a 2 tons elevator is supported by a cable that can safely support 6400lbs.?

Question

what is the shortest distance in which the elevator can be brought to its top with its decending with a speed of 4ft/sec square?

Answer 1

now your making me paranoid

Answer 2

This is actually quite straightforward.

The cable will snap, presumably, at 3.2 tons force on it; so the additional force f = 6400 lbs - W = 2400 lbs before the cable goes belly up. Good old f = Ma = 2400 tells us the deceleration of the elevator cannot exceed a = 2400/M without breaking the cable. And since the elevator's weight = W = Mg, we can find M = W/g; so that a = 2400g/W = 2400*g/4000

I am guessing you meant u = 4 ft/sec and not 4 ft/sec^2 because the units you gave were for acceleration, not "speed." Thus, v^2 = 0 = u^2 - 2aS; where u = 4 fps and a = (24/40)g = .6 g and g = 32.2 ft/sec^2 at Earth's surface. Setting S = u^2/2a = u^2/(.6g)2 = 4^2/1.2g you can do the math to solve for S the shortest distance one can stop the elevator (giving v = 0 final velocity) without snapping its cable.

Answer 3

F=ma. 1 gee = 9.81 m/sec^2 causing its weight. How much more acceleration can be loaded to have its weight on the cable be 6400 lbs less an epsilon? Then s = (at^2)/2, v = at

It cannot descend at 4 ft/sec^2. L/T^2 is not velocity.

Answer 4

Fnet = m * a

1 metric ton = 2,204.6 lbs

Fnet = (2204.6)*(4 ft/sec^2) = I'm lost now too