1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lever with spread load: formula

  1. Aug 22, 2012 #1
    Hi Guys,
    How much force will I need to lift 5000 kg block and how much force will be acting on the pivot point?
    I know how to calculate a normal lever but couldn't find any formula for lever with load evenly spread as per the attached diagram.
    see the attached diagram
    Thanks
     

    Attached Files:

  2. jcsd
  3. Aug 22, 2012 #2
    Do you know about "the principle of lever " ? Minimum force you'd require will balance the load of 5000 kg. Hence lever will be in equilibrium. Clockwise and anticlockwise moments will add up to 0.
     
  4. Aug 22, 2012 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Assuming the 5t load is uniform, you can take it as acting through its centre of mass. Can you post the equation you get?
     
  5. Aug 23, 2012 #4
    Hi,
    Thanks I know this one..... I am not that stupid :D
     
  6. Aug 23, 2012 #5
    So what was your attempt?

    Posting this is a requirement of the forum and Haruspex has already offered a hint.

    By the way a simpler and smaller diagram would make things easier for you.
     
  7. Aug 23, 2012 #6
    I kinda thought that. It would make sense but I need to be 100% sure.
    The equation will be (5000 * 1250) / 800
     
    Last edited: Aug 23, 2012
  8. Aug 23, 2012 #7
    Is that an equation?
     
  9. Aug 23, 2012 #8
    Sorry for the size of my diagram.... it's a printed PDF from Acad
     
  10. Aug 23, 2012 #9
    yes if you check my diagram you will see. I would post the picture on here but it won't allow me.
     
  11. Aug 23, 2012 #10
    Well, like I said, I think your diagram makes it more difficult for yourself.

    I assume that the mechanism is turned through 90 degrees and is actually some sort of beam balance with force F applied by some sort of bent arm?

    You have correctly (I hope) taken the appropriate perpendicular distance for the moment arm for F about the pivot.

    However if you wrote a proper equation (an equation has an equals sign!!:wink:) it would (should) become apparent that you have omitted something on the other side.
    Do we measure force in kg?
     
  12. Aug 23, 2012 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That looks right, but don't forget g.
     
  13. Aug 23, 2012 #12
    That's exactly what it is :)
    Yes the force will be in KGs
     
  14. Aug 23, 2012 #13
    I was checking formulas for levers and they don't mention the gravity......
     
  15. Aug 23, 2012 #14
    What units do you use for force in your school?
     
  16. Aug 23, 2012 #15
    newtons of course :)
     
  17. Aug 23, 2012 #16
    So do you understand what haruspex was getting at?
     
  18. Aug 23, 2012 #17
    to be honest; no I don't
     
  19. Aug 23, 2012 #18
    OK I have sketched your arangement as I understand it.

    5000kg is 5 tonnes ie quite a heavy block so I have drawn the pivot mounted near the edge of something solid and the restraint as the bent arm being tied to the vertical face of the support.

    Now moment = force time distance from the line of action of the force to the pivot.

    So the block exerts a moment equal due to the downward force of gravity on its mass.

    This downward force is given by multiplying its mass (5000kg) by g to get a force in newtons.

    So the block exerts an anticlockwise moment of 5000g newtons.

    This is balanced by a clockwise moment of F times y newtons.

    I mentioned that I hope you understood the correct distance since in your diagram x = y.
    Would you have got it correct if x did not equal y?

    An equation contains an equals sign so an equation is

    5000 x g x 1.25 = F x 0.8

    I am using consistent units here (length in metres) . If you work in millimetres you may get the wrong answer if asked for the force.

    In the second diagram I have shown a different block that does not extend the whole length of the left hand side.

    This is to show that the principle is the same.

    Can you see this?
     

    Attached Files:

  20. Aug 23, 2012 #19
    Thanks you are a star :)

    Another question is what will be force acting on the pivot point. My guess is that the force on the lifting arm will counteract the weight of of the block so it will be my result from the lever calc? Is that correct?
    Thanks
     
  21. Aug 23, 2012 #20
    Vertical equilibrium tells you that the upward reaction at the pivot must equal the weight of the block.

    Can you tell me what horizontal force must act to balance the tie force F?
    I haven't shown it on my diagram
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lever with spread load: formula
  1. Lever problem (Replies: 1)

  2. Lever formula help (Replies: 5)

  3. Levers and torque (Replies: 6)

Loading...