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Moments on a hanging beam

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The weight of the beam is 2000N, the masses: m1=200kg, m2=300kg.
    What are the tensions in the ropes.

    2. Relevant equations
    Moments: ##M=F\cdot L##

    3. The attempt at a solution
    Moments around A:
    $$\mbox{and}\left\{\begin{array}{ll} 1.5\cdot T_B+1.2\cdot T_B+0.6\cdot T_A=1962\cdot 0.6+2000\cdot 0.8+2943\cdot 1.2 \\ 2T_A+2T_B=1962+2000+2943\rightarrow T_A=3452.5-T_B \end{array}\right.$$
    $$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=-986.4$$
    The sign isn't correct and it should be TB=1013.3
     

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  3. Mar 15, 2015 #2

    NascentOxygen

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    Homework help requests typically include a question seeking an answer....

    The masses here are partly supported by their ropes, so each does not press down on the rod with its full weight.
     
  4. Mar 15, 2015 #3

    haruspex

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    I agree with all your working except the final step. You should get about 2000. The given answer seems too low, even from a cursory consideration, since it should be greater than TA.
    A couple of things look odd. W is off-centre, and W is given in N but the masses in kg, but juggling with those still does not give me around 1000.

    Edit: If you are only wanting TB there is a slightly quicker way. If you pick the reference axis for torque carefully then TA doesn't feature.
     
    Last edited: Mar 15, 2015
  5. Mar 15, 2015 #4

    haruspex

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    Karol has taken that into account.
     
  6. Mar 15, 2015 #5

    NascentOxygen

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    Ah, yes.

    I made a spreadsheet, and with the weight of the beam set to 200N (instead of 2000N), I found a first interval of 0.5m (instead of 0.6m) gave tension TA of 1012N and TB of 1541N. xxxxx (There may be errors.)

    :smile:
     
  7. Mar 16, 2015 #6
    In order to eliminate TA i combine the 2 TA forces into one which acts in the middle between them and then i take moments around that point:
    $$1.2\cdot T_B+0.9\cdot T_B=2943\cdot 0.9+2000\cdot 0.5+1962\cdot 0.3=4237.3$$
    $$2.1\cdot T_B=4237.3\rightarrow T_B=2017.8$$
     
    Last edited: Mar 16, 2015
  8. Mar 16, 2015 #7

    haruspex

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    Sure, but I'm saying that if TA is of no interest then you can take a shortcut to finding TB. If you pick your axis for torque carefully TA does not appear in the equation, so the one equation will do it. You don't need the vertical linear balance at all.
     
  9. Mar 16, 2015 #8
    That's what i did but i got a totally different result for TB
     
  10. Mar 16, 2015 #9

    NascentOxygen

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    It's the same answer as you'd have in post #1 had you not made an error in post #1, according to my spreadsheet.
     
  11. Mar 16, 2015 #10
    $$2.7\cdot T_B+0.6(3452.5-T_B)=6308.8\rightarrow T_B=2017.8$$
     
  12. Mar 16, 2015 #11

    haruspex

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    Yes.
     
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