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Static fraction force coefficient

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data
    We have a car with mass of 1200kg driving at 55km/h. It suddenly starts breaking. The breaks "work" on every of the 4 wheels with the constant torque of 12Nm. The radius of the wheels is 0,21m.
    Question 1: How long does the car travel before it stops if the wheels have perfect grip?
    Question 2 the more important one: How big has to be the minimal value of the coefficient of the static fraction force between the wheels and surface for the wheels to really have the perfect grip.
    g=9.81m/s^2

    2. Relevant equations
    As hard to believe as it might seem a haven't found a relevant equation. There is one where the coefficient is tanß but it is not proper. I also found one where c=v^2/(rg) but the result is innacurate.
    There is M=F*r' but not helping..and others that all contain ß..

    3. The attempt at a solution
    I tried with the previously written equation but I got c=0,08 where it should be 0,19.
    I also looked at some solved problem from un minessota but couldn't find anything useful.
    Result for Q1 is 613 metters.
    English is not my primary language so bare with me :)
     
  2. jcsd
  3. Nov 25, 2012 #2

    SteamKing

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    What happens to the kinetic energy of the car while it is braking (not breaking)?
     
  4. Nov 26, 2012 #3
    well because it depends on the velocity in in the equation we have Wk =(m*v^2)/2. So when the car brakes it lowers down and when it stops it is equal to zero.
    It is my rookie year on the university so we are not supposed to solve it with kinetic energy.
    Yes of course braking my bad. :)
     
  5. Nov 26, 2012 #4
    The work, W, done by a torque, [itex]\tau[/itex], is
    [itex]\;W=\int \tau \: d \theta[/itex] ?
     
  6. Nov 26, 2012 #5
    I also found this equation but i don't know how to use it because I am not looking for the value of Work or do not have any other value for theta that 90degrees which is the angle between the surface and the force in the equation
    F(sf)=c(sf)*Fn
    where Fsf is force of static friction; c(sf) coefficient of static friction and Fn the normal Force which in our case eaquals gravitational force??
     
  7. Nov 26, 2012 #6

    SteamKing

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    to calculate the work done by the braking torque, theta is the number of radians the wheels turn while the brakes are applied. It is not the angle between the tire and normal force.
     
  8. Nov 26, 2012 #7
    I see what you are aiming to but can not figure out how many times does it turn or for how many radians does it turn.
    Is there no way I can use the standard torque equation or a static fraction force equation?
     
  9. Nov 26, 2012 #8
    had an epiffany (let's say I spelled that correctly). You divide torque (48Nm) by radius (0.21m) and get braking force, that divide that by mass and get braking acceleration.
    Than easily length. s=v^2/(T/(m*a))
    and for the coefficient you divide braking force by gravity force and get 0,19.
    thanks for the help though :))
     
  10. Nov 26, 2012 #9

    haruspex

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    Spiphing.
     
  11. Nov 27, 2012 #10
    Do you use integration/differentiation in your Physics course?
     
  12. Nov 27, 2012 #11
    Do you use integration/differentiation in your Physics course?
     
  13. Nov 27, 2012 #12
    spiphing ?? ok I get it now :D

    Yes, we do. I guess it just wasn't needed.
     
  14. Nov 28, 2012 #13
    You can only get away so long by using 'bad' physics.
    I don't think this how its is suppose to be solved.
    You should put some effort into this calculus thingy - you are going to need it in the future.
     
  15. Nov 28, 2012 #14
    I agree with you. I will look at it again.
    thank you for your help:)
     
  16. Nov 29, 2012 #15
    In this case we have an definite integral, F, of the form

    [itex]F =\int ^{b}_{a} y(x) \; dx[/itex]

    where
    y(x) is [itex]\tau ( \theta )[/itex]

    but the torque is constant and not a function of the angle through which the wheel turns, so we can bring it out of the integration which gets us to

    [itex]W = \tau \int ^{ \theta }_{0} \; d \theta[/itex]

    so we are calculating the work done by the torque on the wheel as it turns through an angle of 0 radians up to θ radians
     
  17. Nov 29, 2012 #16
    ok than you get a result in Joules and Joule is newton*meter so you divide the work by the g force?? or what.
    I still don't completely understand this theta part. If theta eaquals 2*Pi for one wheel turn than you get the result W=24Pi Joules and if you divide it by gravity force you get a result smaller than 1.
    And could you please tell me why this is an example of bad physics. I would really like to now so I can understand it better.
    :)
     
  18. Nov 29, 2012 #17

    haruspex

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    I'm puzzled. What was wrong with elstudent's analysis?
     
  19. Nov 30, 2012 #18
    It is not a good idea to just look at the units to solve a problem. You should apply your physics knowledge, be it principles, given equations or ....
    Look at this simulation for the relationship between the distance travelled and the total amount of radians the wheel turns through. Is this what was troubling you?
    http://media.pearsoncmg.com/bc/aw_young_physics_11/pt1a/Media/RotMotionStatics/DiscBlockRace/Main.html
    Click on the #1 little block with the squiggly green line and then on run in the simulation.
    Here is another link for the work done by a torque:
    http://physics-help.info/physicsguide/mechanics/work_energy.shtml
    At the bottom of the page you will find "Work and energy for rotational motion"
    It says that the work done by a torque is
    [itex] W = \tau ( \theta _{f} - \theta _{i}) [/itex]
    Does this look familiar?
     
    Last edited: Nov 30, 2012
  20. Nov 30, 2012 #19

    haruspex

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    I agree, but it doesn't seem to me that elstudent did that:
    You divide torque (48Nm) by radius (0.21m) and get braking force,​

    Correct, and easily justified.
    divide that by mass and get braking acceleration.
    Correct and obviously justified.
    Than easily length. s=v^2/(T/(m*a))​

    Unassailable, though the formula seems written out a little wrongly.
    and for the coefficient you divide braking force by gravity force and get 0,19.​

    No problem there either.
     
  21. Dec 3, 2012 #20
    The fact that the the student does a calculus based physics course indicates that he/she should solve the problem with a stronger mathematical approach to the problem. I thought we should also help her/him to solve it that way, rather than revert back to knowledge gained only in the more basic chapters of physics. I though this approach should get him/her only 1/4 of the alloted marks in an exam.
     
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