1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanics of Wheel and Forces Involved

  1. Apr 21, 2012 #1
    If an axle is being pulled with a tractive effort E, can this be resolved into a couple and forwards force E at the point of contact with a surface? Is it correct to add a limiting friction force F acting backwards against the direction of motion and if this is exceeded by E, does the wheel slip?

    Attached Files:

  2. jcsd
  3. Apr 21, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi Eugbug! :smile:
    Yes, that's completely correct …

    the two alternative descriptions are completely equivalent. :smile:

    But I don't see how that helps you with the friction force :confused:
  4. Apr 21, 2012 #3
    The inspiration for this question came about when I was coming down a mountain at top speed on my mountain bike and was thinking about how braking works on a wheel and whether a skid can occur if enough braking force is applied and the wheels don't actually lock.
    To make things simple, take a single wheel rolling along a level surface. The wheel is weightless so there is no moment of inertia to take into account, just say there is a mass M attached to the axle. If a braking force is applied at the perimeter of the wheel and this produces a sliding friction force Fb acting against the direction of rotation of the wheel, what are the forces involved and how do you work out the deceleration of the wheel? The velocity of the wheel is v.

    The way I would approach things is to resolve Fb into a couple Fb and a force Fb acting against the direction of motion at the axle and then the equation of motion becomes:

    -Fb = Ma where a is the deceleration of the wheel. Is this correct?

    Attached Files:

    Last edited: Apr 21, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook