An elevator moving with constant velocity

Question

While an elevator moves up, it moves up with a constant velocity. I read this post and understood that it's because of inertia. However, I'm not really convinced.

So what happens which I have understood is the upward tension($T$) on the rope on the load side imparts an upward acceleration which being greater than the weight $L$ of the elevator itself, causes net force to act upwards. The effort $E$(effort in the sense that the elevator is a pulley) imparts a downward force on the other side of the rope, which is greater $T$, hence causing net force and acceleration downwards.

In an elevator exhibiting dynamic equilibrium, as soon as the acceleration is imparted on the elevator, the effort is made to cease to act such that the net force acting on the effort side is 0, but since it already is in motion, it continues to be in motion because of Newtons First Law, and so is the case for the elevator or load itself.

But how is the effort controlled in such a way so as to make $L=T$ in an elevator? Does it mean that the $E$ is not caused by gravity? Even if it's not, when the net force acting on is 0, won't $mg$ cause the effort to move down with an acceleration again? Or is there another device resisting $mg$?

I just want to understand the mechanism behind how the forces are obtained equal in an elevator, as I had learnt that in an Atwood's Machine, $E>T>L$, so I can't really grasp situations where they are equal.

Answer 1

There is no physical observable called Effort. From your comment reply I think you're just just thinking of force.

The counterweight is supplying most of the force. The rest of the force is supplied by the motor, such that, for counterweight mass $m_c$, elevator mass including cargo $m_e$, motor force $F_m$, elevator acceleration $a_e$, counterweight acceleration $a_c$, and neglecting friction

$$m_c(g-a_c) + F_m = -T = m_e(g-a_e)$$

Note that $g$ is a negative number (gravity points down) and although I don't know how elevators are engineered I suspect $a_c = -a_e$ (when the elevator goes down, the counterweight goes an equal and opposite amount up).

For constant velocity, $a_e = a_c = 0$.

Note that $T$ is the tension on the part of the cable connected to the elevator. Somewhere between the elevator and the counterweight (probably at the pulley itself, but I don't know how elevators are designed), the motor is bearing some of the load, so the cable connected to the counterweight experiences a different tension, $T_c = T+F_m$. ($F_m$ is negative if it's helping to support the elevator, or positive if it's helping to support the counterweight.)

Answer 2

Elevators have motors which control the tension. What you call "E" is not solely caused by gravity.

You have probably noticed that if there is a weight on one side and an empty elevator on the other side, and a person enters the elevator on the ground floor, the weight cannot be heavy enough to lift the elevator and the person alone, since the elevator alone was heavier and brought itself to the ground floor. Hence, elevators cannot only rely on counterweights.

The role of the counterweight is to reduce the energy needed by the motor to operate the elevator. Without the counterweight, the motor must add enough force to carry the elevator as well. With the counterweight, the motor only has to add enough force to carry the passenger/cargo.