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Balancing a plank on springs (Torques)

  1. Apr 9, 2014 #1
    Hello, I would like some help with the following question. I highly appreciate your help ! I got an answer that doesn't make sense so I know there must be a concept that I don't get.

    There is a complicated piece of machinery sitting on a plank of metal which
    is 2 m long. The metal and machinery have a combined mass of 430 kg. The center
    of mass is approximately 1.2 m from the left side of the plank. We need to attach 2
    springs to the plank to lift it off the ground, but it needs to be level. One spring has
    a spring constant of 2200 N/m and is attached to the right edge of the plank. If the
    other spring has a spring constant of 1500 N/m, where should we put it to make sure
    everything is level?


    So what I did is that I wanted to balance torques and forces.

    Forces:

    mg= Fs1 + Fs2
    mg= 2200Δx + 1500 Δx ( Since it is level, then I assumed that delta x is the same)
    mg = 3700 Δx
    430*9.8= 3700 Δx
    Δx = 1.4

    Then, I started balancing torques.
    I have set the left end to be my pivot.
    mg(1.2)= Fs2y ( Where y is the distance of the second spring from the left end)
    mg(1.2) = KΔx y
    430*9.8*1.2 = 1500*1.4 y

    y = 2.4 , which is obviously impossible since the length of the plank is just 2 metres.
     
  2. jcsd
  3. Apr 9, 2014 #2

    haruspex

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    What about the spring at the right-hand end?
     
  4. Apr 9, 2014 #3

    BvU

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    Hello small fan,

    Could you please use the template ? There's good reasons for it being required in PF. Read the guidelines if you don't believe me.

    I don't understand how you can find Δx = 1.4 from 430*9.8= 3700 Δx ?

    Your torque balancing makes a strange impression on me.
    Why is Fs1 missing ?
    Did you make a drawing ?

    Ha Haru! Please take over, bedtime for me!
     
  5. Apr 9, 2014 #4
    Can't I just use one pivot ?

    Hey ! I was wondering why do I need to consider both ends? The object has one pivot.
     
  6. Apr 9, 2014 #5
    Hey !

    I thought I am following it. I am sorry if I didn't. Ok .. I multiply 430 by 9.8 and divide the product by 3700.
    Δx is the compression of the springs.
     
  7. Apr 9, 2014 #6
    Also, Fs1 is missing because it passes through the pivot. So the torque it's producing is equal to zero.

    If you want to show me a better approach please do. Thanks :)
     
  8. Apr 9, 2014 #7

    haruspex

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    Not according to what you posted. You have one spring at the right hand end, one spring at some unknown distance from the left hand end, and a pivot at the left hand end.
     
  9. Apr 9, 2014 #8
    My bad ... got it. Thanks :)
     
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