Vertical forces of 20 and 80 N are applied to the ends of a homogeneous rod 1 m long.

Vertical forces of 20 and 80 N are applied to the ends of a homogeneous rod 1 m long. The pivot point of the rod is located so that the rod is horizontal. Determine your shoulder lengths. The rod is considered weightless.

Given:

F1 = 20 Newton – force applied to one end of the rod;

F2 = 80 Newton – force applied to the other end of the rod;

l = 1 meter – rod length.

It is required to determine the lengths of the arms x1 and x2 (meter) of the rod.

By the condition of the problem, the horizontal bar is and has no weight. Then x1 + x2 = l, or x2 = l – x1.

F1 * x1 = F2 * x2;

F1 * x1 = F2 * (l – x1);

F1 * x1 = F2 * l – F2 * x1;

F1 * x1 + F2 * x1 = F2 * l;

x1 * (F1 + F2) = F2 * l;

x1 = F2 * l / (F1 + F2) = 80 * 1 / (80 + 20) = 80/100 = 0.8 meters.

x2 = l – x1 = 1 – 0.8 = 0.2 meters.

Answer: a shoulder of 20 Newton is equal to 0.8 meters, a shoulder of 80 Newton is 0.2 meters.