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Posted: 12/13/2020 11:37:32 AM EDT
Is it ok to ask a maybe dumb question here
 ?Lets say there’s a beam on a fulcrum with two weights, one on the end of each side. The beam is 7 feet long. The beam is not centered on the fulcrum. One weight is 5lbs and the other weight is unknown. The fulcrum is 2’ away from the 5lb weight and 5’ from the unknown weight. The beam is perfectly balanced this way. How many pounds is the unknown weight? What is the method for solving this question?  
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Posted: 12/13/2020 11:41:39 AM EDT
[#1]
Simple multiplication relationship. Weight times distance from fulcrum.
Since the beam is balanced, the two products must be equal. 5 lbs at 2 feet equals 2 lbs at five feet Weight x distance = weight x distance. This is, of course, fantasand physics where the beam has no mass. Otherwise you have to take its COG and distance from the fulcrum as well in your formula. |
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Posted: 12/13/2020 11:46:38 AM EDT
[#2]
Quote HistoryQuoted: Simple multiplication relationship. Weight times distance from fulcrum. Since the beam is balanced, the two products must be equal. 5 lbs at 2 feet equals 2 lbs at five feet Weight x distance = weight x distance. This is, of course, fantasand physics where the beam has no mass. Otherwise you have to take its COG and distance from the fulcrum as well in your formula. Awesome. Thanks 
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Posted: 12/13/2020 4:08:02 PM EDT
[#3]
The mass of the beam is the critical unknown.
The answer would be different for a beam made of hollow bamboo or aluminum pipe vs. one made of solid steel bar stock. |
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Posted: 12/14/2020 1:41:03 PM EDT
[#4]
First post nails it and all.
Its the "Law of the Lever" and is how Steelyards (scales) work. To balance, weight A at x distance = weight B at Y distance. Or, just look up steelyards and how they work. |
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Posted: 12/17/2020 2:44:47 PM EDT
[#5]
Quote HistoryQuoted: The mass of the beam is the critical unknown. The answer would be different for a beam made of hollow bamboo or aluminum pipe vs. one made of solid steel bar stock. As long as it is balanced the mass of the beam does not actually enter the calculation. |
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Posted: 8/28/2022 2:03:16 AM EDT
[#6]
[Deleted]
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