
#1
Jul1209, 09:40 AM

P: 1

I am trying to solve a reallife design issue involving balance points. It's been a very long time since I've taken calculus and physics. I've search all equations for finding a fulcrum, and all of them assume the mass of the board used to balance is negligible.
Here is my scenario: I have a 48" bar weighing 16 oz, so 1/3 oz per inch. At the far end, I want to place a 16 oz. weight. First part of the question: what equation would I use to find my current fulcrum. The simple equations tell me it will be directly under the weight, and I know this isn't true. The second part is if I want the fulcrum to be 6 inches from the end of the bar, and I want the counter weight to be spread out over 24 inches starting from the opposite end, what is the weight/inch of the material I should use? Equations again would be helpful, as I understand my specs might have to change based on what I find. Oxxxxxxxx So assuming that "" is 3 inches with a weight of 1 oz., and "x" is 3 inches with a weight of 1 oz. plus w, and O has a weight of 16 oz., how do I find w so that the fulcrum is in the correct place? I really appreciate any help on this. This is a project that I'd like to get correct the first time without having to go out and buy new materials after tons of trial and error. Thanks again! Dio. 



#2
Jul1509, 08:50 AM

P: 10

Do you want the board to be balanced? A fulcrum is just used to amplify force, like when prying something. If you want it balanced just find the center of gravity and place the support there.




#3
Jul1609, 06:22 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

I believe that is what he is asking for!
diodagoat, You can think of the board as having its 16 oz weight concentrated at its center point. Suppose the fulcrum is placed "x" inches from the left end of the board and another 16 oz weight is placed at the right end, 48 inches from the left end. Since the total weight of the system is 32 oz, the fulcrum is exerting 32 oz upward at x inches from the left end, a torque of 32x (ounceinches!) "counterclockwise" around the left end. The board itself is exerting 16 oz downward at, as I said, the center, 24 inches from the left end. That is a torque of (16)(24)= 384 clockwise around the left end. The additional weight is 16 oz 48 inches from the left end, a torque of (16)(48)= 768 clockwise around the left end. So there is a total torque of 32x counterclockwise and 384+ 768= 1152 clockwise. Since the board is not rotating those must be equal: 32x= 1152 so x= 35.37. The fulcrum must be 36 inches from the left end which is 12 inches to the right of the center point and 12 inches from the weight at the right end. As soon as I did all that calculation, I realized: just put the fulcrum halfway between the center of the board and the additional weight! For example, if you spread the weight over the entire 6 inches, you can think of it as concentrated at 3 inches from the fulcrum while the weight of the board, at the center, is 18 inches from it. For those to "balance" we must have 3w= 18(16) so w= 6(16)= 96 oz. 


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