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How to find Fa .to find other unknowns ?

  1. Apr 1, 2008 #1
    [SOLVED] how to find Fa.....to find other unknowns...?

    1. The problem statement, all variables and given/known data

    A mountain bike with mass 13.5 kg , with a rider having mass 63.5 kg, travelling at 32 km/h when the rider applies the brakes, locking the wheels. How far does the bike travel before coming to a stop if the coefficient of friction between the rubber tires and the asphalt road is 0.6?


    2. Relevant equations



    3. The attempt at a solution


    Given:

    m1 = 13.5 kg
    m2 = 63.5 kg

    Velocity Initial = Vi = 32 km/h
    Velocity Final = Vf = 0 km/h

    friction coefficient = uK = 0.6






    Force Natural = mg = (13.5 + 63.5) x 9.81 m/s^2 = 755.37


    Force Friction = 755.37 x 0.8 = 453.222


    Fn must equal force of gravity because there is no vertical motion.


    Applied force must be less than force of friction, because bike is slowing down.


    I need to find applied force, so I can use F=ma to find acceleration. Once I get acceleration, I can find displacement. Is my reasoning correct?


    thnkx
     
  2. jcsd
  3. Apr 1, 2008 #2
    instead of giving an answer, can someone tell me what concept I'm missing?


    I have a bunch of questions similar to this, and feel like there's something fundamental I'm missing.
     
  4. Apr 1, 2008 #3

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    It is generally called the "normal" force; hence the symbol Fn.

    In the horizontal direction, applied force is the force of friction F, which is slowing the bike.

    So, you need to know the relation between Fn and F, which is F=(uk)*Fn.

    EDIT: I saw your 2nd post after I submitted mine. Tell me what you feel you are missing.
     
  5. Apr 1, 2008 #4


    Fg = Fn = mg = (13.5+63.5) = 77N

    Ff = 61.1 N


    but how does Fa equal Ff if the bike is slowing down? Doesn't Ff have to be greater to slow it down?


    thnkx for your time shooting_star

    regards vietjon
     
  6. Apr 1, 2008 #5
    I'm very good at math, but for some reason, I can't conceptualize physics problems to solve them mathematically. Maybe that is it?

    Or maybe it's Newton's laws I'm having trouble with. I just started my first physics class yesterday, and read the three laws 10+ times now, and attempted each of my questions 10+ times also.
     
    Last edited: Apr 1, 2008
  7. Apr 1, 2008 #6

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    The bike has obviously zero total force acting on it in the vertical direction, since it has no acceleration in that direction. The only force or forces acting on the bike must be in the horizontal direction, opposite to the direction of motion, slowing it down.

    Tell me what do you visualize by the phrase "applied force" here? What is the agency applying that force? Remember, the rider is not propelling the bike any more, and the brakes are locked.
     
  8. Apr 1, 2008 #7

    at first I had a FBD, with a Fa arrow pointing in the direction of movement. I visualized it as, there's a force moving the bike. ie the rider.


    but I think that's wrong.

    It's momentum, right? the bike would keep moving....but force of friction is slowing it down to a stop, so there is no applied force any more.


    I know what I'm doing wrong now. I'll post the answer......give me a few minutes.....
     
    Last edited: Apr 1, 2008
  9. Apr 1, 2008 #8
    You've almost got it. All you have to do now is relate the force of friction to the acceleration of the car. Any equations you know of that relate acceleration to force??
     
  10. Apr 1, 2008 #9
    Finding Acceleration

    F= ma
    a = Fnet/m = -61.6N/ 77 kg = -0.8m/s^2



    Finding Time


    a= (Vf - Vi) / time

    t = (vf - Vi)/ a = (0 - 8.88m/s) / -0.8 m/s^2 = 11.1 seconds



    Finding Displacement


    Vf^2 = Vi^2 + 2ad

    therefore d = (Vf^2 - Vi^2) / 2a

    sub = (0 - 8.88 m/s ) / 2*(-0.8m/s)^2 = 6.9 meters




    I got 6.9 meters, but my books says 6.7 meters. ?
     
  11. Apr 1, 2008 #10
    Because you did it wrong and the answer you got was remarkably close by coincidence. There is no applied force. You have a frictional force which causes a deceleration, which you can calculate. You can then use that to find the distance travelled since you have the initial velocity, final velocity, and deceleration.
     
  12. Apr 1, 2008 #11

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    How did you get this value of Ff?

    Go back to post #3 and read it once more carefully.

    Find Ff from Fn. Then apply F=ma. It's as simple as that. (The value of 'a' may be -ve or +ve.)
     
  13. Apr 1, 2008 #12
    opps....careless error. I think I get the concept now though.


    I just re-worked it and got 6.7 meters



    Force of Friction


    Fg = mg = Fn = 77*9.81 = 755.37 N


    Ff = ukFn = 0.6 * 755.37 = 453.222 N


    Acceleration


    a = Fnet/m = -453.222/77 = -5.886 m/s^2


    Time


    t = v/a = -8.9/-5.886 = 1.5136 seconds


    displacement


    d= (Vi*t) + (.5*a*t^2)

    = 13.47104 + (-6.7423)


    = 6.7 meters
     
  14. Apr 1, 2008 #13
    You didn't have to find time. You could've just used:

    [tex]V_f^2=V_i^2 + 2ad[/tex]
     
  15. Apr 1, 2008 #14
    ^thnkx.....ill remember that.




    thankx everyone for the help....much appreciated!!!!
     
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