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Bridge Static Equilibrium

  1. Sep 22, 2008 #1
    [​IMG]
    [​IMG]

    Sum(T_z) [pivot at L] = 2 F_r - 4 mg = 0
    2 F_r = 4 mg
    F_r = 2 mg
    F_r = 2 * 120 * 10
    F_r = 2400

    Sum(F_y) = F_l + F_r - mg = 0
    Sum(F_y) = F_l + 2400 - 1200 = 0
    F_l = -1200

    Answer: 1200 N downward. Is this right? This picture seems to portray an impossible equilibrium, so I'm confused. I think the right end should actually bring the whole thing down by overwheighing...
     
  2. jcsd
  3. Sep 23, 2008 #2

    tiny-tim

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    Hi akan! :smile:

    I can't get your picture links to work. :cry:

    Can you describe the bridge? :smile:
     
  4. Sep 23, 2008 #3
    Why can't you get it to work? It's on Image Shack. :S.
    Anyway, my description will be weird, but so is the picture - so bear with me.

    Problem statement:
    A uniform bar of length 8.0 m and mass 120 kg is supported by two vertical posts spaced by 2.0 m, see the figure. Calculate the force on the leftmost support (magnitude and direction!).

    Note: please use g = 10 m/s^2 for simplicity. Show all work.

    Picture description:
    There is a horizontal bridge, whose length is 8 meters. The leftmost end is supported by an upright post support. There is another post support 2 meters to the right from the left one. The force of gravity acts at the center of mass, so I understand it is 4 meters from the leftmost end (or, likewise, the rightmost one). There are no other supports besides these two, so I don't know how there is an equilibrium. But that's the whole problem, as it is stated. Thanks. :)
     
  5. Sep 24, 2008 #4

    tiny-tim

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    Hi akan! :smile:

    hmm … picture works fine now … it shows up as part of the post … didn't yesterday … mystery :confused:

    good description, anyway! :smile:


    Yup … 1200N is correct …

    though it would have been a lot quicker if you'd just taken moments about the right-hand post, wouldn't it? :wink:

    I agree the question is badly worded … "support" begins with "sup", which is the same as "sub", from the Latin meaning "under". :mad:

    The equilibrium is as expected … you have equal forces (1200N) at equal distances from the right-hand post, so the whole thing is balanced on that post! :smile:
     
  6. Nov 9, 2008 #5
    I'm trying to solve a similar problem using this example, but I'm confused as to where the 4 comes from in the first equation:

    Sum(T_z) [pivot at L] = 2 F_r - 4 mg = 0
     
  7. Nov 9, 2008 #6
    Nevermind, it is the downward force due to gravity that is causing a torque force rotating about the axis denoted by L.

    Just tired this afternoon... :)
     
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