# Empty elevator with a max speed going down, find tension on cable.

My apologies for asking two questions in a short period of time, this is the last question I have. Hopefully one day I'll be skilled enough to answer everyone elses! :)

## Homework Statement

An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor).

Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend. Take the acceleration due to gravity to be 9.81 m/s2. Remember to include units with your answer.

## Homework Equations

Kinematics:
y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]

Newton's Second Law
F = ma

I read somewhere that tension would be Ft = m(a+g). Does this seem right?

## The Attempt at a Solution

Using kinematics I find the acceleration from which the elevator is at rest, to the point it reaches 6 m/s.

I need to solve for t to replace in another kinematic equation. I use the distance equation for it. (I am using y+ in from top to bottom).

15.25 = .5at^2
30.5 = at^2
sqrt(30.5/a)=t

I plug that into Vf = Vo + at

6 m/s = a * sqrt(30.5m/a)
36 m^2/s^2 = a^2 * 30.5m/a
36 m^2/s^2 = a * 30.5m
(36 m^2/s^2) / 30.5 m = a = 1.18 m/s^2

Then I plug that into
Ft = m(a+g)

Ft = 722kg(1.18 m/s^2 + 9.81 m/s^2) = 7934.78 N

I am really bad at tension problems and want to double check that I got this correct. Thanks :)

rock.freak667
Homework Helper
That looks correct to me.

However the reason the tension is in that form is due to the fact that the Tension (T) acts upwards while the weight (mg) acts downwards. The resultant of these two is 'ma' such that

T-mg= ma or T = ma +mg = m(a+g)

ehild
Homework Helper

## Homework Statement

An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor).

Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend. Take the acceleration due to gravity to be 9.81 m/s2. Remember to include units with your answer.

## Homework Equations

Kinematics:
y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]

Newton's Second Law
F = ma

I read somewhere that tension would be Ft = m(a+g). Does this seem right?

The tension depends on the situation. It is not a law that "T=m(g+a)"

The lift accelerates downward. What forces act on it, in what direction?

## The Attempt at a Solution

Using kinematics I find the acceleration from which the elevator is at rest, to the point it reaches 6 m/s.

I need to solve for t to replace in another kinematic equation. I use the distance equation for it. (I am using y+ in from top to bottom).

15.25 = .5at^2
30.5 = at^2
sqrt(30.5/a)=t

I plug that into Vf = Vo + at

6 m/s = a * sqrt(30.5m/a)
36 m^2/s^2 = a^2 * 30.5m/a
36 m^2/s^2 = a * 30.5m
(36 m^2/s^2) / 30.5 m = a = 1.18 m/s^2

That is correct, but what is the direction of the acceleration?

Then I plug that into
Ft = m(a+g)

Never "plug in" without thinking. Imagine that the rope is cut and the lift falls free with acceleration g. There is no tension, but your favourite formula would give T=m(g+g) is it true?

Newton's law states that the resultant force is equal to ma, both the magnitude and the direction. What is the resultant force? What is its direction?

ehild

ehild
Homework Helper
That looks correct to me.

However the reason the tension is in that form is due to the fact that the Tension (T) acts upwards while the weight (mg) acts downwards. The resultant of these two is 'ma' such that

T-mg= ma or T = ma +mg = m(a+g)

It is wrong, without taking the direction of acceleration into account.

ehild