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Calculating Effort of a Lever

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the effort required to lift the load.

    2. Relevant equations

    F=mg
    large distance × small effort = small distance × large load

    3. The attempt at a solution

    The load of the crate=

    F=mg
    F=200*9.81
    F=1962N

    large distance × small effort = small distance × large load

    Now this forumla does not represent what is happening as the load it has a large distance (5m) and the effort is small distance (1m) away.

    Is there another forumla for calculating this problem or another method?

    Thanks, Joe
     

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  2. jcsd
  3. Aug 24, 2011 #2

    tiny-tim

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    Hi Joe! :wink:
    (that formula is written for a type 1 lever, this is a type 2 lever)

    it really ought to be written as a general https://www.physicsforums.com/library.php?do=view_item&itemid=64" equation:

    effort distance × effort = load distance × load​

    try that :smile:
     
    Last edited by a moderator: Apr 26, 2017
  4. Aug 25, 2011 #3
    Thankyou so:

    effort distance × effort = load distance × load

    1*effort=5*1962

    effort=5*1962/1

    effort = 9810N

    I hope this is right:)
     
  5. Aug 25, 2011 #4

    tiny-tim

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    Looks good! :biggrin:
     
  6. Aug 25, 2011 #5
    Thanks, the next question was calculate the mechanical advantage which is:

    MA=load/effort

    MA=1962/9810

    MA=0.2

    It just seems quite low and as far as i know there is no unit for MA
     
  7. Aug 25, 2011 #6

    tiny-tim

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  8. Aug 25, 2011 #7
    Thankyou very much.

    Calculate the velocity ratio for this system.If the work put in is 3000J, calculate the usefull output.

    Velocity Ratio = distance moved by effort/distance moved by load

    1/5=0.2 not sure if this is right, the course material is a bit shoddy with this

    If it is right then:

    Efficiency=MA/VR

    Efficiency=0.2/0.2

    1*100=100%

    Doesnt seem right to me?

    I noticed you said this was a second order lever, my course suggests this is a third order since the effort is greater than the load?

    Cheers, Joe
     
  9. Aug 25, 2011 #8

    tiny-tim

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    Hi Joe! :smile:
    Yes, that looks ok. :smile:
    Can't see anything wrong with it.

    The efficiency of a lever should always be 100% unless there's some friction or other obstruction.

    (See http://en.wikipedia.org/wiki/Mechanical_advantage#Efficiency")
    oops! :rolleyes: I should have checked! :redface:

    Yes, looking at http://en.wikipedia.org/wiki/Lever#Classes", you're right, it is a Class 3 lever. :smile:
     
    Last edited by a moderator: Apr 26, 2017
  10. Aug 25, 2011 #9
    Thanks, i guess the 3000J was thrown in there to distract
     
  11. Aug 26, 2011 #10
    Ive been looking over the picture and i dont see how it is a lever, especially since the pivet point is at the end?

    i thought a lever had to have two ends, the image looks like one with a pivet point on the end?

    Wouldnt this make my maths wrong?

    Shouldnt the pivet point be about the hydraulic ram?

    Would you be able to explain this better to me? sorry for being such a pain

    Thanks, Joe
     
  12. Aug 26, 2011 #11

    tiny-tim

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    Hi Joe! :smile:
    No, a lever needs only three things, a load, an effort (or force), and a fulcrum (or pivot) …

    they can be anywhere

    see "Professor Beaker's" explanation, with diagrams of the three types, at http://www.professorbeaker.com/lever_fact.html" [Broken] :wink:
    no the pivot point (the fulcrum) is whatever is fixed (ie doesnt move)
     
    Last edited by a moderator: May 5, 2017
  13. Aug 26, 2011 #12
    Thanks,

    That means that the load is 6m away from the fulcrum(pivet point) and the effort from the ram 1m from pivet point.

    This would make my maths wrong then?

    Since : effort distance × effort = load distance × load

    I guess the distances are from the pivet point?

    I have edited the original picture to include load and effort hopefully putting them in the correct place.

    This means the idea of the ram is to increase the angle of the lift to allow the crate to slide down?

    Thanks a lot for your help, i know there has been a lot of silly questions
     

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  14. Aug 26, 2011 #13

    tiny-tim

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    oops!

    oops! I should have enlarged the diagram! :redface:

    Yes, it should have been 6 instead of 5 each time.
    No, I'm pretty sure the aim is to raise the box, not to move it sideways. :wink:

    (btw, placing the ram where it is has no mechanical advantage, in fact it has a mechanical disadvantage, as you've calculated …

    so noone in their right mind would put the ram there if they didn't have to … it's usually because there's limited space, and that's the closest the ram can get without interfering with everything else in the warehouse :biggrin:)
     
  15. Aug 26, 2011 #14
    Thanks, it will raise the box but it will also increase the angle since it will move about the pivet point
     
  16. Aug 26, 2011 #15
    The new answers should be:

    effort distance * effort = load distance * load

    1m * effort = 6 *1962

    effort=6*1962/1

    effort = 11772N

    MA=Load/Effort
    MA=1962/11772
    MA=0.17 (2dp)

    Calculate the velocity ratio for this system. If the work put in is 3000J, calculate the usefull energy output.

    VR=distance moved by effort/distance moved by load
    VR=1/6
    VR=0.17(2dp)


    Should be right now, the only one im not 100% about is the last one, why would they ask a question that gives the same answer as the mechanical advantage?

    Thanks for all your help, couldn't of done this without you
     
  17. Aug 26, 2011 #16

    tiny-tim

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    You mean the useful energy output, or the efficiency ratio?

    I don't know … since the efficiency seems to be 100%, the energy out should be the same as the energy in, so the question relating to it seems pointless. :confused:
     
  18. Aug 26, 2011 #17
    Maybe its trying to get you to think about if the efficiency is 100% like you said then effectively the usefull energy output will be also 3000J
     
  19. Aug 26, 2011 #18

    tiny-tim

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  20. Aug 26, 2011 #19
    The next question is:

    The system has an efficiency of 65%. Calculate the velocity ratio of the system.

    Im guessing its:

    Since VR=0.17 when efficiency is 100%

    0.17/100 = 0.0017
    0.0017*65 = 0.1105
     
  21. Aug 26, 2011 #20

    tiny-tim

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    I don't understand why you're guessing …

    weren't these topics covered in your lectures or your book? :confused:
     
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