# Solving for Motion Information on a elevator and some lights

1. Feb 6, 2012

### Dko

1. The problem statement, all variables and given/known data
A decorative light fixture in an elevator consists of a 2.0 kg light suspended by a cable from the ceiling of the elevator. From this light, a separate presses the emergency stop button. During the stop, the upper cable snaps. The elevator engineer says that the cable could withstand a force of 40 N without breaking.

2. Relevant equations
v2 = v1 + a12(t2-t1)
r2 = r1 + v1(t2-t1)+.5(a12)(t2-t1)^2
ƩF=ma

3. The attempt at a solution
My biggest problem is getting a value for t2, r2 or a12. I can't seam to solve for any of them. If it had said that the force that did break the cable WAS 40 N I would be fine. But the wording makes it sound like the actual force could be any number higher then 40.

But here is what I have tried. I knew I was going to have problems)
First I tried solving for a12 using v2 = v1 + a12(t2-t1)
0 = 4 + a12(t2-0)
-4 = a12(t2)
a12 = -4/t2

Then went for r2 using r2 = r1 + v1(t2-t1) +.5(a12)(t2-t1)^2
r2 = 0+4(t2-0) + .5(-4/t2)(t2-0)^2
r2 = 4t2 - 2t2(t2^2)
r2 = 4t2 -2t2
r2 = 2t2

My next try was trying to see if pluging into ƩF = ma for both lights would help
The bottom one
-F B.Cable + F Gravity = ma12
-F B.Cable +(0.8)(9.8) = 0.8a12
-F B.Cable + 7.84 = 0.8a12
a12 = (7.84- F B.Cable)/0.8

The top one
-F T.Cable + F B.Cable + F Gravity = ma12
-F T.Cable + F B.Cable + 2(9.8) = 2 ((7.84 - F B.Cable)/0.8)
-F T.Cable + F B.Cable + 19.6 = 2.5(7.84 -F B.Cable)
-F T.Cable + F B.Cable + 19.6 = 19.6 - 2.5 F B.Cable
F T.Cable = 3.5 F B.Cable

So what else should I do? Should I assume that F T.Cable = 40 N and solve from there?

2. Feb 6, 2012

### PhanthomJay

You have left out a lot of information in the problem statement, including the question itself. Please re-post.

3. Feb 6, 2012

### Dko

That is it. There is no real question given other then blanks for the values of t1, t2, r1, r2, v1, v2 and a12. I have given everything I was given to solve the problem

4. Feb 6, 2012

### PhanthomJay

Looks like you assumed vi = 4, where did you get that number? Then you say m =2, but somewhere you used m = .8? then you say' top' and 'bottom'...2 lights? and what does
'From this light, a separate presses the emergency stop button' mean? You left out a lot of stuff......

5. Feb 6, 2012

### Dko

Opps I did leave out a little. It was late when I wrote this. It should say "From this light, a separate cable suspends a second 0.80 kg light. The elevator is moving downward at 4.0 m/s (where I got v1) when someone presses the emergency stop button."

6. Feb 6, 2012

### PhanthomJay

Oh Ok, it looks like as you say the wording makes it sound like the actual acceleration could be any number higher than due to the 40 N limiting force in the upper cable. But it should be asking what was the minimum acceleration during the stop (that is, assume the tension in the upper rope is 40 N and calculate the acceleration accordingly, along with the other values). Your approach is correct.