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Breaking static friction w/ a circle

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A crate (35 kg) is located in the middle of the flat bed of a pickup truck as the truck rounds an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 80.0 m. If the coefficient of static friction between crate and truck is 0.410, how fast can the truck be moving in m/s and miles per hour without the crate sliding? (1 mile = 1609 m)


    2. Relevant equations



    3. The attempt at a solution

    I'm just doing some practice problems before my midterm next week....I have NO idea how to approach this one

    I have mass (35kg)
    coefficient of F_S is .410

    we havent gone over something like this yet in class...but I want to practice any problems I can...

    where do I even begin to start with this one?
     
  2. jcsd
  3. Mar 20, 2013 #2

    ehild

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    Think: how does the crate move with respect to the ground? What force keeps the crate on the flat bed of the trunk?

    ehild
     
  4. Mar 20, 2013 #3
    well the crate will move depending on the movement of the truck...the truck turns too far left...the crate will move to the right

    gravity is keeping the crate on the bed of the truck....and friction is (until it's overtaken) stopping the crate from moving
     
  5. Mar 20, 2013 #4

    haruspex

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    If the crate does not slide, what will be its acceleration (magnitude and direction)?
     
  6. Mar 20, 2013 #5
    if the crate does not move...there is no acceleration therefor no change in direction/magnitude...so 0
     
  7. Mar 20, 2013 #6

    haruspex

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    It doesn't move relative to the truck, but the truck is rounding a bend. The truck is moving at constant speed but its velocity (and therefore the crate's velocity) is changing. Any change in velocity is an acceleration.
     
  8. Mar 20, 2013 #7
    okay...understood...since velocity does depend on direction...and if direction is changing so is velocity

    so then if it involves a circular motion...has a radius...and a velocity

    would I be using

    F_c (centripetal force) = m(v ² / r)?
     
  9. Mar 20, 2013 #8

    ehild

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    Yes, the crate needs a force equal to the centripetal force. What force is exerted on it by the bed of the truck?

    ehild
     
  10. Mar 20, 2013 #9
    perpendicular to the surface would be the normal force....

    and just for the future when I need it

    Ffriction = μ_s * N

    and we know μ_s is .410


    but to answer your question...normal force
     
  11. Mar 20, 2013 #10

    ehild

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    Yes, there is the normal force from the bed which cancels with gravity. But friction is also a force from the bed, in horizontal direction. It prevents slipping. Without friction, the crate would slip radially outward, so the force of static friction points radially inward. It supplies the centripetal force.

    As you have written, the maximum static friction μsN, and N=mg as the crate does not accelerate vertically. And the force of static friction is equal to the centripetal force. So what is the possible maximum speed when the crate does not slip?

    ehild
     
  12. Mar 20, 2013 #11
    hmmm so what you're saying is that the force of static friction would be equal to the centripetal force

    so

    F_s = F_c?

    μ_s N = m(V²/r)

    μ_s * mg = m(V² / r)

    m's cancel so

    μ_s * g = V² / r

    so

    .410 * 9.8 = V² / 80

    so V² would equal (.410 * 9.8) * 80?

    and V would equal √((.410 * 9.8)*80) ?

    is that correct?
     
  13. Mar 20, 2013 #12

    haruspex

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    Yes, the units being...?
     
  14. Mar 20, 2013 #13
    well lets see..

    the square root....of meters/second² times meters

    so meters ² / seconds ²

    square root would make it m/s

    so the answer I'm getting would be

    17.93m/s

    we want this in miles per hour so

    17.93 m/s (1 mile / 1609 meters)(3600 seconds / 1 hour)

    this comes out to

    40.11 m/h

    look right?
     
  15. Mar 20, 2013 #14

    haruspex

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    Yes, that looks right. I approve of the way you embedded the units conversion in the equation. I wish more would do it like that.
     
  16. Mar 20, 2013 #15
    that's how I've always done it...it keeps me better organized! ...but thank you for your help!!
     
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