To a rod 2 meters long and weighing 40N, parallel forces are applied to the bottom at the left 100N and at the right 20N.

To a rod 2 meters long and weighing 40N, parallel forces are applied to the bottom at the left 100N and at the right 20N. At what point should the support be placed to keep the rod in balance?

Two moments of force act on the rod relative to the fulcrum, it is necessary to place the fulcrum so that these moments are balanced, for this an equation can be drawn up, since it is known that a force of 100 N is applied to the left end of the rod, and 20 N to the right end, while the length of the rod is 2 m.
Let’s denote by x the distance from the left end of the bar to the pivot point:
100x = 20 (2 – x);
100x = 40 – 20x;
120x = 40;
x = 0.33 m.
Answer: the support of the rod for finding the rod in equilibrium should be at a distance of 0.33 m from the left end of the rod.