Leverage
If you have a 20 inch wheel and fill it completely with weight with no spaces between the weights, using 10 one oz weights attached on the left of the rim and 10 one oz weights on the right side of the rim that are out an inch further from center than the weights on the left, you get leverage. Lets use 10 gm of leverage for this case. "this measure being done by putting the weights that are further out at the 12-6 O'clock positions and measuring the leverage at the 3 O'clock position."
If you go to a 40 inch wheel and fill it completely with weight with no spaces between the weights, using 20 one oz weights on the left, then 20 one oz on the right that are all still 1 inch further from center than the weights on the left and measure the leverage, will the leverage go up or remain the same?
I have tested and shown that if you have only two weights, one on the left, then one on the right that's 1 inch further out, whatever leverage you get will stay the same if you move to a bigger wheel but still keep the weight on the right one inch further out than the one on the left.
I would like to understand this with a completely filled wheel, two different sizes.
Please explain in laymen's terms, I'm not a physics major.
Thank you
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