Elevator Torque and Angular Momentum

[usnberry]

Homework Statement

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?

Homework Equations

I=1/2*875*.75^{2}=246.1 kg*m^{2}
a=(2*6.75)/(3^{2})=1.5m/s^{2} from x=(x_{i})+(.5)(a)(t^{2})
alpha = a/r = 2 rad/s^2 from a = R * alpha
\Sigma\tau=I*a

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.

T_{1}=m_{elev}*g-m_{elev}*a
T_{2}=m_{counter}*g+m_{counter}*a

From this I get T_{1} = 19056 N

\Sigma\tau = T_{1}*r-T_{2}*r

So...

T_{2} = 18400 N

So should be...
m_{counter} = 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.

 

[learningphysics]

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.