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Why is friction proportional to velocity?

  1. Jun 18, 2012 #1
    We all know in phenomenological equations we like to add a [itex]-\gamma v[/itex] effective force term to Newton's equation to include frictional effects with the environment. Why this specific shape? I understand it's not fundamental, and that it is not always appropriate, but still it seems to be a good approximation in many cases. Why is this? I understand the intuitive idea that a particle moving gets more collisions head-front and as such will experience a backwards force, but why directly proportional to velocity (and not, say, a square root or a quadratic or... etc)?
     
  2. jcsd
  3. Jun 18, 2012 #2

    jedishrfu

    Staff: Mentor

    I think it fits as a good first order approximation and makes the DE more tractable.

    Similar to the idea with a pendulum system where we assume the angle of oscillation is fairly small so that sin(theta) = theta and the DE become easier to solve.
     
  4. Jun 18, 2012 #3
    My guess would be that if velocity doubles the amount of surface encountered by an object doubles and therefore friction doubles.
     
  5. Jun 18, 2012 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Why is friction proportional to velocity?

    In general, it's not. Friction of one dry object against another is fairly constant over a fairly large velocity range. For a solid object moving at a high speed through some fluid, friction is roughly proportional to the square of velocity.

    However, friction is roughly proportional to velocity for a solid object moving at a low speed through a fluid. Why proportional to velocity at low speeds, but the square of velocity at high speeds? One key difference is laminar versus turbulent flow.
     
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