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Torque and static equilibrium

  1. Oct 23, 2014 #1
    1. The problem statement, all variables and given/known data
    You are designing the crosspiece for the A-frame structure in the figure below. Beams AB and AC are 3.35 m long and have a mass of 350.0 kg each. How much tension must the crosspiece EF withstand? Assume that the mass of the crosspiece and the friction at points B and C are negligible.
    http://puu.sh/cnU4U/5763fd520c.png [Broken]

    2. Relevant equations
    torque = (F) (r) (position)
    F = ma

    3. The attempt at a solution
    i took one section of the ladder thing. i chose the pivot point at force of the pin (Fp) and balance the forces on the free body diagram. the 0.8 is from the horizontal distance between the tension and pivot point
    http://puu.sh/cnUsg/52dd539959.png [Broken]
    then i calculated everything and got the answer
    http://puu.sh/cnUJ3/a9d00e4b98.png [Broken]

    anyone can explain what i did wrong??
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 23, 2014 #2

    haruspex

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    You seem to have assumed that the force from the other beam acts, upwards, at right angles to the beam being analysed. What would that imply if you were to draw the diagram for the other beam?
     
  4. Oct 23, 2014 #3
    it's not at right angle with the beam, i assumed it would be acting in the direction with the same as the direction of the extension of the AC bar. and i would assume that there would be another Force of pin which i would name Fp' acting on the second bar.... would the reaction force Fp be acting on the bar too??? or would it be cancelled by the pin itself?? i am a little consufed.
    http://puu.sh/cnXAf/ecb8aba5c6.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  5. Oct 23, 2014 #4

    BvU

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    Hello wise one,
    I agree with the last line on the first solution picture. On the second you then follow that with something completely different ?!

    Last line on the first solution picture:
    You have N = mg and T rolls out as mgx'/0.8 without further ado.

    Not that it matters, but in the figure I would expect Fp to be in the x direction; what do you think ?
     
  6. Oct 23, 2014 #5

    billy_joule

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    Forget about Fp, you don't need it, you don't need to consider what is happening at point A at all.

    Look at the entire structure to get the normal reaction forces at points B and C.

    Then take moments about point A for bar AB (or AC), there are three terms:
    clockwise moment due to normal force - known
    Anti clockwise moment due to self weight - known
    And Anti clockwise moment due to tension in EF - unknown.

    It's important to recognise the most convenient point to take moments about to reduce your work.
     
    Last edited: Oct 23, 2014
  7. Oct 23, 2014 #6
    Both of you are completely correct, thanks for pointing out the mistake. I really didn't need the force of pin. As for the direction of force of the pin, i honestly wouldn't know which way is the right direction, in my book "nelson, Physics for scientists and engineers" it said the direction of the unknown force didn't matter and i just thought it would be convenient to pick that direction.
     
  8. Oct 24, 2014 #7

    BvU

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    The obvious answer I was fishing for is that Fp is horizontal follows from the force balance.
    Usually B and C are anchored to the side walls and EF serves to reduce the sideways push

    Greg Bernhardt: It's funny how so many posts crossed: I completely missed 2 and 3 and it took me a long time to realize #5 wasn't aware of #4.
     
    Last edited: Oct 24, 2014
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