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Two boxes on two slopes

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img15.imageshack.us/img15/407/draw.th.png [Broken]

    Initially m and m' are at rest and at the same height. The planes are perpendicular. Find the minimal distance that separates m and m'.

    2. Relevant equations

    3. The attempt at a solution
    Well the distance that separates them is the length of the vector that describes m 's position relative to m'. But then I fail to find any idea on how to get to that vector.

    Thanks for your help.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 21, 2009 #2
    square root of (a cos (alpha)2 + a sin ( alpha) 2)
  4. Feb 21, 2009 #3


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    Make you origin at the intersection of the two planes; choose the y axis along the line from O to m' and the x axis along the line from O to m. Then m is located at
    [tex] - a cos(\alpha) \mathbf{i} [/tex]

    and m' is located at

    [tex] a sin(\alpha) \mathbf{j} [/tex]

    Then you can figure out the vector from m to m'.

    I'm not sure how the previous post helps you since the a factors out and

    [tex] sin^2(\alpha) + cos^2(\alpha) = 1 [/tex]

    leaving you with something you already knew.
    Last edited by a moderator: May 4, 2017
  5. Feb 21, 2009 #4
    Thanks for the replies.
    I'm sorry, but I think I didn't explain myself correctly.
    The thing is the boxes move down the slopes, so at some point either
    - m hits O
    - m' hits O
    (whichever happens first)
    So at that moment mm' has a certain length, say d. What I'm looking for is that length which probably is a function of α and a.
  6. Feb 21, 2009 #5


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    First of all find the component g along the inclined planes, and the lengths of the inclined planes. From the figure it is clear that m hits O first. From the component of g and the
    length of the inclined plane, find time taken by m to reach O. Using this time find how much m' moves downward. Knowing the length of the inclined plane containing m', you can find the distance mm'.
  7. Feb 22, 2009 #6


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    This is not the right way to solve this problem. First of all, one does not know which block hits O first since the angle \alpha is not specified.

    To springo:

    1. It does not matter which block hits O first.

    2. Can you write and equation that describes the position of m (and hence, also m') as a function of time, t?

    3. If you have expressions for the positions of m and m', then you should also be able to deduce an expression for the distance between m and m' as a function of t.

    4. What is the last thing you need to do?
  8. Feb 22, 2009 #7
    Well, I guess we could find the expressions of OM and OM' as a function of time using the acceleration or the speed and integrating. Let (O, i, j) be (O, uOM ,uOM')
    OM = (a·cos(α) - (1/2)·g·sin(α)·t2) j
    OM' = (a·sin(α) - (1/2)·g·cos(α)·t2) i
    Is that what you suggested?

  9. Feb 22, 2009 #8


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    Good. That's exactly correct so far.

    Now, isn't there a very simple geometric rule you can use to find the distance (MM') between the two blocks?

    PS: Alternatively, if you wish to stick with vectors, you can arrive at this same result by finding the vector MM' out of OM and OM' and then computing its magnitude.
  10. Feb 22, 2009 #9
    Yeah, sure:
    d2 = (a·sin(α) - (1/2)·g·cos(α)·t2)2 + (a·cos(α) - (1/2)·g·sin(α)·t2)2
    d2 = a2 - a·g·sin(α)·cos(α)·t2 + (1/4)·g2·t4
    I had written that down before, but since it wasn't leading me anywhere I thought something was wrong.
  11. Feb 22, 2009 #10


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    I have an extra factor of 2 on the middle term - can you check that?
  12. Feb 22, 2009 #11


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    Next up, you want to find the minimal value of d^2. The way you tackle that depends on whether or not you've had differential calculus. Have you?

    If you have, what do you think you should do?
  13. Feb 22, 2009 #12
    d2 = a2 - a·g·sin(2α)·t2 + (1/4)·g2·t4
    OK, so since d > 0, the minimum value for d is also the minimum value for d2. So we could take the derivative of that.
    d(d2)/dt = 2·a·g·sin(2α)·t + g2·t3
    But then I get three roots, the interesting one leaves us with:
    gt2 = 4·a·sin(α)·cos(α)
    We replace in the first equation and simplify:
    d2 = a2·cos2(2α)

    d = a·cos(2α)

    Is that it? Thanks!
  14. Feb 22, 2009 #13


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    Looks good.

    But just for good form you may need to do one final thing to demonstrate that this is indeed a minimum and not a maximum (well, it's obviously not a maximum since acos(2alpha) < a, which was the initial separation, but it's good to develop the practice of identifying the type of optimal value). Do you know how you would do this?
  15. Feb 23, 2009 #14
    The next derivative is f'' (so 2nd, which is even) and it's negative, so it's a maximum. Is that correct?
    Thanks again!
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