There is a box weighing 50 kg in the lift cabin, the weight of which is 100 kg. Determine the force of pressure of the box

There is a box weighing 50 kg in the lift cabin, the weight of which is 100 kg. Determine the force of pressure of the box on the floor of the cabin, moving upward under the action of a counterweight weighing 250 kg.

mp = 100 kg.

mя = 50 kg.

g = 10 m / s2.

m = 250 kg.

P -?

From the side of the hoist, the weight of the hoist with a load acts on the cable, which we express by the formula: P1 = (mp + my) * g. From the side of the counterweight, its weight acts, the value of which is expressed by the formula: P2 = m * g.

Let us express the acceleration a, with which the hoist rises, and the counterweight falls according to Newton’s 2 law: a = (P2 – P1) / (mp + my + m) = (m * g – (mp + my) * g) / (mp + mа + m) = ((m – mp – mа) * g) / (mp + mа + m).

a = (250 kg – 100 kg – 50 kg) * 10 m / s2 / (250 kg + 100 kg + 50 kg) = 2.5 m / s2.

The lift with the box will move upward with an acceleration of a = 2.5 m / s2.

mя * a = N – mа * g – 2 Newton’s law for a box in a lift.

N = my * a + my * g = my * (a + g).

According to Newton’s 3 law: P = N.

P = my * (a + g).

P = 50 kg * (2.5 m / s2 + 10 m / s2) = 625 N.

Answer: the box presses on the floor of the lift with a force of P = 625 N.