Young modulus problem.

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Homework Statement



The cables of length 70m supporting a lift consist of two steel ropes each of 100 strangs giving a total cross sectional area of [tex]1.0 x 10^-4 m^2.[/tex] Consider a full lift to carry 8 passengers of average mass 75kg. Calculate by how much an empty lift moves down when it is entered by 8 passengers.



Homework Equations



Young modulus of steel in this example is taken to be 2.06 x 1011.

[tex]Young modulus = \frac{Stress}{Strain} = \frac{(Force)(Length)}{(Area)(Extension)}[/tex]


Therefore,


[tex]
Extension = \frac{(Force)(Length)}{(Area)(Young)}[/tex]

Load = 8 people, mass 75kg
Force = 8(75)(9.81)(0.5) (?)

Young mod = 2.06x1011

Extension = unknown

Length = 70m

Area = 0.01m (?)

The Attempt at a Solution



Slightly confused as to what the cross sectional area of this rope would be, I assume that it can't possibly be 0.1mm so I think that it must be 100 times that which is 0.01m or 1cm which sounds OK.

For the attempt, take x = extension, f=force, y = young mod., l=natural length, a = cross-sectional area.

[tex]x = \frac{fl}{ay}[/tex]


[tex]x = \frac{(75x9.81x0.5x8)(70)}{2.06x10^11(0.01)}[/tex]


[tex]x = 1.00 x 10^{-4} m[/tex]

This is wrong, when I use the force as 8(75g) I get 2.00 x 10-4 m and when I use the area as 1.0 x 10-4m with that, I get the right answer of 0.2m or 20mm. But I thought that the load for 2 cables would be spead evenly and also that the cables could not possibly be 0.0001m2?
 

Answers and Replies

  • #2
PhanthomJay
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the area of both ropes is already given; you don't need to calculate it. The weight is shared equally by each cable, so either you divide the total load by 2, and use 1/2 the total area, to solve for the extension, or you use the total load and total area, to get the same value. The ropes are parallel to each other.
 
  • #3
tiny-tim
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The cables of length 70m supporting a lift consist of two steel ropes each of 100 strangs giving a total cross sectional area of [tex]1.0 x 10^-4 m^2.[/tex]

Young modulus of steel in this example is taken to be 2.06 x 1011.

This is wrong, when I use the force as 8(75g) I get 2.00 x 10-4 m and when I use the area as 1.0 x 10-4m with that, I get the right answer of 0.2m or 20mm. But I thought that the load for 2 cables would be spead evenly and also that the cables could not possibly be 0.0001m2?
Hi Gregg! :smile:

I suspect that "total cross sectional area" means for both cables combined … so you can leave out the 0.5 :wink:

And 2.06 x 1011 in what units? :confused:
 
  • #4
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Hi Gregg! :smile:

I suspect that "total cross sectional area" means for both cables combined … so you can leave out the 0.5 :wink:

And 2.06 x 1011 in what units? :confused:
Ah I see. The units for the young modulus is Pascals.

So I can use the total load of 75x8x9.81 for Force, and cross sectional area of 10-4 is crazy, though.

[tex] x = \frac{fl}{ay} [/tex]

[tex] x = \frac{412020}{(2E11)(10^{-4})} [/tex]

[tex] x = 0.020601 m [/tex]

The book says 20mm so it's close to correct. I can understand from the replies that there was no need to divide the load into 2 therefore I could use the total cross sectional area. What was strange though was the way it said that the cables consisted of 100 wires giving a total area of 0.0001m2. This is crazy, a wire could not be that thin and hold more than 3kg let alone 600kg.
 
  • #5
tiny-tim
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What was strange though was the way it said that the cables consisted of 100 wires giving a total area of 0.0001m2. This is crazy, a wire could not be that thin and hold more than 3kg let alone 600kg.
hmm … I'm no engineer …

but that's a centimetre squared … isn't that enough? :redface:
 
  • #6
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hmm … I'm no engineer …

but that's a centimetre squared … isn't that enough? :redface:
yes 100 x 0.0001 is 1cm squared but for the calculation i used 0.0001 because 0.01 seemed not to work. Ill try it once more

[tex] \frac{ (75)(8)(9.81) . (70)}{(2x10^{11}).(0.01)} [/tex]

[tex] \frac{ 412020 }{2000000000} [/tex]

[tex] 2.0601x10^{-4} m [/tex]

[tex] 0.00020601m [/tex]

The answer is 0.020 m so im a factor of 100 out, dividing by something 100x smaller seems the only way to fix this. But 0.1mm as an area? thats a 0.178mm radius. Can't be right? I'm confused, i'll upload a picture of the problem.
 
Last edited:
  • #7
Delphi51
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1 meter squared is like 100 cm by 100 cm or 10^4 cm squared.
Or 1 x 10^-4 meters squared equals 1 cm squared. I think you are only seeing one of the two factors of 100 involved. Working too hard - just use 1 x 10^-4 without any thought.
 
  • #8
tiny-tim
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The answer is 0.20 m so im a factor of 100 out …
hmm … I get 20 mm
… the right answer of 0.2m or 20mm.
Which? They're not the same. :confused:
 
  • #10
tiny-tim
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Edit: 10E-4 works, i'm not sure why though.
Because the question gives you 10-4

what are you using? :confused:
 
  • #11
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Because the question gives you 10-4

what are you using? :confused:
It says 100 strands of 10E-4. I though that was 10E-2. If not, it seems too small a quantity for a cross sectional area of 2 cables which are supporting 8 people. We did a young modulus experiment at college and a piece of copper wire with a cross sectional area of more than that broke after 2kg. Also, if the its constructed of 200 strands of total 10E-4 how small are these 'strands' smaller than a hair? :confused: It's clear though, that I do need to use 10E-4. But I'm not sure why.
 
  • #12
tiny-tim
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It says 100 strands of 10E-4. I though that was 10E-2.
No … same reason as for leaving out the 0.5 …

the "total cross sectional area" in the question means you ignore the "2 ropes" and you ignore the "100 strands" …

sometimes they put extra information in the questions so that they can check that you know what to leave out! :wink:
 
  • #13
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No … same reason as for leaving out the 0.5 …

the "total cross sectional area" in the question means you ignore the "2 ropes" and you ignore the "100 strands" …

sometimes they put extra information in the questions so that they can check that you know what to leave out! :wink:
Thanks, seems obvious now. Quite thin ropes though aren't they.
 
  • #14
PhanthomJay
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Thanks, seems obvious now. Quite thin ropes though aren't they.
As an aside, steel comes in various breaking strengths; using an average grade steel, each rope can handle over 1000 kg before breaking. That's probably strong enough, considering each rope supports 75*4 = 300 kg, plus half the weight of the empty lift.
 
  • #15
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Jay has it. Although his average grade steel is is low balling the case. For any lift carrying people the steel will be of a higher tensile strength. Jay's example gives a safety factor of 3.3. For lifts carrying people the safety factor will be about 10.
 
  • #16
Delphi51
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A STEEL rope with 1 cm squared area should be REALLY strong. I've seen a D6 cat pulled out of the mud with one of those!
 
  • #17
PhanthomJay
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A STEEL rope with 1 cm squared area should be REALLY strong. I've seen a D6 cat pulled out of the mud with one of those!
Yes, it sure would. The problem statement used a rope with an area of .00005 m^2, (0.5cm^2). Even that rope, as oldmancan points out, using a high strength steel, would take about 3000kg before breaking. In the US, that converts to over 6000 pounds breaking strength. And with a diameter of just over 1/4".
 

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