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Moments question

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Two workers are moving a 20kg , 10m scaffolding pole. One stands at the end, the other stands 2.0m from the other end. Calculate the force exerted by the worker standing at the end in holding the pole.
    Assume the mass is distributed evenly and g = 9.8 m/s2.

    2. Relevant equations
    Σclockwise = Σanticlockwise
    3. The attempt at a solution
    20kg * 9.8m/s2 = 196 N
    Taking the centre of the pole as pivot, one worker is 3m from the pivot and the other is 5m. Therefore:
    3F1 = 5F2
    With the mass distributed evenly, that's 2 kg/m through the pole.
    I don't know the rest, my teacher is making a habit of throwing us in the deep end without much instruction. I can see the point, though. Hope it pays off.
     
  2. jcsd
  3. Jan 7, 2016 #2
    You should always pick the unknown force as the pivot if you have more than two unknown forces. By doing so, you eliminate one of the forces and you are left with one unknown to find. Try the problem again but use the person standing 2.0m from the other end as the pivot, this way you can find the force exerted by the person at the very end of the pole directly!
     
  4. Jan 7, 2016 #3

    Doc Al

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    Staff: Mentor

    Good. That's one of the conditions for equilibrium. What's another?

    Good. Hint: What must F1 and F2 add to?

    As faradayscat suggests, you can save effort by wisely picking your pivot point. But you should be able to solve it using any pivot point.
     
  5. Jan 8, 2016 #4
    Didn't even know you could move the pivot. I should look this stuff up on YouTube.
    The moment to the left of that guy is going to be 2m(4kg * 9.8ms-2) = 78.4 Nm anticlockwise. The moment to the right generated by the pole is 8m(16kg * 9.8ms-2) = 1254.4 Nm clockwise. All we want is the force so 1254.4 / 8 = 156.8 N. Unsure how to continue but I'll do my own reading and come back to this.

    All forces must balance.
    196 N.
     
  6. Jan 8, 2016 #5

    Doc Al

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    Staff: Mentor

    Careful. Only the very end of the board will be 2m away from the pivot.

    Hint: What point along the board can you consider its full weight to be acting?
    Good. Express that mathematically. Combine that with your other equation and you can solve for F1 and F2.
     
  7. Jan 10, 2016 #6
    Right, centre of mass, 1m. So 1m * 4kg * 9.8ms-2 = 39.2 Nm anticlockwise.

    Total anticlockwise moment = 39.2 Nm + Force exerted by guy at the end.
    Total clockwise moment = 0 + (4m * 16kg * 9.8ms-2) = 627.2 Nm
    39.2 + F = 627.2
    F = 587.9 Nm
    587.9 / 8 = 73.5 N

    Marked incorrect by the computer, redid the calculations with more sig figs until it was marked correct. Thanks!
     
  8. Mar 2, 2016 #7
    What will be the weight of the pole being carried by the worker 2m from the end of the pole
     
  9. Mar 2, 2016 #8

    haruspex

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    2016 Award

    Strange... since the data given are only to two sig figs, it should have been happy with rounding to 74N.
     
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