The mass of the elevator with passengers is 1000 kg. Find, with what acceleration and in which direction

The mass of the elevator with passengers is 1000 kg. Find, with what acceleration and in which direction the elevator moves, if it is known that the tension of the cable supporting the elevator is equal to ..

Given:

m = 1000 kilograms – the mass of the elevator with passengers;

T1 = 8000 Newton – the value of the cable tension force;

T2 = 100 kN = 100000 – the value of the cable tension force.

It is required to determine a (m / s2) – the acceleration of the elevator with the cable tension forces T1 and T2 and the direction of movement of the elevator in both cases.

Let’s say the elevator moves vertically upward.

For the case T1 = 8000 Newton:

According to Newton’s second law:

T1 – m * g = m * a (where g = 10 Newton / kilogram is an approximate value).

m * a = T1 – m * g;

a = (T1 – m * g) / m = (8000 – 1000 * 10) / 1000 = (8000 – 10000) / 1000 = -2000 / 1000 = -2 m / s2.

Since the acceleration happened with a negative sign, it means that the elevator is moving down.

For the case T2 = 100000 Newton:

T2 – m * g = m * a (where g = 10 Newton / kilogram is an approximate value).

m * a = T2 – m * g;

a = (T2 – m * g) / m = (100000 – 1000 * 10) / 1000 = (100000 – 10000) / 1000 = 90000/1000 = 90 m / s2.

Since the acceleration is with a positive sign, it means that our assumption is correct and the elevator moves up.

Answer: at T1 = 8000 Newton the lift moves downward with an acceleration of 2 m / s2, at T2 = 100 kN the lift moves upward with an acceleration of 90 m / s2.