A 2.25×10 4 N elevator is to be accelerated upward by connecting it to a counter

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?

Explanation / Answer

1. vf = at + vi
12 = a(4) + 0
a = 3 m/s/s
m = 2296 kg
so the force needed is 12.8 X 2296 = 29388
and the mass is 2999 kg call it 3000 kg

2. 29400 N down

3. Yup - you are correct - but in same direction. Gravity is always down

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