SOLUTION: Given: W1•x = W2(L - x), where W1 and W2 are weights on either side
of a balance bar, x is the distance from W1 to the
folcrum point, and "L" is the length of the balance
Algebra ->
Equations
-> SOLUTION: Given: W1•x = W2(L - x), where W1 and W2 are weights on either side
of a balance bar, x is the distance from W1 to the
folcrum point, and "L" is the length of the balance
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Question 872623: Given: W1•x = W2(L - x), where W1 and W2 are weights on either side
of a balance bar, x is the distance from W1 to the
folcrum point, and "L" is the length of the balance bar.
What weight must be applied to the right end of an 8 foot long
balance bar (of negligible weight) to balance a 70 pound child on
the left end who is located 1 foot away from the fulcrum point?
You can put this solution on YOUR website! That's a very strange see-saw, and will not be much fun for that child, but I guess the numbers were chosen to make the arithmetic easier, and I like that. -->-->-->
So there you have it. A 10 pound weight at the end of the long arm of that see-saw, will allow that 70 pound child to bounce up and down by about 1 foot. Weee! --->
