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Fluid Powers: Determine what class lever I'm using

  1. Oct 22, 2015 #1
    I need help figuring out which formula to use to find what class lever I am using in a 10 ton log splitter circuit. The lever is 16 inches long and I am applying 100 pounds of force on the lever which is connected to a cylinder 5 inches in diameter, 24 inch stroke, and a 2 inch rod diameter. My teacher says it's in our book but didn't say specifically which formula it is and I just can't figure it out by looking at them all.
     
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  3. Oct 23, 2015 #2

    Simon Bridge

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    Welcome to PF;
    Where is the pivot in relation to the load and the applied force?
    Look at the diagrams for different class levers in your book and compare.
     
  4. Oct 23, 2015 #3
    Ok so it looks like a class 3 lever. The rod of the piston comes up and touches the middle. not quite sure if you can tell from the picture but I'm pretty sure it looks like a class 3
     

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  5. Oct 23, 2015 #4

    Simon Bridge

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    I've personally never used the class system for levers... I just use the sum of torques: then I only need to remember one equation.
     
  6. Oct 23, 2015 #5
    OK. My teacher didn't actually go over all of this with us so we're flying blind trying to learn for ourselves. Thanks for the help
     
  7. Oct 24, 2015 #6

    Simon Bridge

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    If you know about torque (moments) you can check the equation.
    Locate the pivot, sketch the force vectors in.
     
  8. Oct 25, 2015 #7
    I also have to figure out the pressure at 3 different points and all the 3 points are between 2 cylinders one of which I believe is an intensifier if I'm looking at the problem correctly. I just need an equation to go off of.
     
    Last edited: Oct 25, 2015
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