The length of the lever is 2 m. At its ends, weights of 20 tons and 140 tons are balanced. Find the lever arms

Let’s designate the length of the entire lever as l, and the first arm is l1, then the second arm is l2 = l – l1.

By condition, our lever is in equilibrium, which means that the moment of force acting on the first shoulder is equal to the moment of force acting on the second shoulder. I.e:

M1 = M2 or F1 * l1 = F2 * l2 since M = F * l.

Substitute the expression for l2 into this formula, we get:

F1 * l1 = F2 * (l – l1). Let’s expand the brackets on the right side of the equality:

F1 * l1 = F2 * l – F2 * l1. Let us express the first shoulder from this formula:

F1 * l1 + F2 * l1 = F2 * l;

l1 * (F1 + F2) = F2 * l;

l1 = F2 * l / (F1 + F2).

Let’s substitute the numerical data, and do the calculations:

l1 = 140 t * 2 m / (20 t + 140 t) = 280/160 m = 1.75 m.

l2 = 2 m – 1.75 m = 0.25 m.

Answer: l1 = 1.75 m, l2 = 0.25 m.