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Torque and Angular Momentum

  1. Aug 28, 2007 #1
    A 22,500N elevator is to be accelerated upward by connecting to it a counterweight using a light cable passing over a solid uniform disk-shaped pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875kg and it is 1.50 m in diameter. (a) How heavy should the counterweight be so that it will accelerate the elevator upward through 6.75 m in the first 3.00 s, starting from rest? (b) Under these conditions, what is the tension in the cable on each side of the pulley?

    2. Relevant equations
    I=1/2*875*.75[tex]^{2}[/tex]=246.1 kg*m[tex]^{2}[/tex]
    a=(2*6.75)/(3[tex]^{2}[/tex])=1.5m/s[tex]^{2}[/tex] from x=(x[tex]_{i}[/tex])+(.5)(a)(t[tex]^{2}[/tex])
    [tex]alpha[/tex] = a/r = 2 rad/s^2 from a = R * [tex]alpha[/tex]

    3. The attempt at a solution

    So I get from this is that for the elevator to rise the 6.75 feet in 3.00 seconds the total torque must be 492.2 N*m.


    From this I get T[tex]_{1}[/tex] = 19056 N

    [tex]\Sigma[/tex][tex]\tau[/tex] = T[tex]_{1}[/tex]*r-T[tex]_{2}[/tex]*r


    T[tex]_{2}[/tex] = 18400 N

    So should be...
    m[tex]_{counter}[/tex] = 1628 kg * 9.8 = 1.6*10^4 N

    Unfortunately, according to the book I am wrong on all three counts, the tensions and the mass. The answers are supposed to be 3.16*10^4 N mass counter, 2.60 * 10^4N Tension 1 and 2.67 * 10^4N for Tension 2.

    Any help would be greatly appreciated. Thanks in advance.
  2. jcsd
  3. Aug 28, 2007 #2


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    Homework Helper

    The directions for your forces are wrong I think... the elevator is accelerating upward, so what's the sum of the forces in the vertical direction for the elevator?

    Remember that a is upward for the elevator.

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