Explaining Elevator Tension and Acceleration

[sweet-buds]

Homework Statement

what happens to tension of a rope(connected linearly to 2 bodies) if one of the bodies move closer to slacken the rope? does the tension act on the body in opposite direction to push it? I am confused because as far as I know tension cannot push a body or rope?

Based on this concept, I have a question,

An single cabled elevator receives passengers at the ground floor and takes them to the top floor. where they disembark. new passengers enter and are taken to ground floor. during this round trip,a) when is the tension in the cable equal to the weight of the elevator plus passengers, b)when is it greater c)when is it less?

Homework Equations

The Attempt at a Solution

ordinarily, if the elevator is filled with passengers and is detached from the cord and is at the bottom, then there won't be any tension in the cord. Now in this case N=Mg +mg(mass of elevator, mass of passengers).

when the elevator is moving down(I am assuming the elevator has the cord attached to it from the ceiling of the top floor) Tension T will tighten and the force acting on elevator will be T=Mg+mg. when the elevator hits the bottom floor, normal force comes into play. so N+T=Mg+mg
From the above eqn, T=Mg+mg-N. consequently, Mg+mg>T
Mg+mg=T when N=0. this can happen midway when the elevator is not touching the ground

 

[chaoseverlasting]

No. Actually, you need to take pseudo forces into account here as this is a non inertial frame of reference. Here, the force acts in the direction opposite to that of the acceleration. So, if the lift is going down with an acceleration 'a' then a force, ma, is experienced by the body in the upwards direction.

Using this, you can differentiate between the actual weight of the passengers+lift and the apparent weight.

HINT: The only time the apparent weight will be equal to the actual weight is when the acceleration is zero.

 

[Dick]

No. Actually, you need to take pseudo forces into account here as this is a non inertial frame of reference. Here, the force acts in the direction opposite to that of the acceleration. So, if the lift is going down with an acceleration 'a' then a force, ma, is experienced by the body in the upwards direction.

Using this, you can differentiate between the actual weight of the passengers+lift and the apparent weight.

HINT: The only time the apparent weight will be equal to the actual weight is when the acceleration is zero.

I agree completely, but I think talking about 'pseudo forces', 'non inertial frames' and 'apparent weight' makes the answer sound more mystical than it is. If the tension is greater than the weight, the elevators acceleration is upwards and vice versa. That's it. The direction of the elevators motion is not necessarily the same as the direction of acceleration. You can be moving up while accelerating down - though you might usually call that deceleration.