Question

A 6.0kg mass is hung by a vertical steel wire 0.400 m long and 6.00×10−3cm^2 in cross-sectional area. Hanging from the bottom of this mass is a similar steel wire, from which in turn hangs a 9.00kg mass. find the lower and upper tensile strain, and elongations?

Answer 1

This may help: http://en.wikipedia.org/wiki/Young%27s_modulus

Answer 2

for the lowe wire this is what I've got so far, and still gives me as a wrong answer,...Y=STRESS/STRAIN, which is = to strain=stress/Y, the stress for the lower wire is the weight=(9)(9.8)=88.2,

Answer 3

from there, I applied the formula 88.2/6.0*10^-7/y(for steel wire is 2.00*10^11=7.35*10^12, but it says is wrong :(

Answer 4

Be careful, the stress \(\sigma\) is given by \(\sigma = \dfrac{F}{A}\), which is force per cross-sectional area.

Answer 5

ohhhh, then using this formula Flo/Ao\[DeltaL\] I used this numbers; (88.2)(.400)/6x10^-7(.400), is this correct?

Answer 6

I think the deltaL is .400 because in the problem says is similar wire???

Answer 7

Yeah, seems reasonable.

Answer 8

Nope, the answer I got is wrong :(
Then I really dont know how to work this problem at all, what gets is that it doesnt say the wire elongated??? Please help just one more shot to get the answer right :(

Answer 9

I learned Young's Modulus as \(E\), so I'll write it that way, with stress \(\sigma\) and strain \(\varepsilon\), so that's the notation I'll use.\[E = \frac{\sigma}{\varepsilon}=\frac{FL_0}{A\Delta L} \longrightarrow \Delta L = \frac{FL_0}{AE}\]I think you plugged in your numbers incorrectly... you seemed to have done \(E = .400\), when you should have \(E=200\) for steel.

Answer 10

but im not looking foor deltaL, Im looking for the strain in lower wire, or do I need to get that first in order to get the stress?

Answer 11

Well \(\Delta L\) is the elongation. If you want the strain, it's similar:\[E = \frac{\sigma}{\varepsilon}=\frac{F}{A\varepsilon } \longrightarrow \varepsilon = \frac{F}{AE} \] Just be sure you use the correct numbers.

Answer 12

ok let me see, crossing my fingers >.< :)

Answer 13

I plugged this numbers and 88.2/6.*10^-7 (2.0*10^11)=7.35*10^4

Answer 14

Why \(E = 2.0\times 10^{11}\)? My sources say \(E=200\) for steel. http://www.engineeringtoolbox.com/young-modulus-d_417.html

Answer 15

in our book says 2.00*10^11?

Answer 16

What are the units your book gives?

Answer 17

pascals...well yeah in your table it says 200*10^9= 2.00*10^11

Answer 18

Oh oops...I didn't realize that website had it listed in Gigapascals...yes that looks correct.

Answer 19

you know what I thought, that the F for the lower wire, shouldnt it be the tension of the upper=mg, plus its own tension, meaning (6)(9.8)+(9)(9.8)
Oh I got it wrong btw, but I want to understand the physics cause it might be on my final

Answer 20

I think you mean (6)(9.8)+(6)(9.8), yes? Just double the upper tension? (That reasoning is correct, by the way.)

Answer 21

whoops sorry...REVERSED...the upper wire is double the bottom wire's tension.

Answer 22

>.> so my reasoning is making sense??? uhhh lol

Answer 23

I'd talk with your teacher about it.

Answer 24

well isnt the tension of the first one 6*98, and the lower T1+9*9.8?

Answer 25

Im more of a visual person so I draw this...

Answer 26

Shoot sorry...didn't realize the masses were different. But no, the upper tension should be the greater. The tension is determined by the blocks hanging BELOW the wire.