Question

A 2.25×10⁴ N elevator is to be accelerated upward by connecting it to a counterweight using a light (but strong!) cable passing over a solid uniform disk-shaped pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875 kg and it is 1.50 m in diameter.

How heavy should the counterweight be so that it will accelerate the elevator upward through 6.00 m in the first 4.00 s , starting from rest?

Under these conditions, what is the tension in the cable on elevator side of the pulley?

Under these conditions, what is the tension in the cable on counterweight side of the pulley?

I posted this before but 3000 wasnt the correct answer

Answer 1

The mass of the elevator is 2.25x10^4/g. The acceleration required is derived from the eq for distance covered as a function of a

d=1/2 a t^2.

d = 6 m, 4=3.

6 = 0.5*a*(4)^

a = 0.75 m/s^2

The force required is mass of the elevator * accel plus the force of gravity, 2.25x10^4. that is the answer to B.

The pulley is a different. It's circumfrence is pi*d, pi*6 or 18.8 The acceleration of the elevator times 4 secs gives you the speed at 4 secs. That divided by the circumfrence gives the angular speed at 4 secs. Divide by 3 gives you the angular acceleration. Remember to convert to radians.
The angular accel is going to be opposed by the inertia. Inertia for a thin disc is mr^2/4. m=875, r=.075m
Force to rotate the pulley is derived from angular accel = torque/ inertia, and from F=t/r.
Force required to accelerate the pulley.

The total force required divided by g gives you the necssary mass of the counterweight.