Physics/Engineering question

[Harding91]

Hi all, firstly if this is posted in the wrong section feel free to move it but i couldnt really see anywhere suitable to post it so went with general?

I'm currently studying part time for a degree and im working on one of my assignments and i've confused myself rather a lot.

Hey there, currently working my way through my TMA02 for T174 and ive possibly stumped myself.

Anyone who's good with physics in general able to help? I've put the question below with different figures but the same exact wording and my thoughts/answer so far! Any help appreciated, id prefer it if im wrong just point me in the right direction id rather not have somebody answer it for me as im looking to learn it myself.

Thanks in advance!



Question 5.

g) when the lift is close to the bottom of the lift shaft, the cable has a length of 90m. Show that the mass of the lift fully loaded as decribed in (d) above stretches the cable by about 15mm (4 marks)



Deleted this working out crap because it's wrong now, what i think is the correct answer below!

Edited by Harding91 on Saturday 26th March 11:32

[Harding91]

nevermind im a fool, worked it out now

[2gins]

Do tell then. One hates an unresolved thread.

[Harding91]

Uploaded it as an image as getting it to format correctly on here was a nightmare.

But here's how i solved it i think, or at least it makes sense to me. Will find out when its handed in and marked i guess.

[2gins]

Thanks

I'd be skeptical about using the given value of 15mm to work back to it, all the expressions you're using are linear so you can only ever end up back at 15 mm. Is there a way you can find the extension with the given information without using 15mm as an input?

Also, A = 0.45m, that's a meaty lift cable (0.4m radius or thereabouts). How big is this lift?

On these questions I tend to assume

1 mark for identifying the route to the solution
1 mark for each correct law/expression
1 mark for the answer

Above 4/5 marks I also assume a mark or two for setting out the structure neatly. Gives me an idea of what may be expected.

[V8LM]

You can't use the 15 mm to calculate E and then use that E to calculate the 15 mm extension: you could show the extension was anything. There must be information you've been given previously - E, some value of the extension when the lift is further up the shaft, extension under a different load.

The unit of force is the newton (N), not newton per m.

Edited by V8LM on Saturday 26th March 07:51

[Harding91]

V8LM said:

You can't use the 15 mm to calculate E and then use that E to calculate the 15 mm extension: you could show the extension was anything. There must be information you've been given previously - E, some value of the extension when the lift is further up the shaft, extension under a different load.

The unit of force is the newton (N), not newton per m.

Edited by V8LM on Saturday 26th March 07:51

Yeah using 15mm as the input before was the mistake i made, the second post i put where i got it isnt using that it's using youngs modulus to work it out.

And i know force is newtons as i worked the force out in an earlier question on the assignment but i converted it to newton per mm to input it into the equation to achieve the answers in mm. I should probably make this clear in my workings?

[Harding91]

Also forgot to mention when posting the second post with my solution i worked it out completely differently.

In a previous question they gave me a youngs modulus of the cable to answer that said question, in this question im tackiling now i assumed it would be the same which was a mistake so i had to re-work out the youngs modulus for this question ignoring what i was previously told. Which leads me to 15mm the figure they give me in the question so i'd guess it was right? They give me the 15mm figure in the question and ask me to show how it is 15mm.

I think it's almost impossible i use all the numbers ive gained previously and end up with 15mm if the working is incorrect? Especially considering how long some of the numbers are i doubt it's coincidence that i end up with the figure they've asked me to end up with.

After a nights sleep i re-worked it and looked at it again and im fairly sure my second solution there is correct.

Edit : i also made the mistake of trying to use hookes law initially instead of youngs modulus which was obviously the right way to do it, silly mistake really.

[Harding91]

2gins said:

Thanks

I'd be skeptical about using the given value of 15mm to work back to it, all the expressions you're using are linear so you can only ever end up back at 15 mm. Is there a way you can find the extension with the given information without using 15mm as an input?

Also, A = 0.45m, that's a meaty lift cable (0.4m radius or thereabouts). How big is this lift?

On these questions I tend to assume

1 mark for identifying the route to the solution
1 mark for each correct law/expression
1 mark for the answer

Above 4/5 marks I also assume a mark or two for setting out the structure neatly. Gives me an idea of what may be expected.

