Three boys are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board (which can be disregarded) of L = 3.74 m long.
Two boys are already on either end. One has a mass of m1 = 49.5 kg, and the other a mass of m2 = 32.0 kg. How far from the center should the third boy, whose mass is m3 = 22.2 kg, place himself so as to balance the seesaw?
How would you solve this problem? I have tried, but I am not getting the right answer.
2007-01-11 08:26:17 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You have to balance the torques on either side of the fulcrum. Since you can disregard the mass of the board, you have two torques to calculate initially, m1*(3.74/2)=92.565=t1 and m2*(3.74/2)=59.84=t2. The difference in torque between the two fat kids is going to have to be balanced by the m3 kid. So that equation will be t1-t2=t3=32.725=22.2*r. In this equation, r is the distance from the fulcrum that m3 will be. r=1.4741m.
2007-01-11 08:55:57 · answer #1 · answered by roycefer 2 · 1⤊ 0⤋
49.5*3.74/2 = 22.2x + 32.0*3.74/2
22.2x = 1.87(49.5 - 32.0)
22.2x = 1.87(17.5)
22.2x = 32.75
x = 1.474 m
from the center in the direction of the 32.0 kg boy.
2007-01-11 16:47:16 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