# Math

A one-meter rod is suspended at its middle so that it balances. Suppose one-gram on the rod at the following distances from one end.

20cm 27cm 38cm 44cm 58cm 70cm 83cm 94cm

Where must a one gram weight be hung so that the rod will balances at the 50 centimeter mark
Where must two gram weight be hung so that the rod will balance at the 50 centimeter mark

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1. general principle of a fulcrum:
If a mass of p units is placed x units from the fulcrum and it balances a mass of q units at y units on the oppose side from the fulcrum , then
px = qy

so make a sketch placing the 1 gram weights at the given distances
The the fourth weight on the right side be k g.
then 1(30) + 1(23) + 1(22) + 1(6) = 1(8) + 1(20) + 1(33) + 1(k)
81 = 61 + k
k = 20

so the weight should be placed 20 cm from the centre or at the 70 cm mark.

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posted by Reiny

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