Ameterstick is initially balanced on a fulcrum at its midpoint. you have four identical masses. three of them are placed atop the meterstick at the following locations: 27cm , 38cm , and 99cm .where should the fourth mass be placed in order to balance the meterstick?

The moments to the left of the pivot must be equal to the moments on the right. Let d represent the distance of each mass from the pivot, which is at the 50 cm mark.

md₁ + md₂ = md₃ + md₄
m cancels out from both sides,
d₁ + d₂ - d₃ = d₄
(50 - 27) + (50 - 38) - (99 - 50) = d₄
d₄ = -14
That means the mass should be placed 14 cm to the left of the pivot, at the 36 cm mark.