Solving for Motion Information on a elevator and some lights

AI Thread Summary
The discussion revolves around a physics problem involving a decorative light fixture in an elevator that experiences a cable snap during an emergency stop. The key challenge is determining the values of time, distance, and acceleration after the cable breaks, with the engineer indicating the cable can withstand a maximum force of 40 N. The original poster struggles with the problem due to unclear wording and missing information, particularly regarding the forces acting on the lights and the initial conditions. After clarifying the scenario, it is suggested that the poster should assume the tension in the upper cable is 40 N to calculate the minimum acceleration and other related values. The conversation emphasizes the importance of precise problem statements in physics to avoid confusion.
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Homework Statement


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.

Homework Equations


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

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