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Solving for Motion Information on a elevator and some lights

  1. Feb 6, 2012 #1

    Dko

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    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. jcsd
  3. Feb 6, 2012 #2

    PhanthomJay

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    You have left out a lot of information in the problem statement, including the question itself. Please re-post.
     
  4. Feb 6, 2012 #3

    Dko

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    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
     
  5. Feb 6, 2012 #4

    PhanthomJay

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    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......
     
  6. Feb 6, 2012 #5

    Dko

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    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."
     
  7. Feb 6, 2012 #6

    PhanthomJay

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    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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