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Homework Help: Finding velocity, hinged rhombus

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I don't have the slightest idea on how should i go on making equations, any help would be appreciated.
  2. jcsd
  3. Jul 3, 2012 #2
    Try using the fact that the horizontal distance between the points OA1, A1A2, A2A3 remain in the same ratio, and hence their velocities.
  4. Jul 3, 2012 #3
    How would i go on finding OA1, A1A2 and A2A3?
  5. Jul 3, 2012 #4
    You cant find their exact values, but you sure can find their ratios. Let the first rhombus have length 5l, second 2l, and third 2l. Now, when the outermost hinge is pulled, all their angles will change by the same amount. So you have the distance ratio as....?

    Hint : A tiny bit of trigonometry will help :smile:
  6. Jul 3, 2012 #5

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    Hey Pranav! :smile:

    You're good in drawings aren't you?

    What if you make a drawing with the construction fully folded, meaning A1=A2=A3.
    And another one with the construction fully extended.

    Suppose we measure starting from the wall.

    What are the distances to the wall of each point in both cases?
  7. Jul 3, 2012 #6
    Seems like i am doing something wrong, i am getting the distance ratio as 5:3:2. :uhh:

    Hey ILS! :smile:
    For the second case, the distances will be 10, 6 and 4. I am not able to understand the first case. :confused:
  8. Jul 3, 2012 #7

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    Those are the distances between the points.
    But you need the distances to the wall...?
  9. Jul 3, 2012 #8
    Oops, sorry, they are 10, 16 and 20.
  10. Jul 3, 2012 #9
    Uh yes, I meant the distance ratios from the origin will remain constant. :devil:

    So it should be, 5:8:10.

    Edit : And that agrees with your post above :smile: Now you should be able to find the velocity.
    Last edited: Jul 3, 2012
  11. Jul 3, 2012 #10
    So now what about the velocity? I think i have reached the answer, its d) but i still don't get it why the distance ratio remains constant, how did you come to know that we need to solve it using distance ratio?
  12. Jul 3, 2012 #11

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    Much better! :)

    So suppose we fold the construction completely.
    The distances to the wall will then be 0, 0, and 0.

    Let's call this t=0.
    Now we extend it such that A3 has constant velocity V.
    The means that at some time t the distance of A3 is Vt=20.

    What was the distance to the wall of A2 at that time?
  13. Jul 3, 2012 #12

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    So let's pick the situation where the points are exactly halfway.
    This means the rods make an angle of 45 degrees with the horizontal.

    What are the respective distances to the wall in this case?
  14. Jul 3, 2012 #13
    [tex]x_1 : x_2 : x_3[/tex]

    is constant, so,

    [tex]\frac{dx_1}{dt} : \frac{dx_2}{dt} : \frac{dx_3}{dt}[/tex]


    [tex]v = dx/dt[/tex]

    Last edited: Jul 3, 2012
  15. Jul 3, 2012 #14
    The distance of A2 is 16 from the wall at time t=20/V. Assuming the velocity of A2 as v2 and using the equation v2t=16, substituting the value of t, i get v2=0.8V.

    I get it now, thanks ILS! :smile:

    Any answer to why the distance ratio worked here? If there was a constant acceleration would the distance ratio work?

    EDIT: Infinitum answered both of the question with one equation. Thanks Infinitum. :tongue:
  16. Jul 3, 2012 #15

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  17. Jul 3, 2012 #16
    You both posted at the same time while i was replying to post #11. :rofl:
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