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weights of 40 lbs . and 50 lbs. are placed on opposite sides of the fulcrum and they balance . If one weight is a foot nearer the fulcrum than the other , where is each weight placed?
Let the weight of 50 pounds be placed at a distance of x units from the fulcrum
Then the weight of 40 pounds is placed at a distance of (x-1) units
from the fulcrum
using the concept of order of levers we get 50*(x-1) = 40x
50x-50 = 40x
50x-40x = 50
10x = 50
x = 50/10 = 5
Therefore 50 pounds must be placed at a distance of 5 units and 40 pounds
must be placed at a distance of 5-1 = 4 units from the fulcrum.
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