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Newton's Third Law problem

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Two packages at UPS start sliding down a 20 degree ramp. Package A has a mass of 5.0kg and a coefficient of friction of .20. Package B has a mass of 10kg and a coefficient of friction of .15. How long does it take package A to reach the bottom? The package are positioned next to each other on the ramp, and the distance from package A to the bottom is 2.0m.


    2. Relevant equations
    Newton's Second Law equations.
    [tex]\sum[/tex] (FnetA)x=mAaX

    [tex]\sum[/tex] (FnetA)y=mAaY

    [tex]\sum[/tex] (FnetB)x=mBaX

    [tex]\sum[/tex] (FnetB)y=mBaY


    3. The attempt at a solution
    I tried solving this by putting the horizontal and vertical components in their respective places, but that didn't get me that far, or at least I don't think so. For both y-components of the net force n=mAgcos(20) and n=mBgsin(20), well at least I think so. really I'm not quite sure if I am going in the right direction because I don't know how I am going to solve for time.
     
  2. jcsd
  3. Oct 22, 2009 #2
    Have you guys done the conservation of energy theorem yet in class?

    If so, you can find this by realizing that U1 + K1 + W = U2 + K2
     
  4. Oct 22, 2009 #3
    no I don't think we covered that yet.
     
  5. Oct 22, 2009 #4

    cepheid

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    Solving for time is no problem because you know that package A was moving under uniform acceleration, and you know what distance it travelled. There is a simple kinematics formula that applies to motion under constant acceleration.

    Of course, you need to calculate what that acceleration *is*. I think that the slightly tricky part of this problem is (if I am interpreting the problem correctly) the fact that the two packages are in contact, meaning that the net force on package A in the "down the ramp" direction depends on three things: its weight, friciton, and the contact force from package B pushing on it.
     
  6. Oct 23, 2009 #5
    The problem says that the packages are "next to each other", but seems to left implicit that they are in contact. Also, it lefts implicit that package A is closer to the end of the ramp than package B.
    If I understood the problem correctly, it is necessary to see whether packages A and B will be always next to each other during the motion (if they weren't, then you could simply treat the problem as if package A was alone). Since B's mass is bigger, its weight is bigger, so it would move faster if it were alone in the ramp; so they will be always next to each other during the motion.
    I would solve this problem this way: first, I would draw a sketch of the situation, a free-body diagram representing all the forces acting on package A (four forces: weight, friction, contact force from package B pushing on it --as cepheid said--, and the normal force that the ramp exerts on the package -which has the same direction and opposite orientation to one component of the weight).
    Secondly, I would decompose weight, "cancel" what needs to be cancelled and see what is the net force. Then, find the acceleration and, since I have distance and acceleration, find the time.
     
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