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Magnitude of friction on the driving wheel

  1. Sep 14, 2012 #1
    After discussing this with my friend using our O level knowledge, we came to a conclusion. But we have some queries.

    So firstly, we have our driving wheel on the air and the engine is turned on. So the Free Body Diagram would look like this: http://imgur.com/xVk0w

    So in this case the net force of the wheel is 0N as F2 and F1 are equal in magnitude. However, about F1, there is a net moment which turns clockwise (F1XL).

    Then now we put it on the ground and and experiences a friction oppsoing F1 as friction opposes force in a single plane of contact. So now, the Free Body Diagram would look like this: http://imgur.com/Pw1oh

    From this, since F2-F1=0N, so the net force=friction only. While the wheel still has to maintain its clockwise moment so F1L > FrictionL and since L is a constant, so F1>Friction.
    So from here, it all makes sense to us, but then after thinking again since that friction force that we have been dealing up to now is static friction so shouldn't F1 be equal to friction? But if F1 is equal to the friction, the the wheel would not be able to turn in the first place.

    So we were stuck at this part with no solution in mind. Could someone help us out with this? Thanks :-)
     
  2. jcsd
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