Using a lever, a worker lifts a slab weighing 100 kg. What force is applied by the worker to the larger arm

Using a lever, a worker lifts a slab weighing 100 kg. What force is applied by the worker to the larger arm of the lever, equal to 1.4 m, if the smaller arm is equal to 0.6 m?

According to the balance rule of the lever:
F1 / F2 = l2 / l1, where F1 is the force with which the plate acts on the shoulder l1 (F1 = Ft = m * g, where m is the mass of the plate (m = 100 kg), g is the acceleration of gravity (g = 9 , 8 m / s²)); F2 is the force applied by the worker to the arm l2 (N), l1 is the smaller arm of the lever (l1 = 0.6 m), l2 is the larger arm of the lever (l2 = 1.4 m).
Let us express and calculate the force applied by the worker:
F2 = F1 * l1 / l2 = m * g * l1 / l2 = 100 * 9.8 * 0.6 / 1.4 = 420 N.
Answer: The worker applies a force equal to 420 N.