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Equilibrium: max distance traveled; undisturbed tension

  1. Nov 15, 2015 #1
    • Missing template due to originally being posted in a different forum.
    Hi,

    I need some guidance on the following problem, please.


    upload_2015-11-15_13-19-3.png
    A daredevil attempts to walk the full length of suspended. A 20 kg wrecking ball hangs at the end of this uniform beam of length 4 m and mass 10 kg and is attached to a hinged wall at an angle of 53 degrees. A cable attached to the wall at an angle of 60 degrees is tied to the end of the beam and supports a maximum tension of 525 N.

    How far along the beam can a man of 70 kg walk without the cable breaking?
    (Apologies for the diagram-sketch.)

    ##Fgm = (\frac12)(4m)(\sin 37°)+ Fgm(x \cdot 4m)(\sin 37°) - F_{wall}(4m)(\sin 53°)##

    ##F_w =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°}##

    ##F_{net} = F_{wall} - F_t##
    ##F_{wall}- F_t = 0##
    ##F_{wall} - 525 N =0##
    ##F_{wall} = 525 N##

    ##x = \displaystyle \frac{419.265 - 29.4889}{412.8451} = 0.94412##

    Finally, multiply 0.94412 by 4 m:
    ##0.94412 \cdot 4 = 3.776 m##


    Thanks in advance.
     

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    Last edited: Nov 15, 2015
  2. jcsd
  3. Nov 15, 2015 #2

    PeroK

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    What's the question?

    In your solution you haven't said what you're doing. It's just a lot of undefined terms and some numbers.
     
  4. Nov 15, 2015 #3
    Thanks, PeroK.

    How far along the beam can a man of weight 70 kg walk without the cable breaking?

    ##\displaystyle \tau_{pivot} = \tau_{cw} - \tau_{ccw}##
    d = 4 m = length of beam

    ##0 = Fgm(\frac12 \cdot 4m) \sin 37° + Fgm(x \cdot 4m) \sin 37° - F_{wall} (4m) \sin 53°##
    ##F_{wall} (4m) \sin 53° = (9.8 m/s^2)(10 kg)(\frac12 \cdot 4m) \sin 37° +
    ((9.8 m/s^2)(70 kg)(x \cdot 4m) \sin 37°##

    ##F_{wall} \sin 53° = (9.8 m/s^2)(10 kg)(\frac12) \sin 37° +
    ((9.8 m/s^2)(70 kg)(x) \sin 37°##

    ##F_{wall} =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°} ...Eq(1) ##

    Applying the translation equilibrium condition,
    ##F_{wall} - F_{T} = F_{net} x##
    ##F_{wall} - 525 N = 0##
    ##F_{wall} = 525 N##

    Now, substituting ##F_{wall} = 525 N## into Eq(1),

    ## 525 N =\displaystyle \frac{29.4889 + 412.8451x}{\sin 53°}##

    ## x = \displaystyle \frac{419.265 - 29.4889}{412.8451} = 0.94412##

    Finally, multiply 0.94412 by 4 m:
    ##0.94412 \cdot 4 = 3.776 m##
    So a man of 70 kg can walk 3.776 meters along the beam before the cable breaks.
     
  5. Nov 15, 2015 #4

    PeroK

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    I can guess at some of the terms you're using and I can guess you're taking moments about the pivot. I don't see anything I can recognise as tension in the cable nor the influence of the 20kg mass.

    I've no idea what F is.
     
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