingroup$ the effect is same while walking on sand where there is no surface tension,do we need to put extra information in newtons third law. $\endgroup$ Jan 17, 2018 at 17:02
  • $\begingroup$ The same principles apply. If sand is tightly packed in a solid it will be able to resist your weight more. If sand is loose the weight of your body can displace the particles, but Newton's 3rd law still applies in all these situations. $\endgroup$
    – LUPHYS
    Jan 17, 2018 at 18:16
  • $\begingroup$ You could ask the same question about air. When you step up or step down, why doesn't the air hold you up? It's not because of a "failure" of Newton's 3rd Law. It's because you aren't actually exerting a force of mg on the molecules. They accelerate out of the way long before the sum of forces reaches mg. On the other hand, if you move fall fast enough through the air, the net force can become mg and you won't accelerate anymore. This also happens to race cars such as top fuel dragsters where there are enough collisions per second to produce a net 0 acceleration forward. $\endgroup$
    – Bill N
    Jan 17, 2018 at 19:59

3 Answers 3


As we know from Newton's second law, F=ma. When you step on the water you are accelerating the water away from your foot, this means there is less reaction force felt by your foot because most of the force is going into displacing the water.

  • $\begingroup$ so does it mean that newtons third law is valid only when both bodies are at rest. $\endgroup$ Jan 17, 2018 at 17:00
  • $\begingroup$ No, imagine you are swimming through sand. You would push the sand backwards and the sand would push you forwards. $\endgroup$
    – LUPHYS
    Jan 17, 2018 at 18:18

while standing on water i m giving a force mg to water

Actually, this isn't happening.

You might have a reasonably strong arm and be able to push on a scale with a force of 10 pounds. Then I give you a single feather and ask you to apply 10 pounds of force to it. You won't be able to. The feather will accelerate so quickly, that it doesn't supply such a strong force/resistance to your arm.

The same thing is happening with the water. When you try to push down on it with your feet, the water accelerates away instead of pushing back (very strongly) to you.

So the third law is working and both you and the water are exerting a force on each other. The problem for you walking is that the force is less than $mg$.


Answer lies in the fact that every matter has a property of elasticity and rigidity, then if we consider a solid road on which we walk then too we exert some deformative force but that solid is strong enough to hold it. Now when we put our feet on water then since water is not rigid enough, its molecules get displaced from their mean position. And yes every action has an equal but opposite reaction which is applicable here also considering the kind of surface you are taking, for example if you jump from a height over sea water it would behave as an equally strong road as water has from a lamina over at that place.

  • $\begingroup$ if that is the case we should have got same pain when we hit a wall and hit a sponge surface. $\endgroup$ Jan 17, 2018 at 19:12

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