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Trying to saw a 1/2 in Al rod in half with only 4 AA batteries as the power source

  1. Mar 4, 2017 #1
    Hi there

    So we have a group design project in my mechanical systems 2 class and it basically goes as follows:
    design a mechanism to cut a half inch aluminum bar in half using a standard hacksaw blade and 4 AA batteries. You have to use a motor w/ input > 5000 rpm for the 'main' component

    we conceptualized a system which implements a chuck to hold the bar while the motor rotates it at a certain ratio and it linear translates back and forth across the saw blade which has increasing depth against the bar....

    I got a lot of work done, so far. such as deriving the approximate number of linear translations necessary such that the cut debris does not completely clog the blade (2.5 times)...

    I'm stuck on force though. I'm trying to figure out the proper amount of normal force (or any kind, I guess) to apply to either the blade or the rod against the blade ...

    I found some equations / theories (in multiple places) that relate the force to the volume of debris generated, the 'sliding distance' and the hardness of the softer material... but nowhere can I find any elaboration on this hardness. Does anyone know? is it a hardness ratio between the two materials?

    the equations are
    V = K*F*l/H
    and
    d (depth of wear) = K*F*l/(H*A)


    I basically just want to mathematically calculate and optimize the theoretical requirements but I feel like I haven't yet learned exactly what I need to do.

    If someone would help me, that would be so awesome :)

    thankss
     
  2. jcsd
  3. Mar 10, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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