# Lift Problem involving forces, acceleration and speed

1. Aug 29, 2011

### Pho3nix

1. The problem statement, all variables and given/known data
A Lift in a mine shaft takes exactly one minute to descend 500m. It starts from rest , accelerates uniformly for 12.5 seconds to a constant speed which it maintains for some time and then decelerates uniformly to stop at the bottom of the shaft.

The mass of the lift is 5 tonnes and on the day in question it is carrying 12 miners whose average mass is 80 kg.
(i) Sketch the speed-time graph of the lift.

During the first stage of the motion the tension in the cable is 53 640 N
(ii) Find the acceleration of the lift during this stage.
(iii) Find the length of time for which the lift is travelling at constant speed and the final deceleration.

2. Relevant equations
mg-T=ma
Positive down
g=9.8m/s^2

3. The attempt at a solution
(i) Don't know how to get a graph up so i'll just describe it; a linear increase in speed from rest for 12.5s. Then a horizontal line representing a constant speed from some time. After that, a deceleration at an unknown time producing a linear decrease in speed to rest, at 60s.
(ii) (5000+80*12)g - T = (5000+80*12)a
a=0.8m/s^2
(iii) This is the part I can't do. I don't know where to start without any information on the time or distance at the start of deceleration. The answer I get from the back of the book is 40s and 1.33m/s^2.

2. Aug 29, 2011

### Tomer

right.

About the second part -good job.

The third part is indeed a little harder.
As I've said, you can find the final speed of segment 1, up until the lift doesn't accelerate.
You can calculate the distance the lift passed in this first segment.
Call "t" the time in which the lift moves with a constant velocity. What is the distance (as a function of t) the lift passes at segment two?
From the data you've given, you know now the distance of segment 3 (as a function of t) and the time of segment 3 (as a function of t). You also know the initial velocity and the final one (0). Use this information and 2 formulas relating the distance, time, acceleration and velocities to get your acceleration and time. These are two variables with two equations.

Good luck!

3. Aug 31, 2011

### Pho3nix

Ty for your response and hints to part 3. I've worked out part 3 using the speed-time graph (which I guess the question was hinting at) and it's quite simple, which fits in with a small part 3 question in an M1 book.

I used area=half base times height. Solving for the top side gives 40s; the time the lift was traveling at constant speed. The rest can be worked out easily.

EDIT: My bad, its area = 1/2(base+a)*maxv

Last edited: Aug 31, 2011