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Number displayed on an electric scale when an object hits it

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data
    We have an object, whose mass is accumulated in its center. The mass is m.
    We let an object positioned at the height of h over the scale free fall.
    Let R be the number that will be displayed on the scale right when the item hits it.
    Find R(h).

    2. The attempt at a solution
    [tex]v_{terminal} = \sqrt{2gh} [/tex]
    [tex]p = m \sqrt{2gh}[/tex]
    [tex]R = \frac{F}{g}[/tex] where [tex]F = \frac{dp}{dt}[/tex]

    Since the force the object will exert on the surface of the scale is the change in momentum over the change in time, how am I supposed to determine the time, during which the momentum was "transferred"?
     
  2. jcsd
  3. Oct 20, 2016 #2

    Bystander

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    You're certain you've transcribed the problem statement completely and correctly?
     
  4. Oct 20, 2016 #3
    have you seen a fault?
     
  5. Oct 20, 2016 #4
    To be honest if you, I also feel as if some information were lacking here. Anyway, can you understand the point of this problem? Have I explained it precisely enough?
     
  6. Oct 20, 2016 #5

    PhanthomJay

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    The problem asks for the scale reading when the object first hits it. What is the scales spring deformation at that instant?
     
  7. Oct 20, 2016 #6
    There will be hardly any deformation, because the scale is electric.
     
  8. Oct 20, 2016 #7

    Bystander

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    Point out to the "poser" of the problem that there's something "missing," and hope for the best.
     
  9. Oct 20, 2016 #8
    What if we assumed that the scale will slightly deformate once the object hits it? Would it, then, be solvable?
     
  10. Oct 20, 2016 #9

    PhanthomJay

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    There won't be any deformation even if the scale is like my old fashioned non trustworthy bathroom spring scale. Either info is missing or if you take the problem as written, the scale reading per F = kx. Is_______?
     
  11. Oct 20, 2016 #10
    [tex]kx / g [/tex]
    ?
     
  12. Oct 20, 2016 #11

    haruspex

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    As others have noted, that does not strictly make sense. But from a practical point of view, what you would see is a large value displayed almost immediately. So I would interpret it as the maximum value displayed.
     
  13. Oct 20, 2016 #12

    haruspex

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    That is the mass shown when the spring compression is x, but you do not know what x is. See my preceding post.
     
  14. Oct 20, 2016 #13
    [tex] m\sqrt{2gh} / dt = kx [/tex]
    [tex]x(t) = \frac{m}{t} k\sqrt{2gh}[/tex]

    [tex]R = \lim_{t \to 0} \frac{x(t)}{g}[/tex]
    Is it the potential solution?
     
  15. Oct 20, 2016 #14

    PhanthomJay

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    I gather electric scales give max readings at the moment of impact? I think mine reads 888 for some reason. Anyway, imagine it is an old fashioned scale that reads 0 at no load , like a grocery scale for weighing bananas. F= kx and at the very instant of impact, x is _____ so F is?
     
  16. Oct 20, 2016 #15
    [tex]x = \sqrt{\frac{2mgh}{k}}[/tex]
    And hence
    [tex] F = \sqrt{2mghk} [/tex]
    I think this should work.
     
  17. Oct 20, 2016 #16

    haruspex

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    This cannot be exactly right since it does not tend to 2mg as h tends to zero. In the conservation of energy equation, you need to take into account the total distance which the mass descends (it's a bit more than h). And the final answer should be in terms of the indicated mass, not in terms of the force.
     
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