Young's Modulus and Interatomic Bond Length PLEASE HELP

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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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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
 
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?
 
So i redid it and got Ao= .01539

and the force should be -950.6
 
it didn't work with those numbers idk what to do anymore but thank you
 
no its in meters because its supposed to be in meters.

i got 12021839.34 N/m and it was wrong
 
i converted it to meters because everything we do is always in meters
 
ok i redid my calculations again and got -264308724.8 N/m^2
 
would -264308724.8 be correct or should it be positive
 
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.