# A balanced lever has two weights on it, the first with mass 5 kg  and the second with mass 3 kg. If the first weight is  6 m from the fulcrum, how far is the second weight from the fulcrum?

Feb 25, 2017

10 meters

#### Explanation:

Imagine a see-saw where two people sit on either end.

Assuming the two smileys weigh the same amount and are the same distance from the center of the see-saw (the fulcrum), the see-saw is balanced.

Now imagine, the pink smiley and her best friend sit together on her end of the see-saw. What happens? The pinks' side will go down, while the boy's side goes up!

Next, let's say the pink's scoot up the see-saw so that they are sitting half-way between the end of their side and the fulcrum. What happens now?

They're balanced again! So, we've changed two variables: weight and distance. The ability to rotate around a fulcrum, torque, is dependent on force and distance.

If a system is not rotating, clockwise and counter-clockwise rotation are balanced:

${F}_{1} \cdot {d}_{1} = {F}_{2} \cdot {d}_{2}$

In our case:
${F}_{1} \cdot {d}_{1} = {F}_{2} \cdot {d}_{2}$
${d}_{2} = \frac{{F}_{1} \cdot {d}_{1}}{F} _ 2$
${d}_{2} = \frac{5 k g \cdot 6 m}{3 k g}$
${d}_{2} = 10 m$