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Friction coeffecient

  1. Nov 4, 2004 #1
    I need to find the coeffecient of kinetic friction for the following problem... all the stuff after the dashes is what I've done... The answer is approximately half the answer I keep getting, so I think I'm missing something...


    The driver of a 1200kg truck travelling 45km/h [E] on a slippery road applies the brakes, skidding to a stop in 35m. Determine the coeffecient between the road and the car tires.

    ------------------------

    45km/h = 12.5m/s

    find the time it took to stop to find acceleration:
    v=d/t 12.5 = 35/t t=2.56s

    find acceleration:
    a=v/t a=12.5/2.56
    a=4.88 m/s^2

    find force:
    a=f/m 4.88= f / 1200 f=5856N[E]

    find normal force:
    mg = 1200 (9.81) = 11772N

    find friction coeff.
    Force of friction = coeff * Normal Force
    5856N = x * 11772
    x = .49

    The answer in the book is .23... I don't understand how to find friction force. I substituted the force of the truck for the friction because I remember seeing that somewhere with a similar question. I'd appreciate any help.
     
  2. jcsd
  3. Nov 4, 2004 #2

    Doc Al

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    Staff: Mentor

    speed not constant

    This would only be true if the speed were constant, but it's not. Instead, you could use the average speed, then you'd be OK. Since the truck's speed ranges from 12.5 m/s to 0, what's the average speed?
     
  4. Nov 4, 2004 #3
    It worked! Thank you! ... I just have one last question. When I solved for the Force (a = f/m) , what was I solving for in relation to the truck? And why do I put that force as the force of friction when I'm trying to solve for the coeffecient? If the force of friction was equal to the force of the truck, then the truck shouldn't have been moving right?

    Thanks again!
     
  5. Nov 4, 2004 #4

    Doc Al

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    Staff: Mentor

    I think you are confusing yourself a bit. When you apply Newton's 2nd law (F = ma) you are solving for the net force on the truck. Which, in this case, happens to be the frictional force since that's the only horizontal force acting on the truck. The term "force of the truck" has no meaning.

    In summary: The truck was moving along at some speed (net force = zero). The brakes were applied, thus applying a frictional force (the ground pushing on the truck) that accelerated (de-accelerated) the truck, slowing it to speed zero. You calculated the acceleration, then the force of friction [itex]F_f[/itex]. To calculate the coefficient of friction, you applied [itex]F_f = \mu N = \mu mg[/itex].

    (Tip: If you can resist the temptation to plug in numbers too soon, you can often save yourself a bit of calculating. In this case [itex]F_f = \mu mg[/itex] ==> [itex]ma = \mu mg[/itex] ==> [itex]\mu = a/g[/itex].)
     
  6. Nov 4, 2004 #5
    I get it now =)
    Thanks a lot!
     
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