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Homework Help: Third Order Lever Problem

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    The 200Kg Crates are placed on a hydraulic lifting beam: (Diagram attached)

    A. State the order of lever used here
    B. Calculate the effort required to lift the load
    C. Calculate mechanical advantage for this system
    D. Calculate the velocity ratio for this system. If the work put in is 3000J.Calculate the useful energy output.
    E. System has efficiency of 65%. Calculate the velocity ratio for this system.

    2. Relevant equations

    Effort Required = large distance x small effort = small distance x large load

    Mechanical advantage = load/effort

    Velocity ratio = distance moved by effort/distance moved by load OR
    Velocity ratio = Mechanical Advantage/Efficiency

    Efficiency = output/input

    Efficiency = mechanical advantage/velocity ratio

    3. The attempt at a solution

    A. Third Order as the effort lies between to the fulcrum and load

    B. Force Effort (Fe) x Distance Effort (De) = Force Load (Fl) x Distance Load (Dl)

    Fe = (Fl x Dl)/De
    Fe = (200 x 6)/1
    Fe = 1200N

    C. Mechanical Advantage (Ma) = Load / Effort

    Ma = 200/1200
    Ma = 0.16 Recurring

    D. This is where i have become slightly confused do i assume the system is 100% efficient for this part of the question? In which case i could use the following to calculate the valocity ratio

    Velocity ratio (Vr) = Mechanical Advantage (Ma) /Efficiency (E)
    Vr = 0.16/100

    As such the equation Velocity ratio = distance moved by effort/distance moved by load would be the one to use. The problem is i have not been given the distances moved. I know i need to take the work put in somehow to get my answer but not sure how to use this. Could someone please point me in the right direction?

    E. I will calculate this once i've solved for part D

    Attached Files:

  2. jcsd
  3. Apr 17, 2012 #2

    I guess i worded this question wrong. Anyway have spend some more time on it and now have a slightly different question for this system. The force load is 200KG but as there is gravity acting on the crate will the force load be (200 x 9.81) making it 1962N rather than 200N?

    I have now figured out how to use the work put in to calculate the displacement. One last question as the effort displacement is 5m from the load effort will the displacement be 5x larger or is it 6x as this is the distance from the fulcrum. I believe it will be 5x?
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