Fluid Powers: Determine what class lever I'm using

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Bsmith072
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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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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.
 
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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  • Handwritten sketch in a notebook of a hand pump lever mechanism: labels read 100# force, hand pump, 1 1/2 dia, 3/4 stroke and 3/4 rod dia, with a lever arm, pivot triangle and a piston rod connecting to a rectangular block below.
    IMG_20151023_200909.jpg
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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.
 
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
 
If you know about torque (moments) you can check the equation.
Locate the pivot, sketch the force vectors in.
 
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.
 
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