# Resisting force depends on velocity? [duplicate]

Why does resisting force depend on velocity? I think there is no relation between resisting force and velocity of object. Please speak about it logically.

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## marked as duplicate by Waffle's Crazy Peanut, Dilaton, dmckee♦Jul 14 '13 at 13:29

The air resistance, a force sometimes known as a drag, always increases with the speed but the precise dependence on the speed differs in various situations, especially depending on the presence of turbulence etc.

For low speeds, the air resistance is proportional to the velocity, $\vec F = -b\vec v$. This is known as Stokes' drag. A direct proportionality is the only natural, smooth function of a vector mapped to another vector. The unit (or constant-length) vector in the direction of motion isn't natural as it would be discontinuous for $\vec v =0$.

When you're moving by speed $v$, a higher number of particles hit you on the front side. The difference between collisions from the front and from the back go like $v$ and because each collision with a (very fast) particle effectively changes your momentum by the same amount, the change of the momentum per unit time – the force – will be proportional to the velocity, too.

At higher velocities, the drag increases as a higher power, typically quadratic power, of the velocity. That's for a higher Reynolds number, for turbulence. It takes some effort to quickly explain the scaling but it can be done. Well, let me say something: $$F_D\, =\, \tfrac12\, \rho\, v^2\, C_d\, A,$$ This is the formula for the drag. It doesn't depend on the velocity at all (because the drag is due to universal turbulence and "vortices" and not just some surface friction). Because both sides have to have the same units, $F$ has units of mass times distance per squared time (mass times acceleration), the velocity is the only factor that can give us powers of a second (the unit of time), and it has to be squared to produce the right power. We call this argument "dimensional analysis".

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I find your explanation of the drag proportional to $v^2$ a bit strange. Since you are saying that the velocity is only part of the expression for drag to suffice to the unit of force. – fibonatic Jul 14 '13 at 12:20
Dear fibonatic, it's enough to look at the velocity because it's the only quantity that contains a nontrivial power of a second. The task to calculate the powers of $\rho,v,A$ is equivalent to a set of linear equations but the equation for the right powers of one second simply tells you that the exponent above $v$ has to be $2$ because the force contains $1/s^2$. – Luboš Motl Jul 14 '13 at 18:50
I agree that the units should be the same on both sides of the equal sign, however the unit $s^{-2}$ also could have been part of the constant $C_D$. I think it would be better if a model would be based on experiments or calculations, instead of just matching units. – fibonatic Jul 14 '13 at 21:48
Nope! $C_D$ is a dimensionless coefficient, the drag coefficient, that only depends on the shapes etc. It has to be dimensionless because the fluid etc. simply has no other dimensionful parameter that would determine the relevant mechanical properties in this situation. So such a dimensionful $C_D$ couldn't emerge from any legitimate calculation. It's the whole point of dimensional analysis that we classify all dimensionful quantities whose powers may enter the equations and here it's just the viscosity and the dependence on viscosity must disappear in the turbulent limit. – Luboš Motl Jul 16 '13 at 9:54