A balanced lever has two weights on it, the first with mass $1 kg$ and the second with mass $7 kg$ . If the first weight is $8 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2018-03-12T18:35:08

$~1.14$ meters

When a lever is balanced, the torque exerted on either side of the fulcrum is equal, so the lever is held in equilibrium.

$tau = rFsintheta$ where $tau$ is the torque on one side of the lever and $theta$ is the angle that the side makes with the fulcrum.

$tau_1 = (1*9.8) * 8 * 1$
$tau_2 = (7*9.8) * r * 1$

Knowing that $tau_1$ and $tau_2$ are equal, we can combine the two equations to solve for $r$ .

$9.8 * 8 = 68.6r$
$78.4 = 68.6r$
$r = 1.14$

A balanced lever has two weights on it, the first with mass 1 kg 1 kg and the second with mass 7 kg7 kg. If the first weight is 8 m 8 m from the fulcrum, how far is the second weight from the fulcrum?