Young's Modulus and Interatomic Bond Length PLEASE HELP

AI Thread Summary
The discussion focuses on calculating Young's modulus for a titanium alloy wire under a 97 kg load, which stretches the wire by 1.49 cm. Participants clarify the necessary variables: original length (2.9 m), change in length (0.0149 m), force (due to gravity), and cross-sectional area derived from the wire's diameter (0.14 cm). The correct formula for Young's modulus is established as stress divided by strain, but there is confusion regarding unit conversions and calculations. Ultimately, the calculated Young's modulus values vary significantly, indicating potential errors in the participants' computations. Accurate calculations and consistent unit usage are emphasized for determining the correct Young's modulus.
luckyg14
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A hanging wire made of an alloy of titanium with diameter 0.14 cm is initially 2.9 m long. When a 97 kg mass is hung from it, the wire stretches an amount 1.49 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm3.

Based on these experimental measurements, what is Young's modulus for this alloy of titanium?
Y = CANNOT FIGURE OUT

As you've done before, from the mass of one mole and the density you can find the length of the interatomic bond (diameter of one atom). This is 2.60 10-10 m for titanium. As shown in the textbook, the micro quantity ks,i (the stiffness of one interatomic bond) can be related to the macro property Y:
ks,i = CANNOT FIGURE OUT
 
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Pls quote the equations you are aware of that relate to the definition of Young's modulus and (from your textbook) the relationship between interatomic bond stiffness and the modulus.
 
young's modulus is stress/ strain which is (F*Lo)/(Ao*delta_L) I do not know how to get the numbers and I don't understand how they relate to interatomic stiffness
 
okay so i know that Lo= 2.9m and delta_L= 0.0149m but I'm not sure how to get F and Ao
 
luckyg14 said:
okay so i know that Lo= 2.9m and delta_L= 0.0149m but I'm not sure how to get F and Ao
You're given the wire diameter, so what do you think the cross-sectional area will be?
What force will be exerted by a suspended 97kg mass?
 
Ao= 0.9079

I know force= m*a and I know that m is 97 but i don't know a
 
luckyg14 said:
Ao= 0.9079
What units? But I don't think that's right whatever units you chose. Pls show how you calculated it.
I know force= m*a and I know that m is 97 but i don't know a
What does it mean to say something is hanging? What is the nature of the force that makes the wire stretch?
 
To find Ao i converted cm to m and then divided it by 2 to get the radius then used the area of a circle formula

and the force would be gravity
 
  • #10
luckyg14 said:
To find Ao i converted cm to m and then divided it by 2 to get the radius then used the area of a circle formula
Correct procedure, but the answer was way off. Please post all that working.
and the force would be gravity
Yes, so what force will the 97kg mass exert?
 
  • #11
So i redid it and got Ao= .01539

and the force should be -950.6
 
  • #12
it didn't work with those numbers idk what to do anymore but thank you
 
  • #13
luckyg14 said:
So i redid it and got Ao= .01539
That's in sq cm, right?
and the force should be -950.6
Newtons. Always state the units.
So, plug those numbers in, be careful with the units, and post your working.
 
  • #14
no its in meters because its supposed to be in meters.

i got 12021839.34 N/m and it was wrong
 
  • #15
N/m^2
 
  • #16
luckyg14 said:
no its in meters because its supposed to be in meters.
Then you calculated it wrongly. The diameter is 0.14cm, not 0.14m.
 
  • #17
i converted it to meters because everything we do is always in meters
 
  • #18
ok i redid my calculations again and got -264308724.8 N/m^2
 
  • #19
would -264308724.8 be correct or should it be positive
 
  • #20
luckyg14 said:
would -264308724.8 be correct or should it be positive
It should be positive. If you are consistent, ΔL has the same sign as the force, which you've taken to be negative.
But I don't think that number is right yet. Radius = 0.07 cm, so area ≈ 0.015 sq cm = 1.5E-6 sq m. L = 2.9 m.; F = 97g ≈ 950N; ΔL ≈ 1.5 cm = 1.5E-2m.
950*2.9/(1.5E-6 * 1.5E-2) ≈ 1.2E11. That's roughly 500 times your number.
 
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