Looking for help with Elevator Problem

1. Oct 18, 2012

Scene

I'm not sure if I approached this problem correctly and would like some assistance.
1. The problem statement, all variables and given/known data
An elevator with a total mass of 800 kg is moving downward at 6 m/s. It slows to
a stop with constant acceleration in 3 seconds.
a) Find the tension T in the supporting cable while the elevator is brought to
a rest.
b) A 50.0-kg man stands on a bathroom scale while riding on the elevator.
What is the reading on the scale as it is brought to a halt?
c) After the stop, the elevator starts to move down – initially with 1 m/s2
acceleration. What is the reading of the weighing scale?

2. Relevant equations

ƩF=ma

3. The attempt at a solution
For a), I wrote ƩF=ma and I considered going up to be positive My equation was T - mg = ma. I ended up with T = m(a + g) for finding the tension. I saw that the elevator was being brought to rest so I figured that the acceleration would be zero and ended up with T = mg which was 7,840 Newtons.
For b) I drew a free-body diagram of the man and the forces acting on him which were the normal force and mg. I wrote ƩF=ma and my equation was N-mg = 0 which gave me N = mg which was 490 Newtons.
For c), I did pretty much the same thing as b), except my equation was N = m(a + g). I plugged in the given a and g and ended up with N= 440 Newtons.

Just looking to see if I approached this problem correctly. I can't help but think I did it wrong because I didn't utilize the given 6 m/s at all. Any help is greatly appreciated.

Last edited: Oct 18, 2012
2. Oct 18, 2012

Simon Bridge

Acceleration cannot be zero because the speed of the elevator is changing.
They want the situation after the elevator has started accelerating but before it has come to rest.
This is where you needed the 6m/s.

How can the forces be equal - the man is accelerating: there has to be a net unbalanced force on him.

When was the last time you saw bathroom scales that gave readings in "Newtons"?

You have probably ridden in elevators before ... when the elevator is going down, do you feel lighter or heavier?

3. Oct 18, 2012

Scene

I used a kinematic equation of Vf = Vit + at to find the acceleration, which I got to be -2m/s2. I used this to get a tension of 6,240 N.

You are right. I used the acceleration I found earlier in my equation N = m(g + a) and ended up with 390 N for the scale reading which would be 39.8kg.

I got confused and thought I needed a reading smaller than 390 N, but he's already moving at 7.8 m/s2 downwards in that reading. When he's standing still, his reading would be 490 N and in part c) he's moving at 1 m/s2 downward so his reading is 440 N. I think I have it down, now.

Last edited: Oct 18, 2012
4. Oct 18, 2012

mishek

What does it mean when it's negative? In which direction is your acceleration pointing?

5. Oct 18, 2012

Simon Bridge

Better - but your equation is incorrect ... you can see that the first term on the right, Vit, which I take to be $v_i t$, has units of length ... but the second term on the right, $at$ as well as $v_f$ on the left have units of speed. This is inconsistent as well since, from your answer, it looks like you may have used the right one anyway.

Also - you have stated earlier that you have chosen +y (upwards) to be the positive direction. The initial velocity of the elevator is in the -y direction. I don't think you took that into account.

When you get a result you should always be thinking "what does this mean?" Try to see if your answers are consistent with what you know.