Answer to Question #134422 in Quantum Mechanics for Jade

Question

When a boy is sitting 1 meter from the center of a see-saw, another boy must sit on the
opposite side 0.5 meter farther from the center compared to the first boy to maintain balance.
When the first boy carries an additional weight of 10 kg and sits 1.5 meters from the center, the
second boy must now move to 3 meters from the center to maintain balance. What is the weight
of the heavier boy?

Answer 1

Let the weight of the heavier boy be M and lighter boy be m

Since in the first condition to be in equilibrium their moment must be equal

$M\times0.5=m\times1\\M=2m\ \ \ \ .......(1)$

in the next condition

$(m+10)\times1.5=M\times3\\1.5m+15=3M \ \ \ ..............(2)$

from (1) and (2)

we get m=3.33 Kg and M=6.66 Kg