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Had a question on a physics test, a person is at rest on a frictionless surface and can not throw anything, can they move?

I said yes, by "swimming" against the air and got docked because "it wouldn't generate enough force."

So that makes me wonder, how much force can a human generate by flapping their arms?

Edit: So it will be the mass of the air moved times the velocity it is moved at, i.e.: the momentum created in the air. This is going to be related to the size, shape, and velocity of the arm, but where do I go from here to figure out how to calculate this?

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  • $\begingroup$ By the way, if there is some way to calculate this, I'll be more than happy to do the legwork myself if someone can point me in the right direction of how I could go about calculating this. As it stands, I can't even find where to start. $\endgroup$ – Pevinsghost Apr 20 '16 at 1:33
  • $\begingroup$ they can move by pushing air. The force to move something without friction can be as small as you want. although the movement would be small as well. $\endgroup$ – Peter R Apr 20 '16 at 1:34
  • $\begingroup$ @PeterR Granted, but is there some way I could calculate the force created? $\endgroup$ – Pevinsghost Apr 20 '16 at 1:36
  • $\begingroup$ Yes, without friction, F=ma, where m is your mass and a is the acceleration from a force F. What the equation is telling you that for any force F, however small, there will be an acceleration. If there was friction, the force would have to overcome the static frictional force before you began to move. $\endgroup$ – Peter R Apr 20 '16 at 1:49
  • $\begingroup$ OK, found a start, basically, it will be the mass of the air moved times the velocity it is moved at, i.e.: the momentum created in the air. This is going to be related to the size, shape, and velocity of the arm. Since it involves angular motion from an axis at the shoulder, this will probably play a factor as well. $\endgroup$ – Pevinsghost Apr 20 '16 at 1:50
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You can calculate the air drag along your arms using the formula for air drag, $F = \frac{1}{2} \rho u^2 c_D A$, takin $u$ the speed at which they'll effectively move relative to air, and $A$ their projected area on the plane perpendicular to motion. $\rho$ is the density of air, that's why it's so much less efficient than in water.

$c_D$ will be close to 1. But you'll have to subtract the drag on your body, and its $c_D$ will be also close to 1 with a larger area. So because the projected area of your arms is little, they'll have to move fast to hope to make a difference...

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