Fluid Powers: Determine what class lever I'm using

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Discussion Overview

The discussion revolves around identifying the class of lever used in a 10-ton log splitter circuit, including the dimensions and forces involved. Participants explore the application of lever classes, torque calculations, and the need for specific equations related to pressure in hydraulic systems.

Discussion Character

  • Homework-related, Technical explanation, Exploratory

Main Points Raised

  • One participant seeks assistance in determining the class of lever for a log splitter, providing specific measurements and forces involved.
  • Another participant asks for clarification on the location of the pivot in relation to the load and applied force, suggesting that diagrams in the textbook may help.
  • A participant suggests that the lever appears to be a class 3 lever based on their observations of the setup.
  • One participant mentions their preference for using the sum of torques instead of the class system for levers, indicating a different approach to the problem.
  • Another participant encourages the use of torque equations and sketching force vectors to analyze the situation further.
  • A participant raises an additional concern about calculating pressure at three different points in the system, mentioning the potential presence of an intensifier.

Areas of Agreement / Disagreement

There is no clear consensus on the classification of the lever, as participants express differing approaches and understanding of the topic. Some participants agree on the need to analyze torque, while others focus on the class system.

Contextual Notes

Participants have not fully established the definitions or assumptions regarding lever classes and torque calculations, leading to potential ambiguities in their discussions.

Who May Find This Useful

Individuals interested in mechanics, particularly those studying levers and hydraulic systems, may find this discussion relevant.

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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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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