# A balanced lever has two weights on it, one with mass 8 kg and one with mass 3 kg. If the first weight is  9 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 31, 2016

This problem can be solved by recognizing that the product of force and length ($F \times l$) is a constant.

#### Explanation:

If a lever is balanced, then the torque is the same on both sides. Torque is a twisting force that is the product of the force applied and the radial distance from the fulcrum (point of rotation):

$\tau = F \cdot r$
which you may also see written $\tau = F \cdot l$ or $\tau = F \cdot d$

Since we are talking about the length of a lever arm rather than say, a gear, it is easier to thing in terms of the length of the arm.

The force applied is the weight (force of gravity) on each person. And while mass (in kg) is not actually a force, in this case we can use it as though it is, because: ${F}_{g r a v i t y} = m g$, so the weights would be about 30N and 80N, which is the same ratio as 3 kg and 8 kg.

To solve, we know the torques are equal:

${\tau}_{1} = {\tau}_{2}$
${F}_{1} {l}_{1} = {F}_{2} {l}_{2}$
$8 k g \cdot 9 m = 3 k g \cdot {l}_{2}$

solving for ${l}_{2}$
${l}_{2} = \frac{8 k g \cdot 9 m}{3 k g}$
${l}_{2} = 8 \cdot 3 m = 24 m$

Why one would build a lever with such long arms for such small masses is beyond me...