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Elevator cable Tension problem

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    An elevator cable can stand a maximum tension of 19000N before breaking. The elevator has a mass of 520 kg, and a maximum acceleration of 2.16 m/s^2. Engineering safety standards require that the cable tension never exceed two-thirds of the breaking tension. How many 65-kg people can the elevator safely accomodate? round down to the nearest integer


    2. Relevant equations
    ma=T-mg


    3. The attempt at a solution
    Ok so this is what I did, I tried to come up with an equation for the total mass the elevator can handle

    ma=T-mg
    T=mg +ma
    m=T/(g+a)

    T= (2/3)(19,000N)= 126667.67
    g=9.8m/s^2
    a=2.16m/s^2

    So i plugged everything in:

    m= [(12,6667.67)/(9.8+2.16)] / 65.0kg and it came out to be around 16.3 kg, which i rounded down to 16 people, and apparently this answer is wrong. Can anyone help me out?
     
  2. jcsd
  3. Oct 1, 2009 #2
    Okay, you are sort of on the right track. The reason you are getting the wrong answer is because you only have one m in your equation. When you write ma=T-mg, the m represents the mass of the elevator plus the total mass of the people. Since you know the total mass of the elevator, you can go through all the work and solve for the total mass of the people and then divide by 65. This should give you the correct answer. So you can check, my equation looks like this: T-(me+mp)g=(me+mp)a, where me is the mass of the elevator and mp is the total mass of the people.
     
  4. Oct 1, 2009 #3
    ah, yes, i forgot the mass of the elevator, i got 8.3, which i rounded down to 8 ppl, thanks
     
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