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Consider the following scenario: a body is dropped and it hits the air molecules below it with a force equal to it's weight. By Newton's third law, the air molecules below should exert the same force on the body and force due to weight should cancel out and object should not fall. Why doesn't this happen?

Apoorv
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  • Think about the mass of the respective objects and the resultant acceleration – ChrisM Jan 19 '15 at 12:27
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    This is a duplicate of many questions already in here such as: http://physics.stackexchange.com/questions/39215/newtons-3rd-law-how-can-i-break-things or http://physics.stackexchange.com/questions/45653/with-newtons-third-law-why-are-things-capable-of-moving – Martin Jan 19 '15 at 12:29
  • This is not a duplicate to the marked question. The OP's misunderstanding is not Newton's 3rd law - he's correct that the air molecules and the body put equal and opposite forces on each other - his mistake is thinking that when a falling body hits an air molecule, that force must be the same as the body's weight. It's a misunderstanding of the force objects put on each other when they have a collision. – Brionius Jan 19 '15 at 20:29
  • @Brionius Can you please elaborate further? – Apoorv Jan 20 '15 at 20:21
  • When the object hits an air molecule, it does not put a force equal to its weight on the air molecule. The force that one object puts on another in a collision is determined by relative speed, impact angle, and the characteristics of the objects' surfaces. It's determined by conservation of energy, and the energy lost due to inelasticity of the objects. – Brionius Jan 20 '15 at 20:30
  • The force that a book puts on an air molecule is much, much less than its weight. All the air molecules on the bottom of the object collectively exert around 15 pounds force per square inch, and there are a lot of them hitting any macroscopic object! – Brionius Jan 20 '15 at 20:33
  • @Brionius I'd be grateful if you could point me to a link that discusses such force in detail – Apoorv Jan 28 '15 at 03:46
  • NOT a duplicate. Body accelerates until m.g = drag. It then continues at that velocity with no net force so Newton is happy. Basic drag equation is mg = 0.5 x Cd x Rho x A x V^2 where Cd= drag coefficient (1 for flat plate drag) Rho = air density (1.2 kg/m^3 mksa at sea level), A = frontal area and V = velocity. | If velocity rises above this point drag > m.g and it slows. If v slows drag drops and m.g -d drag accelerates it. – Russell McMahon Jan 28 '15 at 04:16

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It does happen, but often air turbulence makes such floating objects difficult to see. The most common example, are clouds, which float at a certain height.

You can also get a mist layer (photo).

iantresman
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  • This really doesnt ans the question. The question comes from a misunderstanding of Newtons 3rd law. – Paul Jan 19 '15 at 13:54
  • I think the question comes from a misunderstanding of how impact forces work - he thinks the force of the impact of the body on the air molecules must be the same as the weight of the body, which is not true. – Brionius Jan 19 '15 at 20:27