A rod of constant cross-section consists of two parts: iron and lead. In the place of soldering of these parts

A rod of constant cross-section consists of two parts: iron and lead. In the place of soldering of these parts, it is mounted on a support and balanced (see Fig.). The length of the lead part is 50 cm. How long is the iron part? The density of lead is 11376 kg / m², the density of iron is 7900 kg / m².

Given:

s1 = s1 = s – constant section bar;

ro1 = 11376 kg / m3 (kilogram per cubic meter) – lead density;

ro2 = 7900 kg / m3 – iron density;

l1 = 50 centimeters = 0.5 meters – the length of the lead part of the rod.

It is required to determine l2 (meter) – the length of the iron part of the rod.

According to the condition of the problem, the rod is in a state of equilibrium. Then, according to the rule of the lever:

F1 * l1 = F2 * l2;

m1 * g * l1 = m2 * g * l2, where m1, m2 are the mass of the left and right parts of the rod, g is the acceleration of gravity;

m1 * l1 = m2 * l2;

ro1 * V1 * l1 = ro2 * V2 * l2, where V1, V2 are the volumes of the left and right parts of the bar;

ro1 * s1 * l1 * l1 = ro2 * s2 * l2 * l2;

ro1 * s * l12 = ro2 * s * l2 ^ 2;

ro1 * l1 ^ 2 = ro2 * l2 ^ 2;

l2 = (ro1 * l1 ^ 2 / ro2) ^ 0.5 = l2 * (ro1 / ro2) ^ 0.5;

l2 = 0.5 * (11376/7900) ^ 0.5 = 0.5 * 1.440.5 = 05 * 1.2 = 0.6 meters (60 centimeters).

Answer: the length of the iron part of the rod is 60 centimeters.