The ends of a homogeneous rod 1 meter long are applied vertical forces of 20 and 80 Newton points

The ends of a homogeneous rod 1 meter long are applied vertical forces of 20 and 80 Newton points of support of the rod are located. So the rod is horizontal. Determine the length of the shoulders of the rod to read weightless.

According to the problem statement, the rod is weightless and in equilibrium. Then the following relation is valid for the rod:
F1 * l1 = F2 * l2 (1)
where F1 and F2 are lesser and greater forces, respectively, l1 and l2 are shoulders of lesser and greater forces to the point of balance.
Since the total length of the rod is 1 meter, then l1 + l2 = 1, or:
l2 = 1 – l1 (2).
Let’s write the formula (1) taking into account the formula (2):
F1 * l1 = F2 * l2;
F1 * l1 = F2 * (1 – l1);
F1 * l1 = F2 – F2 * l1;
F1 * l1 + F2 * l1 = F2;
l1 * (F1 + F2) = F2;
l1 = F2 / (F1 + F2) = 80 / (20 + 80) = 0.8 meters;
l2 = 1 – l1 = 1 – 0.8 = 0.2 meters.
Answer: the shoulder of lesser strength is 0.8 meters, the shoulder of greater strength is 0.2 meters.