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Unable to solve type of question (dynamics): finding coefficient of friction

  1. Jul 22, 2012 #1
    So I've been busting my head for the past 2 days trying to figure out what I was doing wrong and I just can't. I have a whole bunch of these types of questions and can't solve any of them.
    My own equations and reasoning make sense to "me" but don't lead me to the right answer. I am positive I'm overlooking some huge thing. Please help!

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

    a 1250kg car traveling at 16.67m/s comes to a sudden stop in 35m. Find the coefficient of friction acting on the brakes.

    Vi: 16.67m/s
    Vf: 0m/s
    m: 1250kg
    a: (found), -3.969
    g: 9.81m/s2
    d: 35m

    2. Relevant equations

    Fnet(x)=ma
    Fnet(x)= Fa-[itex]\mu[/itex]Fn
    Fa-[itex]\mu[/itex]Fn = ma

    3. The attempt at a solution
    Going insane!:surprised See attached doc...

    Thanks again guys!
     

    Attached Files:

  2. jcsd
  3. Jul 22, 2012 #2

    TSny

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    What does Fa stand for?
     
  4. Jul 22, 2012 #3
    Hi :biggrin: Applied force.
     
  5. Jul 22, 2012 #4

    TSny

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    Hi Apollinaria. Besides the force of friction, is there another applied force that acts horizontally? If so, can you describe it?
     
  6. Jul 22, 2012 #5
    There's nothing else. The wording of the problem I posted is exactly as is in the worksheet I was given. Unless you mean the a(deceleration)... :uhh:
     
  7. Jul 22, 2012 #6

    TSny

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    If friction is the only force that acts horizontally, why did you include a vector labeled FA in your drawing that acts in the direction of motion?
     
  8. Jul 22, 2012 #7

    CWatters

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    The coefficient of friction is the "ratio of the force of friction between two bodies and the force pressing them together"

    http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction

    At first glance I don't think you have enough information to answer that but ... perhaps they meant the coefficient of friction acting on the tyres (or perhaps that's the same as the brakes but I can't think why)...

    Anyway you could calculate the stopping force from the energy balance..

    work = Fa x distance = 0.5 x mass x V2

    Fa = 0.5 x mass x V2/distance

    The vertical force on the tyres is just

    Fg= Mass x g

    So the ratio is

    = (0.5 x mass x V2)/(mass x g x distance)

    mass cancels

    = (0.5 x 16.672)/(9.8 x 35)

    = 139/343
    = 0.405

    Have edited this to get the ratio the right way up!

    If thats right Yippiee. If not I'm sorry but it's been a long day.
     
    Last edited: Jul 22, 2012
  9. Jul 22, 2012 #8
    Because I am struggling with this dynamics unit. So you're saying I can exclude the FA?
    How would I know if there is an FA? Only when it is stated?
     
  10. Jul 22, 2012 #9
    Nice to see you again Watters. And unfortunately that's not the answer but good try anyway lol :biggrin:
     
  11. Jul 22, 2012 #10

    TSny

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    It's important to know exactly what forces are acting for a specific problem. When you draw a force diagram (free body diagram) you should only include forces that actually act on the object. A good thing to do: whenever you draw a force on a diagram, ask yourself where that force is coming from. If you can't answer that for a given force, then you should question whether that force should be there.

    When the car is breaking to a halt, only the force of friction is acting in the horizontal direction (we are neglecting air resistance).
     
  12. Jul 22, 2012 #11

    CWatters

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    Oh heck I give up.
     
  13. Jul 22, 2012 #12
    Thank you so so so much for explaining this.

    I thought there was always an applied force when there was a force of friction; as in, they come in pairs. Kind of like Fg and Fn do when an object is resting on something (but not when it is in the air).

    Thanks again for bringing this to my attention. I will now think many times before drawing :smile:
     
  14. Jul 22, 2012 #13
    You and I both :wink: However, problem solved so it's all good now.
     
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