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Math Help - 3rd class lever

  1. #1
    prs
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    3rd class lever

    a third class lever, 200g load is 6m from the pivot, and applied effort is 1m from the pivot,
    The book asks CALCULATE THE VELOCITY RATIO FOR THIS SYSTEM. IF THE WORK PUT IN IS 3000J, CALCULATE THE USEFUL ENERGY OUT PUT.

    1) load, 200 x 9.81 =1962n
    2) effort to lift the load, load x distance from load to fulcrum / distance from fulcrum to effort. 1962 x 6 / 1 =11772n.
    3) mecanical advantage or force ratio, load/effort. 1962/11772 =0.167 (0.17)

    problems, velocity ratio= distance move by effort / distance move by load, trouble is there is no indication of any movement that i can see,
    also if i use force ratio / efficiency x 100. no machines are ever 100%

    should i have worked out a new applied effort from 3000j if so can somebody guide me on this,

    thanks p,
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  2. #2
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    Quote Originally Posted by prs View Post
    a third class lever, 200g load is 6m from the pivot, and applied effort is 1m from the pivot,
    The book asks CALCULATE THE VELOCITY RATIO FOR THIS SYSTEM. IF THE WORK PUT IN IS 3000J, CALCULATE THE USEFUL ENERGY OUT PUT.

    1) load, 200 x 9.81 =1962n
    2) effort to lift the load, load x distance from load to fulcrum / distance from fulcrum to effort. 1962 x 6 / 1 =11772n.
    3) mecanical advantage or force ratio, load/effort. 1962/11772 =0.167 (0.17)
    This is, of course, 1/6.

    problems, velocity ratio= distance move by effort / distance move by load, trouble is there is no indication of any movement that i can see,
    also if i use force ratio / efficiency x 100. no machines are ever 100%
    If you are referring to the "velocity ratio", which I admit I have never seen in a "lever problem" before, since the load arm is 6 times as long as the effort arm, the load will move 6 times as far as the effort in the same amount of time and so will have 6 times the velocity.

    should i have worked out a new applied effort from 3000j if so can somebody guide me on this,

    thanks p,
    A lever changes force, not energy! Conservation of energy, ignoring friction (which is what would keep a machine from being 100% efficient) not mentioned in this problem, if 3000 J of work were put in, 3000J of useful work will be gotten out.
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  3. #3
    prs
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    thanks for that h, i cant work out why 3000j was ever mentioned in the question, but never mind,
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