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Newtons second law homework

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A cyclist and her bicycle have mass 75kg. she is riding on a horizontal road, and positions herself so that 60% of the normal contact force is on the back wheel and 40% on the front wheel. The coefficient of friction between the tires and the road is 0.8. What is the greatest acceleration she can hope to achieve?
    Whilst riding at 6ms^(-1), she applies both brakes to stop the wheels rotating. In what distance will she come to a stop?

    2. Relevant equations

    Newtons second law

    3. The attempt at a solution

    I think the friction force of the back wheel will be the forward force of the bicycle so

    0.48R=75a ,i dont consider the frictional force of the front wheel ?

    I have no idea for the next.
     
  2. jcsd
  3. Jun 25, 2010 #2
    Re: mechanics

    You can calculate friction force of both wheel. Friction force have to do work on distance to stop. Conservation of energy is the easiest way to solve this problem.

    regards
     
  4. Jun 25, 2010 #3
    Re: mechanics

    thanks but isn't the friction force provide the forward force for the bicycle?

    and also is part 1 correct?
     
  5. Jun 25, 2010 #4
    Re: mechanics

    Yes, but only friction from back wheel (which is driven) provide forward force. During braking both wheels "produce" force.

    I dont clearly understand what you mean by "0.48R=75a"?

    regards
     
  6. Jun 25, 2010 #5
    Re: mechanics

    For (1), the friction of back wheel, (0.8)(0.6R) provides the forward force, ma.

    Therefore , (0.8)(0.6R)=ma ,also do i consider the friction of the front wheel here?

    ie 0.48R-0.32R=ma ?
     
  7. Jun 26, 2010 #6
    Re: mechanics

    No. When bicycle accelerated, front wheel did not produce friction. Only back wheel is driven by a cyclist.

    So, in (1) you have 0.8*0.6*m*g=m*a where g is 9.81m/s^2.

    BTW... I think, she can hope to achieve that acceleration, but she does not have enough power.

    regards
    Bartek
     
  8. Jun 26, 2010 #7
    Re: mechanics

    Thanks Bartek,

    For (1) ,isn't that the front tyre is in contact with the ground,so when it rubs against the ground ,fricton is produced?

    For(2), can i just use the motion formulas ,v=u+at

    I don think so because the acceleration is not constant in this case?

    (2)
     
  9. Jun 26, 2010 #8
    Re: mechanics

    When the bike start and accelerates friction force is propelled force. So, only force from back wheel works in this case - because only back wheel is driven by a biker. Front wheel theoretically resist against movement, but it is rolling friction - you can omit it (such as an air resistance etc.).

    When bicycle is braking both wheels are stationary and rub on the road. Both of them produce friction which slow bicycle. You should use the formula for distance in uniformly delayed movement. Or you can use conservation energy law.

    Why do you think that acceleration is not constans?
     
  10. Jun 26, 2010 #9
    Re: mechanics

    Can you be of any help, I need a simple (if that is possible) equation to help me determine the weight increase of a 12 stone occupant of a car doing 30mph coming to a dead stop?
     
  11. Jun 26, 2010 #10
    Re: mechanics

    According https://www.physicsforums.com/showthread.php?t=5374" you have to start new thread, and show that you have attempted to answer your question in order to receive help...

    Try this :-)

    regards
     
    Last edited by a moderator: Apr 25, 2017
  12. Jun 26, 2010 #11
    Re: mechanics

    thanks for your help!
     
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