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2007 F=ma exam number 28 bicycle

  1. Jan 20, 2014 #1
    1. The problem statement, all variables and given/known data
    A simplified model of a bicycle of mass M has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is w, and the center of mass of the bicycle is located midway between the tires and a height h above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude a. Air resistance may be ignored.

    Case 1 (Questions 28 - 29): Assume that the coefficient of sliding friction between each tire and the ground is \mu, and that both tires are skidding: sliding without rotating. Express your answers in terms of w, h, M, and g.

    What is the maximum value of \mu so that both tires remain in contact with the ground?

    the picture and problem is here: http://www.artofproblemsolving.com/...sid=dbbc9ef02e52f1694956d4336f64b1a2#p2853851


    2. Relevant equations
    ∑[itex]\tau[/itex]=I[itex]\alpha[/itex]
    [itex]\tau[/itex]=Fr



    3. The attempt at a solution

    I tried setting the front wheel contact point as pivot point and then having a weight force down on the center of mass and then normal forces on both wheel contact points as well as friction forces. Couldn't get the answer. Then I tried using the center of mass as pivot point and still couldn't get anything. What am I doing wrong? Thank you.
     
  2. jcsd
  3. Jan 20, 2014 #2

    mfb

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    Please show those attempts, otherwise it is hard to see what you did wrong.

    How did you evaluate the condition for the maximal value of ##\mu## to have both tires on the ground?
     
  4. Jan 20, 2014 #3
    Hello mfb! Thank you for the fast reply. I just thought of the bike as not rotating so that if I solve for mu then I have the mu at which the bike does not rotate. I uploaded a picture of my work here: http://postimg.org/image/egoldhcp9/
    sorry about the bad lighting.
     
  5. Jan 20, 2014 #4

    mfb

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    That sketch makes clear where the problem is: you assume that both tires carry the same weight - that is not true, otherwise the bike would not change its velocity.
     
  6. Jan 20, 2014 #5
    So the front wheel would bear more weight because the bike is about to tip over correct? That makes the problem harder to solve :(. How would I solve for mu then if I have two different normal forces and therefore two different frictional forces? Thanks.
     
  7. Jan 20, 2014 #6

    mfb

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    Right
    No, it is necessary to make it solvable.
    What is the condition for the bike to keep ground contact with both wheels, i.e. not tipping over? What can you say about the forces at this point?
     
  8. Jan 20, 2014 #7
    For the bike to not tip over the net torques must be zero and if taking the pivot point to be the front wheel contact point then the normal force on the back wheel times w must equal the weight of the bike times w/2?
    Or in other words:
    Fn(w)=Mg(w/2)?
     
  9. Jan 20, 2014 #8

    mfb

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    That is right, but not the point I was asking about. This has nothing to do with torques. If the force is positive the bike stays on the ground, if the force would be negative the tire loses contact. Where is the point between those two?

    No. As your pivot point is accelerated, there is an additional force you have to consider. It comes from the acceleration of the bike.
     
  10. Jan 20, 2014 #9
    Here what force are you referring to? The net force? Sorry I'm a little slow at this :)
     
  11. Jan 20, 2014 #10

    mfb

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    The vertical force between rear wheel and road.
     
  12. Jan 20, 2014 #11
    Sorry for late reply. I'm getting ready for the snow lol. I just read my textbook and it says pivot points must be taken at center of mass if the object is accelerating so you are right about the pivot point I chose having other forces. Also is the force you talk about above the normal force of the ground on the wheel? What you are saying is that the front wheel takes all of the weight because the normal force on the rear wheel is not positive or negative but zero. If that is true than doesn't that violate the problem which states that both wheels should be in contact? Also you said that the if the force is positive then the bike stays put and if its negative then it rotates but if the normal force on rear wheel is positive wouldn't it push the rear wheel up and therefore rotate the bike? By the way the answer is w/2h which I do get when I use torque equation and the thing you said about the rear wheel not touching so thank you! (I know u said this doesn't have to do with torques but thats the only way I know how to solve it)
    Thanks so much for being patient. Sorry to ask so many questions, its just that I really like to understand the physical phenomena behind physics problems and I tend to ask a lot of questions lol. You are the bomb!!
     
  13. Jan 21, 2014 #12

    mfb

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    I talked about more than one force, this is one of them.
    Right.
    It is "just" in contact - the bike does not rotate, so the wheel does not lift from the ground.
    Put an object on a table. There is a force between the object and the table. Does the object start to float?
    Torques are relevant, just not at the step where I said "torques are not relevant at this step".
     
  14. Jan 21, 2014 #13
    Thanks for being patient with me. Here is a pic of the "correct work"
    I still can't shake the weird feeling that this problem gives me. It doesn't feel right to set the normal force of the rear wheel to zero. I sorta get what you're saying about the wheel "just barely" touching the ground but for something to have contact with something else then there must be a normal force?

    It doesn't float because although the object has a weight force mg, it is countered by the normal force from the table onto the object. But in this question the weight is not at the wheel itself but in between the two wheels which kinda confuses me as to this barely touching business. There is no such thing as negative normal force (if you're making the positive direction up) because normal force can't go back into the ground. if there is a normal force then the bike is touching and if it is zero then it is not. Basically in this problem all the weight at the center of the bike is being countered by the normal force acting at the front of the wheel. This normally would cause the bike to flip upward if you think about torques but there is a friction force at the front wheel to counter that torque. Basically the rear wheel does nothing? Still doesn't make sense to me.

    Also here you say:
    but what do you mean by this? The friction force causes the bike to change velocity. What does that have to do with setting the tires to have different normal forces?

    Promise this will be last question even if I don't understand it lol. Thanks so much.
     
  15. Jan 21, 2014 #14

    mfb

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    No, the force can be zero. This is an ideal case, of course - in a real setup, even the tiniest road bump would change the situation.

    Right, as long as you don't connect ground and wheel with a rope.

    Gravity counters this torque (as seen from the front wheel, acting on the center of mass // as seen from the center of mass, acting as normal force at the front wheel).

    If both normal forces are the same, the frictional forces are the same, and torques are unbalanced. That leads to the "contradiction" you have in your first image.
     
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