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[High school physics] Wheel touching a kerb. What are the forces on the wheel?

  1. Apr 18, 2009 #1
    [Solved] Wheel touching a kerb just about to lift. What are the forces on the wheel?

    1. The problem statement, all variables and given/known data
    In order to lift a wheel over a kerb of height h, a minimum force F is be applied at the axle. The radius of the wheel is R. Draw all the forces on the wheel then find its weight.

    *IMAGE003*

    2. Relevant equations
    T = r X F

    3. The attempt at a solution
    I managed to find the weight of the wheel. But what I am confused about is the drawing of the forces on the wheel.

    My teacher presented a drawing like this below.

    f is the tangential friction on the wheel.
    F is the horizontal force at the axle.
    W is the weight of the wheel.
    N is the normal reaction force at the point of contact.

    *IMAGE003a*

    Since he argued that without friction the wheel will slide and therefore the wheel cannot be pushed up the kerb. Thus we must conclude that the normal reaction force on the point of contact is not passing through the center of the wheel, otherwise that point will move.

    However I do not understand. Because if we imagine the surface of the point of contact is an infinitesimally flat surface, and the friction is parallel to the surface but the normal reaction force would be perpendicular to the surface. Therefore N would pass through the center for geometrical reasons.

    However this is a dilemma. Since if there is friction then that point of contact will move. So I must conclude that there is no friction present. In other words, since there is no force parallel to that surface in contact, there cannot be any static friction.

    So instead my drawing is like below.
    *IMAGE003b*

    Which is correct? My teacher gave an example of the situation on an ice kerb. He says that it will otherwise be impossible for the wheel to be lifted up by the horizontal force. However, I feel that even if there is no friction it is possible for the wheel to be lifted up the kerb.

    Both drawings give the same answer to the weight of the wheel since it is done by taking the torque relative to the point of contact.
     

    Attached Files:

    Last edited: Apr 18, 2009
  2. jcsd
  3. Apr 18, 2009 #2
    I realised my drawings are too large! Sorry for that! :)
     
  4. Apr 18, 2009 #3

    Astronuc

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    For the wheel to lift itself over the curb, it must pivot about the contact point. And friction must prevent slip (no slip implies static friction).

    Applying the weight of the wheel at is center of mass, then what is the moment arm between the CM and contact point? One must consider the angle between the moment arm and weight.
     
  5. Apr 18, 2009 #4
    Yes I have found the weight to be

    W = F [(R -h) / (h(2R - h))^0.5]

    But I still cannot understand why is there friction...
     
  6. Apr 18, 2009 #5

    Hurkyl

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    Reasoning with infinitessimals is tricky -- you should put little faith in such arguments unless you really know what you're doing, or you know how to turn the argument into a sound one without infinitessimals.

    e.g. once we start thinking about very, very small scales, we have to worry about the wheel and curb deforming at the point of contact to accomodate the forces involved. And even an "infinitessimal" change in an "infinitessimal" surface can have a large effect on the direction of the normal.
     
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