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Homework Help: Truck Bed Friction Problem

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data

    The coefficients of friction between the flat bed of the truck and crate are (coefficient of static friction) = 0.8 and (coefficient of kinetic friction) = 0.7. The coefficient of kinetic friction between the truck tires and the road surface is 0.9. If the truck stops from an inital speed of 15 m/s with maximum braking (wheels skidding), determine where on the bed the crate finall comes to rest or the velocity relative to the truck with which the crate strikes the wall at the forward edge of the bed.

    The distance from the crate to the wall at the forward edge of the bed is 3.2m

    Truck is going to the right.

    2. Relevant equations

    Friction Force = Coefficient of Friction * Normal Force.

    3. The attempt at a solution

    Atruck=(.9)(9.81)=8.829 m/s^2 going to the left

    T=1.69895 , time it takes for the truck to stop

    (coeff. of kinetic friction of the crate)*Normal Force=M*Acrate

    acceleration of the crate in relation to the truck = 6.827-8.829 = 1.962 m/s^s going to the right

    V^2 = Vinitial + 2(A)(X-Xinitial)
    V^2 = 0 + 2(1.962)(3.2)
    V = 3.54356 m/s

    What did I do wrong? I couldn't figure out how to use the coeff. of static friction of the crate and truck bed.
  2. jcsd
  3. Jan 29, 2012 #2
    Woops! Wrong section I guess haha. This stuff is advanced for me -_-
  4. Jan 30, 2012 #3


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    Homework Helper

    It seems to me that "Acrate=6.867" is the acceleration relative to the truck. Just what you want. Don't subtract the truck's acceleration.
  5. Jan 30, 2012 #4
    I feel like that's the acceleration of the crate relative to the ground. An answer I found on another site was around 2.46 m/s or something and that would make my number higher rather than lower.

    Not 100% sure though, does anyone else have any ideas?
  6. Jan 31, 2012 #5


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    Sorry, I am mistaken. I see that if there is zero friction then Acrate = 0 and that is certainly not relative to the truck. So your a = 1.962 is correct.
    But your final velocity calc doesn't make sense.
    In the 1.7 s it takes the truck to stop, the crate will move d = ½at² = ½(1.962)*1.7² = 2.835 m. When the truck is stopped, the crate is still moving and has another 0.35 m to go before it hits. Slowing down due to friction now. You have another calc to do.
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