A homogeneous beam weighing 1.2 tonnes lies on the ground. How much force does one end of the beam exert on the ground

A homogeneous beam weighing 1.2 tonnes lies on the ground. How much force does one end of the beam exert on the ground when trying to lift it at the other end?

First, the mass of the beam must be converted into kilograms m = 1.2 t = 1200 kg
In this case, it is necessary to use the rule of moments M1 = M2 where, M1 = F1L1, and M2 = F2L2, then F1L1 = F2L2
The body is affected by the force of gravity (mg) and some force that lifts the beam up (F1), equal to the reaction force of the support, i.e. the force with which the beam acts on the ground.
L1 is the shoulder of the force F1, equal to the length of the beam, L2 is the shoulder of the force of gravity (mg), and L2 = L1 / 2 because mg acts on the middle of the beam.
We get the formula:
F1L1 = mgL1 / 2
L1- we shorten and get:
F1 = mg / 2
g- acceleration due to gravity, approximately equal to 10m / s ^ 2
We substitute and count:
F1 = 1200kg * 10m / s ^ 2/2
F1 = 6000N = 6kN