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A meterstick 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?

Respuesta :

scme1702
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.

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