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Armadillo Problem (Newton's Second Law)

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    For sport, a 12 kg armadillo runs onto a large pond of level, frictionless ice with an initial velocity of 5.0 m/s along the positive direction of an x-axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of 17N in the positive direction of the y-axis. In unit-vector notation, what are the animal's (a) velocity and (b) position vector when it has slid for 3.0 seconds?

    2. Relevant equations

    Fnet=ma

    Kinematic Equations

    Other equations depend on the free body diagram(s).


    3. The attempt at a solution

    I feel like I'm on the verge of getting this one; I just need a hint. ><

    If the ice is frictionless, that means the armadillo is still sliding at a constant velocty of 5.0 m/s after he slips, right?

    I think it's finding the upward velocity on part (a) that I'm kind of having trouble with.

    What I did first was draw a free body diagram for the armadillo at the second position where two forces (I think) are working on the animal: The force of the wind (17N) upward, and mg downward. Was that correct, or am I missing one? I used a = (w - mg)/m (from w-mg = ma), where w is equal to the force of the wind, to find the acceleration, which came out as -8.38 m/s^2, for me. I really don't know if that's right or not. Then I tried to use the kinematic equation v = v0 + at to find the upward velocity. I rearranged the equation to v - at = v0; I remember from free-fall motion that when an object reaches its max upward height, v=0; also, I used the acceleration -8.38 m/s^2. My answer came out to be 4.26 m/s for the upward velocity, which is close to the right answer (to correct sig figs, the answer should be 4.2 m/s in the positive y direction). I still feel like I missed something. Can somebody show me where I might have gone wrong?

    [Edit] Nevermind; I don't think this works after all. D:

    Also, for part (b), can I just use kinematics to find x and y after 3.0 seconds? (I don't want to try it yet until I know I have part (a) right)
     
    Last edited: Oct 17, 2007
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  3. Oct 17, 2007 #2

    Hootenanny

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    I don't think the question is saying that the wind is blowing upwards, rather it is blowing parallel to the pond and perpendicular to the x-axis. In otherwords, think about drawing an x and y axis on the surface of the ice.
     
  4. Oct 17, 2007 #3
    "Pushed by a force in the positive direction of the y-axis" doesn't sound like the wind is blowing parallel to the pond. =/
    Even if it is initially blowing parallel, that doesn't matter; all that matters if the force that's applied to the armadillo.

    Perhaps you could elaborate?

    Better yet, what should my free body diagram for position 2 look like?
     
  5. Oct 17, 2007 #4

    Hootenanny

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    It depends how you orientate your axis; the standard way would be thus;

    [​IMG]
     
  6. Oct 17, 2007 #5
    Well, this is only advanced high school physics, so I'm pretty sure we're sticking with our regular coordinate system where y is vertical and x is horizantal. I know for a fact my teacher wouldn't assign a problem where our axes were different without letting us know.

    This problem doesn't even call for 3-D space; just vector components in the x and y directions (our coordinate system being like I said).
     
  7. Oct 17, 2007 #6

    Hootenanny

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    Slipping over the ice implies that the armadillo is still in contact with the ice and besides, is the wind blowing up through the ice?
     
  8. Oct 17, 2007 #7
    lol. The direction of the wind doesn't matter. It exerts an upward force on the armadillo. Honestly, wind usually spirals all over the place; it doesn't have a definite direction.

    Basically, the armadillo is running with an initial velocity. As soon as it hits the ice, it trips (it goes a little up and a little forward; in other words, its velocity has an x component and a y component). Part (a) is asking for the velocity vector after 3 seconds.

    The correct answer is given to me: (5.0 m/s)i + (4.2 m/s)j

    It's not interpreting the problem that I can't figure out; it's really just setting up my free body diagram and finding the upward velocity.
     
  9. Oct 17, 2007 #8

    Hootenanny

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    I disagree with your interpretation, how about we go with my interpretation and see if we get the right answer?
     
  10. Oct 17, 2007 #9

    All right; I'll agree to that. (:

    Although... getting the right answer doesn't exactly help me. It's knowing how to work the problem. You seem do physics differently from the curriculum I'm being taught.
     
    Last edited: Oct 17, 2007
  11. Oct 17, 2007 #10

    Hootenanny

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    So, if the y axis is horizontal we can ignore gravity and the only force acting is the 17N wind force. So all you need to do is work out the acceleration and plug it into the appropriate kinematic equation to obtain the velocity in the y direction after three seconds.
     
  12. Oct 17, 2007 #11
    I get -4.25 m/s

    Close to what I was getting before. Still, I should get 4.2 m/s (after rounding to correct number of significant digits).

    Meh. I'll just find out the solution in school and then post it back on here...unless someone else still once to try. We really don't do Physics that way in my class. Maybe you living in the UK has something to do with it.

    I appreciate your attempt to help me, though. (:
     
  13. Oct 17, 2007 #12

    Hootenanny

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    Wait, how did you get a negative answer? The force is applied in the positive y direction.
     
  14. Oct 17, 2007 #13
    It doensn't matter. I already know the answers to the problem. The way you're working it just doesn't make sense to me: y has always been vertical. Like I said, I'll post the solution I get from school, and then perhaps you can explain to me this y-being-horizantal buisness and how it gives you the right answer. Does that sound like a decent plan?

    What I would really love someone to do is to look through my attempt and show me my errors.
     
  15. Oct 17, 2007 #14

    Hootenanny

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    He's one:

    You have a negative vertical acceleration, that means that your armadillo would be accelerating vertically downwards, i.e. through the ice!

    Also, you have a negative acceleration and yet a positive final velocity, how does that one work?
     
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