Question

A balanced lever has two weights on it, the first with mass `8 kg` and the second with mass `3 kg`. If the first weight is `2 m` from the fulcrum, how far is the second weight from the fulcrum?

Answer 1

`5.333m`

Explanation:

We apply the second condition for static equilibrium which states that the resultant torque (moments) must be zero,
ie. `sum vectau = 0`,
where `vec tau= vecr xx vecF` is the cross product of the position vector `vecr` from the axis of rotation (fulcrum) to the applied force and the applied force `vecF`.

So application in this case yields :

`r_1F_1-r_2F_2=0`

`therefore (2xx8g)-(r_2xx3g)=0`

`therefore r_2=(2xx8xx9.8)/(3xx9.8)=5.333m`.