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Suppose you are given a 4 m long straight bar of uniform density balanced at its center and a weight of 4.6 kg hanging on the bar at a distance of 12.0 cm from the pivot. Where must we hang a 2.6 kg mass to have the system remain in balance?

Respuesta :

Olajidey

Answer:

21.23 cm

Explanation:

Being that the mass of the bar is equal on both sides, the moment produced by the masses cancel each other out, so you can just look at the hanging masses.

Sum of the moments about the fulcrum = 0... I'll say that moments produced on the 4.6 kg side are positive

0 = 4.6kg * g * 12.0 cm - 2.6kg * g * x

g's cancel. solve for x

x =( 4.6  * 12) / 2.6

x = 55.2 / 2.6

x = 21.23cm

so 2.6kg mass should be pivoted at a distance of 21.23 cm from pivot.

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