# At the end of a thin rod 1m long and 5kg in weight, a 7.5kg weight is suspended.

**At the end of a thin rod 1m long and 5kg in weight, a 7.5kg weight is suspended. At what distance from the free end of the rod should the support be placed to keep the rod in balance?**

Given:

m1 = 5 kilograms is the mass of the rod;

L = 1 meter – length of the rod;

m2 = 7.5 kilograms – weight of the cargo;

g = 10 Newton / kilogram – acceleration of gravity.

It is required to determine dl (meter) – at what distance from the free end of the rod it is necessary to install the support.

We assume that the weight of the rod itself is distributed parallel to its length. Then:

mx * dl * g = (m2 + m1 – mx) * (L – dl) * g, where mx is the mass of the bar to the free end to the support;

mx * dl = (m2 + m1 – mx) * (L – dl);

mx * dl = m2 * L – m2 * dl + m1 * L – m1 * dl – mx * L + mx * dl;

m2 * L – m2 * dl + m1 * L – m1 * dl – mx * L = 0.

Taking into account that mx = m1 * dl / L, we get:

m2 * L – m2 * dl + m1 * L – m1 * dl – (m1 * dl / L) * L = 0;

m2 * L – m2 * dl + m1 * L – m1 * dl – m1 * dl = 0;

m2 * L – m2 * dl + m1 * L – 2 * m1 * dl = 0;

m2 * L + m1 * L = 2 * m1 * dl + m2 * dl;

L * (m1 + m2) = dl * (2 * m1 + m2);

dL = L * (m1 + m2) / (2 * m1 + m2) = 1 * (5 + 7.5) / (2 * 5 + 7.5) =

= 12.5 / 17.5 = 0.71 meters (result has been rounded to one decimal place).

Answer: The fulcrum must be set at a distance of 0.71 meters from the free end.