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I am working on a DIY project, I am trying to replicate a bar compressor which for some reason is only sold in Europe so I thought it would be fun to make my own without having to waste so much material. I simply want to make sure the force going down with the plate somewhat replaces a person to jump in the trashcan. Somewhere above 80+ lbs, since I am no engineer I was wondering how would one go about finding the force applied going down to the trash if about 50lbs is pulling down the lever. I understand that materials are key to this but as an amateur I just wanted to know how to get started. Sorry for posting here but I thought it would suit this page the most for this question. Thank you.

Drawing of beam

Actual use of bar

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  • $\begingroup$ The force multipler is about (20.5+18.5)/18.5 which is roughly double the input. Altough the real value is slightly less than this and varies by angle. This is just statics. $\endgroup$
    – joojaa
    Feb 17 '17 at 10:56
  • $\begingroup$ How is this not just a very basic statics problem from high school? $\endgroup$
    – Olin Lathrop
    Feb 17 '17 at 13:17
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The input force is 18.5+20.5 ft away from the fulcrum and the load is 18.5 away from the fulcrum.

This means the force applied to the load is $\frac{(18.5+20.5)}{18.5} = 2.1$ times greater than the input force or for an input force of 50 lbs you get an output force of 105 lbs.

However this assumes that all forces are applied perpendicular to the line between point and fulcrum. If this is not the case then you need to decompose the forces into tangential and normal components.

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  • $\begingroup$ Its about 0-15% less because of the angle across the work cycle assuming force is down, this is why i said its about double. I would have expected a graph from you :P but yes comparable to jumping on top. Less messy perhaps. $\endgroup$
    – joojaa
    Feb 17 '17 at 15:35
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Balance the moments acting.

So,

EffortForce x 39 x cos 30 = 18.5 cos 30 x resulting force.

Or more accurately,

enter image description here

Sin(90+ theta - alpha ) x 39 x effortForce = cos(theta) x 18.5 x requiredForce

This equation will give you answer for any angle of level.

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The force at the middle would be like 2 times bigger than the input. I calculated for this exact situation and it would be around 94lbs.

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  • $\begingroup$ How did you arrive at this answer? Without showing your working this is a very low quality answer. A large part of this site is the how and the why, not just the final number. $\endgroup$
    – AndyT
    Feb 17 '17 at 9:36

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