A board weighing 10 kg lies on two logs, the distance between which is 1 m. On the first log, the board presses

A board weighing 10 kg lies on two logs, the distance between which is 1 m. On the first log, the board presses with a force of 40 N. With what force does it press on the second log? At what distance from the first log is the center of gravity of the board, if it coincides with the middle boards?

g = 9.8 m / s ^ 2.
l = 1 m.
F1 = 40 N.
F2 -?
r1 -?
The gravity force Ft = m * g acts on the board.
This gravity of the board is distributed over the two logs. Since the force F1 acts on the first log, the force F2 = m * g – F1 acts on the other log.
F2 = 10 kg * 9.8 m / s ^ 2 – 40 N = 58 N.
The product of the force F1 and the distance from the center of gravity on one side r1 is equal to the product of the force F2 and the distance from the center of gravity on the other side r2.
F1 * r1 = F2 * r2.
Since the distance between the logs on which the board lies is 1 meter, then r1 + r2 = 1.
r2 = 1 – r1.
F1 * r1 = F2 * (1 – r1).
F1 * r1 = F2 * 1 – F2 * r1.
F1 * r1 + F2 * r1 = F2.
r1 = F2 / (F1 + F2).
r1 = 58 N / (40 N + 58 N) = 0.59 m.
Answer: a force F2 = 58 N acts on the second log, the center of gravity of the board is at a distance of r1 = 0.59 m from the first log.