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Required power for equally balanced elevator cab-counterweight

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    An elevator system (traction system) consists of a cab, a counterweight, and the motor-powered pulley. The cab and the counterweight is connected by a rope which is attached on the pulley. Consider the cab and the counterweight have equal mass = 1000 kg.
    The elevator is intended to move at 1 m/s. This is a balanced mass on 2 sides of the pulley. Calculate the motor power required.

    mCa = mCo = 1000 kg
    v = 1 m/s

    P (kW) = ?


    2. Relevant equations

    F dt = mv
    P = F * x/t



    3. The attempt at a solution

    The elevator is initially at rest, so initial momentum is simply zero. No information about the acceleration, so let's say the cruising speed of 1 m/s is reached in 1 s.
    F = (mCa + mCo) * v / t
    F = 2000 * 1 / t
    say t = 1
    F = 2000

    The distance required to accelerate to cruising speed is
    x = 0.5 * a * t^2
    since we have stated that 1 m/s reached in 1 sec, then a = 1 m/s^2. Thus,
    x = 0.5 * 1 * 1^2 = 0.5

    The power calculation:
    P = F * x / t
    P = 2000 * 0.5/1
    P = 1000 W = 1 kW

    Conclusion:

    1. When accelerating to it's cruising speed (1 m/s), the system requires 1 kW from the motor regardless the friction.

    2. when it's already in its cruising speed, the power required is ZERO, regardless the friction. Means, the power is only to overcome the friction.


    Am I right, folks? did I do correctly in my analysis of using momentum? This is equally (balanced) masses at both sides, so gravitational forces cancel each other.
    Thanks for your comments..
     
  2. jcsd
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