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Conceptual Help! acceleration of eggs sliding on truck using kinetic friction

  1. Oct 22, 2011 #1
    NEED CONCEPTUAL HELP ONLY!

    A package of eggs (mass 60.0 kg) sits on the flatbed of a pickup truck with an open tailgate. The coefficient of static friction between the package and the truck's flatbed is 0.350, and the coefficient of kinetic friction is 0.250.

    a) The truck accelerates forward on level ground. I found the maximum acceleration the truck can have without the eggs sliding off the truck bed is given by (.350)(9.8 m/s/s) = 3.43 m/s/s by setting [itex]\mu[/itex]s*mg = ma. Easy.

    (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the package sliding along its bed. What is the acceleration of the eggs relative to the ground?
    I found this by setting [itex]\mu[/itex]k*mg = ma, which gives, canceling out m, (.250)(9.8 m/s/s) = 2.45 m/s/s, which is RIGHT.

    Now, what I am having trouble understanding is what is meant in part (b) by acceleration of eggs relative to the ground. How is this different from the acceleration of the eggs with respect to the truck? What would the acceleration of the eggs relative to the truck be? And why, then if acceleration of the eggs is given just by a = [itex]\mu[/itex]k*g, why it is not in the negative direction? I understand the truck is a non-inertial reference frame, so if the truck were accelerating, what would then be the acceleration of the eggs on the truck sliding on the truck relative to the ground?

    Help please!
     
  2. jcsd
  3. Oct 22, 2011 #2

    I like Serena

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    Welcome to PF, anniecvc! :smile:

    You appear to understand most of it anyways.

    To clarify, in (b) the truck accelerates with an acceleration barely exceeding 3.43 m/s/s.
    This is relative to the non-inertial frame of the ground.
    The eggs are accelerated with 2.45 m/s/s relative to the ground, which is less than the acceleration of the truck.

    All the calculations were with respect to the ground because Newton's laws only work in inertial frames.

    Anyway, now you can calculate the acceleration of the eggs with respect to the truck.
    It's the difference of the 2 accelerations.
    This difference is negative, which fits, since standing in the truck you'll see the eggs sliding backwards out on the street.
     
    Last edited: Oct 22, 2011
  4. Oct 22, 2011 #3

    PhanthomJay

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    It means that with respect to a stationary observer on the ground, otr the ground itself, the eggs are accelerating at 2.45 m/s, which means that after t seconds, the eggs will be Vot + (1/2)(2.45)t^2 away from that observer.
    the relative acceleration of the eggs wrt the truck is the difference between the eggs acceleration wrt the ground and the truck's acceleration wrt the ground (watch signage).
    because it is accelerating in the positive direction with respect to th ground. The truck is accelerationg and moving to the right, the eggs are sliding to the left wrt the truck, and thus , the friction force acting on the eggs is to the right, in the direction of the acceleration wrt the ground.
    The truck IS accelerating. And no matter how great its acceleratiion, the acceleration of the eggs relative to the ground will be the same.
     
    Last edited: Oct 22, 2011
  5. Oct 22, 2011 #4
    Wow, thank you so much serena and phantomjay, you're amazing! i can finally wrap my head around it and get some sleep.
     
  6. Oct 22, 2011 #5

    I like Serena

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    You're welcome! Hope to see you around. :smile:
     
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