Calculating Thickness of Steel Cable for Passnger Lift

"The total length of the cable supporting the lift when full of passengers is 800 kg".

You've got me there. I haven't met those units of length.

me to

JohnDG....

The "designer" should do his own friggin' homework and not post it on the internet for others to complete.

This "brain dead" original poster lacks even the normal guile and subtlety to disguise his homework. !!!

Do your own damn engineering home work !!! You may learn something in the process !!

and he is the "designer"?

ok brain dead mybe male ? now whos calling who, Question how do I work it out ? I've just about given up tell me how to work it out then please.

It cannot be worked out. It is an awful question.

800 kg is not a length, and the length of the cable is unlikely to be an issue in simplified instructional analyses like this, because the length of cable being supported changes with lift position.

You would need to at very least, know how the lift operates, specifically what the peak acceleration is when the lift begins to lift the load. (In real engineering, you'd want to consider peaks from oscillation too... and you'd need to know loads of other things re the cable rigging, etc.)

You could assume that the passenger load + the cabin mass + the mass of cable being supported (between an upper pulley? and the cabin) when the cabin is at ground level is 800 kg. But your assumption(s) could be wrong.

The question was perhaps intended to ask what size cable would support a given weight, if you are given a cable material strength rating. But it does not ask that.

You can't answer the question because the 800kg length is non-sensical, and you are not given any of the dynamics.

Best to talk to the teacher to find out what he or she had in mind. Many of us can and do work this kind of thing out in our sleep, (and have for decades) but none of us can give you an answer to a bad question.

Another possible interpretation: Perhaps the lift is stationary, and simply hanging there, and the load on the cable is intended to be some mass (evidently, 800 kg) under the acceleration due to gravity: for example, 8000 N. Then it is a very simple problem to solve. But that's not the question as asked. I hope you don't have to put up with this kind of thing often - it is demotivating to good students.

Somehow I wish I knew his Professor's email addie to let him know what a lazy slug his student is by asking us to do his homework problems.

There is no substitute for properly studying...to circumvent the process will eventually come around and eventually bite one in the arse during one's career.

(aside note: I'd hate to be a passenger in that "lift" designed by the OP!!!! DEAD MEAT comes to mind very quickly!)

1050MN/m²

Thanks. I spaced out and didn't get back in time to edit the question.

No need to apologize "errare humanum est"

since the question appears to have come from a student, let us rather teach them than excoriate them.

Like Fezzik, when we are stumped we should go back to the beginning.

summation of forces equals change in momentum (F dt = d(mv) )

conservation of energy, mass and momentum

and various constituitive relationships. (pv = nrt, f=kx, etc)

keeping this in mind, here is how to solve the problem

1) draw a diagram of the gondola, with cables attached

2) define the direction and locations of all forces(weight, wind, nuclear strong force etc) acting on the gondola

3) write equations for those forces, broken down into components, that satisfy the kinematics constraints of the problem.

Once you know the force then you look at the load carrying capability of the cable. Its is given in terms of a stress, which has units of pressure - force/area. So you divide the load by the alowable stress to get the required area.

That should get you there, I don't think i have compromised anything - just showing someone how to fish.

Sadly, this totally artificial scenario doesn't give any information useful for a kinetic analysis - unless we are to assume that the lift could be in free-fall for (most of the) 14m, and then still apply a safety factor of x10.

Hell, may as well use a ram, and push it up from below.