Young's modulus Help!!!!?

Question

You hang a heavy ball with a mass of 35 kg from a gold rod 3 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod, and find that the rod stretched 0.002653 m.
Using these experimental data, what value of Young's modulus do you get?
Y = ? N/m2.
The atomic mass of gold is 197 g/mole, and the density of gold is 19.3 g/cm3.
Using this information along with the measured value of Young's modulus, calculate the speed of sound in gold.
v = ? m/s.

Answer 1

To calculate the Young's modulus, you must first calculate the stress on the rod. Divide the force generated by the 35 kg mass by the cross-sectional area of the rod:

σ = F/A = 35 kg*9.81m/s^2 / (1.9E-3 * 2.6E-3 m^2) = 69.5 MPa.

Linear strain is change in length over original length:

ε = ΔL/L0 = 0.002653/3 = 8.843E-4

Young's modulus E = σ/ε = 6.95E7/8.843E-4 = 78.6 GPa.

The speed of sound in gold is sqrt(E/ρ) = sqrt(7.86E10/19,300) = 2020 m/s.

Notice that you first have to convert the density to SI units.

You can check your answers against the values posted on the Wiki page below.

Answer 2

This is likely a very simple problem and when treated as a 1-d stress problem, has nothing to do with the poissons ratio of gold. Use statics to determine the stress in the bar

stress = Young's Mod * strain

stress = F/A = m*g/A...

E (or Y) = stress/strain from Hooke's Law

E = m*g/A/strain

then v = sqrt(E/rho)

they give you the density, so I don't think you need the atomic mass. You may also need take the self weight of the bar into account though...which takes a little more work to derive the relation between the applied load and the measured stretch...

Answer 3

LD is spot on. another piece of information - in case you do a YM test attempt utilizing a organic copper cord. in case you get "creep" IE the metallic seems to strech under its own weight very slowly think ofyou've have been given a sturdy set-up. additionally for extra accuracy you're able to make a dial. you'll want some pins and countless extremely stiff cardboard. positioned one pin in a weighted corron reel to apply as a pivot. superglue yet another pin a million meter out of your tie factor on you cord under attempt. Make an prolonged skinny piece of cardboard approximately 5mm via 230mm and precisely 200mm from one end positioned the pin through and into the cotton reel )then positioned a small mark on the top of the long end). this is your pivot factor. flow this into alignment so as that the 30mm end is in touch with the pin on your cord 20mm faraway from the pivot pin. the different long end ought to be resting on a ruler and ideally your piece of cardboard should be at a genuine perspective to the cord being measured. That set up provides the equivalent of utilizing a chew of cord ten circumstances longer than you surely use and therefor if used properly improves your accuracy via one decimal place. i grow to be assuming you basically had room for a a million meter set-up yet once you will get a 10 meter set-up do it and use the pivot rig. That way you're turning out to be the equivalent of one hundred meters.

Answer 4

Atomic weight is irrelevant.
Poisson's ratio would be more useful.

Without Poisson's ratio you can calculate
the speed of sound in the rod, but not in
bulk gold.

Answer 5

Young's Modulus Of Gold