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This question already has an answer here:

Newton's third law of motion states that every action has an equal and opposite reaction. That is the reason we do not sink into the earth, because when our weight exerts a force on the earth it also exerts an equal and opposite force on us.

But when we stand on quicksand or on fluids we can sink in. How is this possible? Does it not exert an equal and opposite force on us? Or are Newton's laws different in the case of fluids and substances of low densities?

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marked as duplicate by AccidentalFourierTransform, stafusa, Jon Custer, Sebastian Riese, Prahar Jun 4 '18 at 12:50

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    $\begingroup$ Possible duplicate of With Newton's third law, why are things capable of moving? $\endgroup$ – AccidentalFourierTransform May 30 '18 at 17:56
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    $\begingroup$ The sand's motion is the "opposite reaction" (if only consider the sliding and not the "falling" part, where the "opposite reaction" is the motion of the Earth). But if you want a good understanding of a physical law, you really need to read it very carefuly. Ignoring even individual words can confuse you - they can carry a lot more meaning than may be immediately obvious. $\endgroup$ – Luaan May 31 '18 at 8:06
  • $\begingroup$ Can't answer yet.. but the force on the sand is equal to force on person, but the sand is accelerated out of the way because of small mass. $\endgroup$ – MarsJarsGuitars-n-Chars May 31 '18 at 22:41
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This is a common confusion when people first learn Newton's Third Law. They get the idea that it implies motion can never begin.

The (wrong) argument is that, since the Earth exerts a force $F$ on you by gravity, the sand must exert an equal and opposite force $N = -F$ on you. Then the total force on you is $N + F = 0$, so you can't fall. Of course this argument can't be right, because it either means that nothing can ever start moving, or that Newton's laws don't apply to sand, as you propose, and neither make sense.

What Newton's Third Law really says here is

  • if the Earth exerts a gravity force $F$ on you, you exert a force $-F$ on the Earth
  • if the sand exerts a normal force $N$ on you, you exert a force $-N$ on the sand

It gives no relationship at all between $F$ and $N$, so you can indeed fall.

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    $\begingroup$ Note that "falling" with constant speed -- well, it's not physically free falling, so let's call it sinking, like sinking into sand or a fluid -- somewhat counterintuitively does not require a force difference, since the motion does not change. Newton's first law itself dictates it: "perseverare in statu suo quiescendi vel movendi..." (bodies "stay in their state resting or moving...") unless an outside (net) force acts upon them. It's just the acceleration in the beginning which needs a force. $\endgroup$ – Peter A. Schneider May 30 '18 at 13:06
  • $\begingroup$ Yes ;-)... Yes, seriously. That distinction is important because it is not intuitively clear that there is zero net force when sinking into sand. $\endgroup$ – Peter A. Schneider May 30 '18 at 13:17
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    $\begingroup$ Also, what Newton's Third Law does mean is that when something starts moving, something else must have started moving in the opposite direction. The trouble is that in the environment familiar to us, very often the something else is the Earth, whose inertia is so immense that its velocity does not appear to change. $\endgroup$ – zwol May 30 '18 at 14:21
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    $\begingroup$ @JimmyJames Yes, it requires a force; no, it does not require a net force. The same force I'm exerting on it is exerted by friction on the marble in the opposite direction; that is why it doesn't accelerate, once it's started moving. It's a similar situation a person experiences falling in the atmosphere at terminal velocity. S/he is in force equilibrium between friction and gravity. $\endgroup$ – Peter A. Schneider May 30 '18 at 15:26
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    $\begingroup$ @zwol Archimedes reportedly said, "Give me a lever and a place to stand and I will move the earth". What he didn't realize is that he already had both. His legs were his lever. He moved the earth every time he got up out of bed. Just not by much. $\endgroup$ – candied_orange Jun 3 '18 at 3:38
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Read the law well!

If body A exerts a force on body B, then body B exerts a force (equal in magnitude but opposite direction) on body A.

So: force on B from A = - force on A from B.

We have two forces on two different bodies.

So your conclusion may be: If the earth pulls on us, we pull on the earth (via gravity! Not via "pushing with out feets". Like the sun on the earth and vice versa). No sand at all so far. Well, we don't fall but the sand $brakes^1$ our motion, so it certainly exerts a force on us but less than the earth does on us! No one ever said, that the force of the earth on us is the same as the force of the sand on us (remember: the law speaks of two forces on two different bodies). On the other hand, it is of course true that we exert the same force as the sand exerts on us on the sand. That's why the sand moves away.

1: This is not the correct physical formulation; constant speed, no force, blabla. Of course, but the point of this answer is to make the distinction between on whom the forces act. And not on an exact description of a non-linear motion.

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Two things.

  1. A uniform motion (like sinking into quicksand with constant speed) does not require a net force; the force onto our foot soles touching the sand exactly counteracts our weight. If that was not so, we would accelerate. In fact, Newtons's First Law forbids a force difference. We accelerated when we started to sink in, and then there indeed was a net force (because the soft sand or fluid was giving and could not carry us) as is required for acceleration.

    But what happens once we have started sinking in? In fluids and probably in quicksand (which I assume behaves similar to a fluid) the friction increases with velocity; at some point a body moving through it will encounter an amount of friction which cancels out its weight. From then on it will move with constant velocity because no net force is exerted on it any longer. For a person falling through air that's the terminal velocity, about 250 km/h. In water the friction is much stronger and the terminal velocity will be much smaller.

  2. Interestingly there is an equilibrium of forces in the complete system even when we accelerate; the earth is drawn towards us by the same force by which we are drawn to the earth, and it accelerates (ever so minutely) towards us, preserving the earth+man system's momentum.

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    $\begingroup$ What are your grounds for claiming that you will sink at constant speed? Separately, even if it were true, I don't think it's a hepful contribution to this question - that's not the aspect of the system that the OP is confused about. $\endgroup$ – Brondahl May 30 '18 at 15:12
  • $\begingroup$ @Brondahl Em... the OP asked "But when we stand on quicksand or on fluids we can sink in. How is this possible? Does it not exert an equal and opposite force on us?" That "sinking" has usually two phases: An acceleration phase and a plateau phase when friction (which in gases, fluids and probably sand since it may behave like a coarse fluid is velocity dependent) equals gravity. So I assume that at some point one will sink at constant speed. This answer answers the question for the plateau phase. The acceleration phase is probably less debated because indeed there is no equilibrium. $\endgroup$ – Peter A. Schneider May 30 '18 at 15:34
  • $\begingroup$ @Brondahl I edited the answer a bit to emphasize the acceleration phase during the time a net force exists. $\endgroup$ – Peter A. Schneider May 30 '18 at 15:36
  • $\begingroup$ Given enough time, gravity doesn't care about friction. You will stop sinking once you've displaced your weight (then it's "from then on"). "the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g." - 'friction increases with velocity; at some point a body moving through it may encounter an amount of friction which cancels out its weight.' but with QS, we're so far from termV, that that's totally negligible. $\endgroup$ – Mazura May 31 '18 at 1:04
  • $\begingroup$ @Mazura You are right about quicksand and displacement; QS is probably a complicated affair (it's probably non-Newtonian, how can you get submerged at all when you are lighter etc.). Let's say I was answering the fluid part after full immersion, or the first inches of sinking into (quick) sand when displacement is still negligible, ignoring suction etc. $\endgroup$ – Peter A. Schneider May 31 '18 at 1:16

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