Thanks for the feedback. I think what i did above was working out the extension or 'stretch' without using the 15mm as the input

And the areas correct as per the figures they gave me of the cable having a circular cross-sectional with a diameter of 24mm so i half that to give me the radius and then it's just pi multiplied by the radius squared giving my area in metres squared which i then just convert to mm squared. It could just be a theoretical area given for the question hence why it seems bigger than it maybe should be, im not sure i just work off the figures im given.

And it'll be laid out neater on my final hand in this was mad scientist scrawlings on not much sleep plus i will probably hand write it instead of fighting with the pc to format it again.

[tapkaJohnD]

Harding,

Have you seen this: https://www.youtube.com/watch?v=SRGaW3maK38

"Writing Math[s - it's American] Equations in Word"

John

[Harding91]

tapkaJohnD said:

Harding,

Have you seen this: https://www.youtube.com/watch?v=SRGaW3maK38

"Writing Math[s - it's American] Equations in Word"

John

I use libreoffice currently and it can also be done in that and i know how im just not quick nor efficient at it and i honestly prefer writing out maths work by hand anyway

But thank you for the link, much appreciated.

[V8LM]

You're asked to show the increase in length is 15 mm. So all you need to show in your answer is the second half - state the equation for the modulus, rearrange to express as deltaL, plug in the numbers.

"Force per mm" makes no sense. Force per length has units of kg s-2, not newtons.

Just stick to the same units and scale (N, m).

Also learn to state only significant figures, not every decimal place your calculator gives.

And as has been asked before, how thick is the cable? 0.45 m^2 cross sectional area is not a cable.

Edited by V8LM on Saturday 26th March 15:46

[Harding91]

V8LM said:

You're asked to show the increase in length is 15 mm. So all you need to show in your answer is the second half - state the equation for the modulus, rearrange to express as deltaL, plug in the numbers.

"Force per mm" makes no sense. Force per length has units of kg s-2, not newtons.

Just stick to the same units and scale (N, m).

Also learn to state only significant figures, not every decimal place your calculator gives.

See what you're saying, as for significant figures we're told to always put the entire number unless specified otherwise. Just following the instructions on that part.

[motco]

V8LM said:

The unit of force is the newton (N), not newton per m.

Edited by V8LM on Saturday 26th March 07:51

Quite so - I was reluctant to comment but it did seem odd.

[V8LM]

I added to my post - how thick is the cable? Did you calculate this, or was it given. Similarly, did you calculate E, or was that given?

An elasticity value of 0.1 GPa is that of rubber. Does thie lift use a 70 cm thick bundle of elastic bands?

Edited by V8LM on Saturday 26th March 15:56

[motco]

The effective 'E' of a steel cable is going to be less than the 30,000,000 psi (200,000 N/mm2) of typical steel. The cords will compact and the cable will seem quite elastic.

[V8LM]

Constructional elasticity, sure.

But the OP's 'cable' is 76 cm thick.

[Goldenballs13]

What information was given in the question?

Also, it doesnt make any difference to the answer, but I suggest always using SI units (m, not mm, N, not kN etc..) purely as its a bit neater, and it avoids any errors when the problems get a bit more complicated. I should know, I'm in my 3rd year of an engineering degree!

[motco]

Goldenballs13 said:

What information was given in the question?

Also, it doesnt make any difference to the answer, but I suggest always using SI units (m, not mm, N, not kN etc..) purely as its a bit neater, and it avoids any errors when the problems get a bit more complicated. I should know, I'm in my 3rd year of an engineering degree!

Sorry, but lb. and sq.in. make sense in the real world and silly mistakes are glaringly obvious - unlike with Newtons for which nobody has a gut feeling. Anyway, my copy of Roark's 'Formulas for Stress and Strain' is an old Imperial one!

[Goldenballs13]

motco said:

Sorry, but lb. and sq.in. make sense in the real world and silly mistakes are glaringly obvious - unlike with Newtons for which nobody has a gut feeling. Anyway, my copy of Roark's 'Formulas for Stress and Strain' is an old Imperial one!

Can use imperial or metric as you wish, but I was talking more about converting to the base unit, like m instead of mm. Same would apply for the silly old-fashioned imperial! However I do have some great machinists books from my Grandad, so old they talk about units such as 'butloads'!