SOLUTION: Joe, Jane, and Bill weigh 40, 70, and 90 lbs. respectively and use a plank 15 ft. long for a teeterboard. Bill sits 3 ft., Jane 5 ft., and Joe 8 ft. from the fulcrum, on the same s

Question 38982: Joe, Jane, and Bill weigh 40, 70, and 90 lbs. respectively and use a plank 15 ft. long for a teeterboard. Bill sits 3 ft., Jane 5 ft., and Joe 8 ft. from the fulcrum, on the same side. How far must their father, who weighs 188 lbs., sit from the fulcrum to balance his children?
(Not from a book) I got stumped on trying to put this equation together. Thank you for your help.
Found 2 solutions by checkley71, MaxW:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
40*8+70*5+90*3=188X OR 320+350+270=188X OR 1560=188X OR X=1560/188 OR X=8.2978732 FEET FROM THE FULCRUM.
Answer by MaxW(2) (Show Source): You can put this solution on YOUR website!
using the equation d1(w1) = d2(w2)
Joe = 40 lbs(8)
Jane = 70 lbs(5)
Bill = 90 lbs(3)
Father = 188 lbs(x)
Therefore
40(8) + 70(5) + 90(3) = 188(x)
320 + 350 + 270 = 188x
940 = 188x divide both side by 188
x = 5
Joe, Jane and Bill's father must seat 5ft from the fulcrum.
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