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How is force distributed orthogonally along a material plane?

  1. Aug 30, 2009 #1


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    I want to simulate a plane thin rectangular board attached to one spring below at each end. (So there are 2 springs in total.) The springs span the entire edges. The spring-board-attachment is a rotatable axis.

    I can't find out how the force acts on the springs. I don't know whether there will be any force on the spring below at the farer end. If I press directly above one end onto a spring only this spring gets compressed while the other does not because the board only rotates around its axis? If I press onto the center of the board, both springs receive equal force. What is with any other point of pressure though? What's the downwards force onto the springs??


  2. jcsd
  3. Aug 30, 2009 #2


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    Welcome to PF!

    Hello pL1! Welcome to PF! :smile:
    To find the force on each spring, take moments about the other spring. :wink:
  4. Aug 30, 2009 #3


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    This sounds good!

    Thank you!

    And thank you for the welcome!

  5. Sep 2, 2009 #4


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    I was actually unable to write down the motion equations for the angular momentum. However I have derusted myself a bit and wrote down the Euler-Lagrange DE system.
    Now I don't seem able to solve it. It may even be unsolvable as the setup incorporates a gravitational pendulum of sorts for which the DE can't be solved analytically.

    This is why I am thinking of solving it numerically. Now I obviously have to do 2 (infinitesimal) rotations around 2 different rotation points in 2D space in one iteration. But the question is: what's the result of 2 such rotations: vector addition of each start-/end-rotation connection vector? I can't make up my mind.

    http://www.alice-dsl.net/l.hansen/spring_board_Euler_Lagrange_DE.PNG [Broken]

    Thus thanks for any help!

    Last edited by a moderator: May 4, 2017
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