# Young modulus problem.

1. Jan 10, 2009

### Gregg

1. The problem statement, all variables and given/known data

The cables of length 70m supporting a lift consist of two steel ropes each of 100 strangs giving a total cross sectional area of $$1.0 x 10^-4 m^2.$$ 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.

2. Relevant equations

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

$$Young modulus = \frac{Stress}{Strain} = \frac{(Force)(Length)}{(Area)(Extension)}$$

Therefore,

$$Extension = \frac{(Force)(Length)}{(Area)(Young)}$$

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

Young mod = 2.06x1011

Extension = unknown

Length = 70m

Area = 0.01m (?)
3. 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.

$$x = \frac{fl}{ay}$$

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

$$x = 1.00 x 10^{-4} m$$

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?

2. Jan 10, 2009

### PhanthomJay

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. Jan 10, 2009

### tiny-tim

Hi Gregg!

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

And 2.06 x 1011 in what units?

4. Jan 10, 2009

### Gregg

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.

$$x = \frac{fl}{ay}$$

$$x = \frac{412020}{(2E11)(10^{-4})}$$

$$x = 0.020601 m$$

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. Jan 10, 2009

### tiny-tim

hmm … I'm no engineer …

but that's a centimetre squared … isn't that enough?

6. Jan 10, 2009

### Gregg

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

$$\frac{ (75)(8)(9.81) . (70)}{(2x10^{11}).(0.01)}$$

$$\frac{ 412020 }{2000000000}$$

$$2.0601x10^{-4} m$$

$$0.00020601m$$

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: Jan 10, 2009
7. Jan 10, 2009

### Delphi51

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. Jan 10, 2009

### tiny-tim

hmm … I get 20 mm
Which? They're not the same.

9. Jan 10, 2009

### Gregg

10. Jan 10, 2009

### tiny-tim

Because the question gives you 10-4

what are you using?

11. Jan 10, 2009

### Gregg

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? It's clear though, that I do need to use 10E-4. But I'm not sure why.

12. Jan 10, 2009

### tiny-tim

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!

13. Jan 10, 2009

### Gregg

Thanks, seems obvious now. Quite thin ropes though aren't they.

14. Jan 10, 2009

### PhanthomJay

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. Jan 10, 2009

### oldmancan

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. Jan 10, 2009

### Delphi51

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. Jan 10, 2009

### PhanthomJay

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".