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Mass Spring System and Torque

  1. Oct 8, 2016 #1

    1.

    001.png


    2. Relevant equations

    // Tension of the spring
    // Where sinL between `bottom right` of the spring and the bar

    Ft = sinL * kx

    // Where x0 is the initial length of the spring

    Ft = sinL * k(x - x0)

    // Force of gravity
    Fg = -mg

    // Find x0
    // sqr = square root

    x0 = sqr(100^2 + 100^2)
    // Converted to meters
    x0 = 1.421

    Ft + Fg = 0
    sinL * k(x - x0) - mg = 0
    sinL * k(x - x0) = mg
    1/1.421 * 1000(x - 1.421) = 10 * 9.8
    (x - 1.421) = 98 / 1/1.421 * 1000
    x - 1.421 = 98 / 703.73
    x = 1.5602579540448752

    // Converted to centimeters and rounded
    x = 156cm

    But for some reason this doesn't seem to be the right answer. ;(

    If someone could help me would be really appreciated.

    Thanks.
     
  2. jcsd
  3. Oct 8, 2016 #2

    billy_joule

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    Does a spring get longer or shorter when you pull on it's ends?
     
  4. Oct 8, 2016 #3
    I think longer, isn't it?
     
  5. Oct 8, 2016 #4
    I meant x to be bigger than x0 after the rigid bar has been starching the spring.
     
  6. Oct 8, 2016 #5

    billy_joule

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    Yes.
    The diagram shows the spring already stretched, with length ## \sqrt2~ m ## as you've found.
    If the bar were removed, the spring would contract to its relaxed length. which must be ## < \sqrt2~ m ##
     
  7. Oct 8, 2016 #6
    I see, so I guess I was trying to solve the wrong problem from the beginning, so the question is: find the resting length if the bar is removed?

    I this case I have to subtract the length caused by the bar on the spring, so that we'd find the original length of the spring without the bard acting on it?

    BTW: Sorry for my silly questions.
     
    Last edited: Oct 8, 2016
  8. Oct 8, 2016 #7
    Updated my comment.
     
  9. Oct 8, 2016 #8

    billy_joule

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    Yes,that's one way of saying it.

    This isn't right.
    There are three forces acting on the bar - it's weight force, the springs force, and a reaction force where it's connected to the wall.

    The simplest way to find T is take moments (torques) about the bar-wall joint, that way we don't need to consider the wall reaction.
     
  10. Oct 9, 2016 #9
    Does the following sound reasonable:


    ##m = 10kg##
    ##g = 9.8m/s^2##
    ##k = 1000N/m##

    ##x0 = \sqrt{2}##
    ##theta = 45°##

    ##T = k * x##

    ##Ft = L * T * sin(theta)##
    ##Fg = L/2 * m * g##

    ##L * T * sin(theta) = L/2 * mg##
    ##2L * T * sin(theta) = mg##
    ##T = 1/2 * mg / sin(theta)##

    ##k * x = 1/2 * mg / sin(theta)##
    ##x = (1/2 * mg / sin(theta)) / k##
    ##x = (1/2 * 98 / \sqrt{2}/2) / 1000##

    ##x = 98 / \sqrt{2} * 1000##
    ##x ~= 0.06929m##

    // Now finally subtract from the original length
    ##diff = x0 - x##
    ##diff = \sqrt{2} - 0.06929##
    ##diff = 1.3449170978168133914017m##
    ##diff = 1.3449170978168133914017 * 100##

    // Rounded
    ##diff = 134cm##
     
  11. Oct 10, 2016 #10
    I think it seems to be ok.
     
    Last edited: Oct 10, 2016
  12. Oct 10, 2016 #11

    billy_joule

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    Science Advisor

    Yes, looks good.
    Except here you call moments forces;
     
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