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Planking is Physics

  1. Aug 14, 2011 #1
    We were assigned to do a planking and then prove why it is in equilibrium by computing its torque, etc. We selected the pool ladder rail to do the planking. Our team member who did the planking weighs 74 kilos and 5'1" tall. The pool ladder rails are 1 m above the ground. Torque=Force x momentum arm.

    I am confused about how to apply the first and second condition of equilibrium.

    How do I get the forces acted by the pool ladder rails on the my teammate?

    Are there any other forces acting downward on my teammate aside from her weight?

    My attempts:

    ΣT = Fx1 + Fx2 + Fx3 ...

    ΣT = (1.5548780488 m x 162.8 lbs)+(F x 1 m)+(F x 1 m)

    Is this correct?

    Thank you so much!
     
    Last edited: Aug 14, 2011
  2. jcsd
  3. Aug 14, 2011 #2

    kuruman

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    I have never "planked" anyone or anything so I don't know what you are talking about. I gather that this is an equilibrium problem and the solution to it should be standard. A diagram should be very useful in helping us to help you.
     
  4. Aug 14, 2011 #3
    I do not know how to make a diagram but here is the link of our taken photo for our assignment:
    http://hippomonstro.tumblr.com/post/8907741055/planking-for-physics

    I hope this helps. Thanks! :)
     
  5. Aug 14, 2011 #4

    I like Serena

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    Aaaah, planking! :smile:

    For your torque you won't need the height of the rungs, but the distance between them.

    Furthermore, you should (first) use the equilibrium condition that says that the sum of the vertical forces is zero.
     
  6. Aug 14, 2011 #5
    Oh, yeah I remember. You are right. Thank you for that! :smile:
     
  7. Aug 14, 2011 #6
    Ah, so R(resultant vector) - lbs(weight of my teammate) - F(Force of the pool ladders) = 0

    Therefore, I have to solve for the F first which is

    F = R-lbs?

    Is this correct? I'm so confused.
     
  8. Aug 14, 2011 #7

    I like Serena

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    R(resultant vector) must be zero, otherwise your teammate would move vertically.
    So the sum of the 2 forces of the pool rungs must be equal to the weight of your teammate.
     
  9. Aug 14, 2011 #8
    I tried to figure this out and this is what I have come up with:

    Since the R and the F of the pool rungs are unknown, I cannot solve for the first condition of equilibrium. So, I used torque to solve for the forces.

    (assuming the values are correct)

    ΣT = Fx1 + Fx2 + Fx3 ...

    ΣT = (1.5548780488 m x lbs)+(F x 1 m)+(F x 1 m)
    = (253.13414634464 m.lbs)+(F x 1 m) + (F x 1m)
    (253.13414634464 m.lbs)/(1M x 1M) = (F x 1 m) + (F x 1m)/(1M x 1M)
    (253.13414634464 m.lbs)/(1M x 1M) = (F x 1 m) + (F x 1m)/(1M x 1M)
    (253.13414634464) = F + F

    Will I just divide 253.13414634464 into half?
     
  10. Aug 14, 2011 #9

    I like Serena

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    Hmm, let's define F1 as the force exerted by the one rung, and F2 the force exerted by the other rung.
    Furthermore let W be the weight of your teammate.
    All these forces are vertically oriented.

    Let the distance between the 2 rungs be "d".
    And let the distance between the first rung and the center of mass of your teammate be "c".

    Equilibrium means: ΣFy = 0: F1 + F2 - W = 0

    Furthermore for the moment sum, we need to pick a fixed reference point.
    Let's take the first rung as reference point.

    Then: ΣT = 0: F1 x 0 - W x c + F2 x d = 0

    If you know W, c, and d, you can deduce F1 and F2.
     
  11. Aug 15, 2011 #10
    I did the equation you have posted and here is what I got:

    ΣT = F1 x 0 - W x c + F2 x d = 0
    I cancelled out "F1 x 0" since it is equal to zero.

    So, ΣT = - W x c + F2 x d = 0

    My value for W=162.8 lbs, for d = 59 cm and for c = 29.5 cm (I divided the d to get this value). Substituting these:

    ΣT = - 162.8 lbs x 29.5 cm + F2 x 59 cm = 0
    - F2 x 59 cm = - 162.8 lbs x 29.5 cm
    - F2 x 59 cm = - 4802.6 lbs.cm
    - F2 x 59 cm/59 cm = - 4802.6 lbs.cm/59 cm
    - F2 = - 81.4 lbs

    So, F2 = 81.4

    To get the F1, the placed the zero in d, and placed the value of d in 0 in ΣT = 0: F1 x 0 - W x c + F2 x d = 0.

    I also got 81.4 for F1.

    Having all of these values,

    ΣT = T1 + T2 + T3
    ΣT = (81.4 lbs x 59 cm) + (81.4 lbs x 59 cm) - (162.8 lbs x 29.5 cm) = 4802.6

    I did not get a zero value for torque, but the force acting downwards - the forces acting upward is equal to zero.

    What is 4802.6? Is my solution wrong?
     
    Last edited: Aug 15, 2011
  12. Aug 15, 2011 #11

    I like Serena

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    Good! :smile:


    No, this is not the way to go.
    You should use the other equilibrium equation for ΣFy to deduce F1......

    Furthermore you already used up the torque formula.
    You have calculated the moments with respect to the left rung.
    That comes out as zero and that's that.
    You're not supposed to recycle it and put other numbers into it.
     
  13. Aug 15, 2011 #12
    Thank you! :biggrin:

    So the next step would be

    R - W - F1 - F2 = 0?

    What is the other equilibrium equation for ΣFy to deduce F1?
     
  14. Aug 15, 2011 #13

    I like Serena

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    Nooooo. As I said before R = 0. :rolleyes:

    Your equation is W - F1 - F2 = 0.
     
  15. Aug 15, 2011 #14
    Therefore, 163.8 - 84.1 - 84.1 = 0

    So I no longer need to compute for the resultant vector? :D
     
  16. Aug 15, 2011 #15

    vela

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    You keep reintroducing R. What exactly do you think this resultant vector is?
     
  17. Aug 15, 2011 #16
    I was just confused. :frown:

    but anyway, thank you for the help. God bless. :smile:
     
  18. Aug 15, 2011 #17

    I like Serena

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    Hmm, I though your teammate had a different weight.
    I also thought you had calculated a different value for F2.
    And finally, what you just wrote does not add up to 0. :confused:

    I guess not, unless you have some nifty way to calculate zero! :biggrin:
     
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