Solving for Motion Information on a elevator and some lights

Click For Summary
SUMMARY

The discussion centers on solving a physics problem involving a decorative light fixture in an elevator, where a 2.0 kg light is suspended by a cable that can withstand a maximum force of 40 N. The key equations utilized include kinematic equations and Newton's second law, specifically ƩF=ma. The main challenge is determining the values for time (t2), position (r2), and acceleration (a12) during the elevator's emergency stop, particularly under the assumption that the tension in the upper cable is 40 N. The correct approach involves calculating the minimum acceleration during the stop based on this tension.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematic equations
  • Basic knowledge of forces and tension in cables
  • Ability to solve algebraic equations
NEXT STEPS
  • Learn how to apply Newton's second law in dynamic systems
  • Study kinematic equations in detail for motion analysis
  • Explore tension calculations in multi-body systems
  • Investigate real-world applications of physics in elevators and safety systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and engineers involved in mechanical systems, particularly those focusing on dynamics and safety mechanisms in elevators.

Dko
Messages
9
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
40
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
1K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
4K
Replies
5
Views
37K