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Force and Work on an Elevator

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data

    An elevator consists of an elevator cage and a counterweight attached to the ends of a cable that runs over a pulley. The mass of the cage (with its load) is 1200kg, and the mass of the counterweight is 1000kg. The elevator is driven by an electric motor attached to the pulley. Suppose the elevator is initially at rest on the first floor of the building and the motor makes the elevator accelerate upward at the rate of 1.5 m/s.

    Mass of elevator = 1200kg
    Mass of counterweight = 1000kg
    Acceleration = 1.5 m/s

    a) What is the tension in the part of the cable attached to the elevator cage? What is the tension in the part of the cable attached to the counterweight.

    b) The acceleration lasts exactly 1.0s. How much work has the electric motor done in this interval? Ignore frictional forces and the mass of the pulley.

    Time = 1.0s
    Interval = 1.5m

    c) After the acceleration interval of 1.0s, the motor pulls the elevator at a constant speed until it reaches the third floor, exactly 10.0m above the first floor. What is the total amount of work that the motor has done up to this point.

    Distance = 10.0m

    2. Relevant equations

    [tex]\Sigma[/tex]F = m*a
    Work = F*D

    3. The attempt at a solution

    Tension on the cable connected to elevator:
    [tex]\Sigma[/tex]F = m*a
    T-mg = ma
    T = mg + ma
    T = 1200kg(9.8m/s^2 + 1.5m/s)
    T = 13,560 N

    Tension on the cable connected to counterweight:
    [tex]\Sigma[/tex]F = m*a
    -T-mg = ma
    -T = ma + mg
    T = -m(a + g)
    T = 8,300N

    This problem confused me. I set the tension to be negative because the motor is pushing down on the cable. This really dosen't make sense. Maybe some force is negative, but it sure isn't tension. Also, since the elevator is going down, why isn't acceleration negative? Keep in mind, I did get the right answer.

    W = F * D
    W = (13,560N - 8,300N) * 1.5m
    W = 7890 J

    I know this answer isn't right. I don't understand what I am supposed to do for the force.

    [tex]\Sigma[/tex]F = ma
    [tex]\Sigma[/tex]F = 0
    F - mg = 0

    This problem is similar to the last one. Once again, I'm not sure how to calculate the force. Do I need to examine just the force on the elevator and the motor or that caused by the counterweight as well? How are the elevator and the counterweight related? Obviously the counterweight reduces the amount of force that the motor needs to apply in order to overcome the inertia of the elevator.
  2. jcsd
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