The distance between the front and rear wheel axles of the cart is 2 meters. The front wheels of the cart act on the ground

The distance between the front and rear wheel axles of the cart is 2 meters. The front wheels of the cart act on the ground with a force of 1200H, and the rear wheels with a force of 1600. What is the mass of the cart? With what forces will the rear and front wheels of the cart act on the ground if the rear axle is displaced from the front by a distance of 2.2 m? Consider that the position of the front wheel axle does not change, the mass of the cart wheels can be neglected, g = 10H / kg

The weight of the cart P is equal to the sum of the forces with which the wheels act on the ground at points A (front) and B (rear):
P = Fп + Fz = 1200 N + 1600 N = 2800 N.
The weight of the cart P is equal to the force of gravity Ft, which can be considered applied to the center of mass of the cart O:
P = Fт = mg,
where m is the mass, g is the acceleration of gravity.
m = Fт / g = 2800 N / 10 m / s2 = 280 kg.
Leverage rule relative to point A before the wheels shift:
Ft * AO = Nz * AB.
Answer: The weight of the cart is 280 kg. After the rear wheel has shifted, the front wheel will hit the ground with a force of 1345 N, the rear wheel with a force of 1455 N.