Question

Last question on elasticity

A 103 kg uniform log hangs by two steel wires, A and B, both of radius 1.20 mm. Initially, wire A was 2.50 m long and 2.00 mm shorter than wire B. The log is now horizontal. What are the magnitudes of the forces on it from
(a) wire A and
(b) wire B?
(c) What is the ratio dA/dB?

Answer 1

I know that I need to use young's modulus of steel and the stress-strain relation :
\[\dfrac{F}{A} = Y\dfrac{\Delta L}{L}\]

Answer 2

Before that, balancing vertical forces I get :
\[103g = T_A+T_B \tag{1}\]

Answer 3

I am not sure yet how to make use of that stress-strain relation here...

Answer 4

Hmm, OK. I haven't studied this so I'll be counting on you for the most part.

If shorter wire stretches by \(\Delta L\) then the longer one must stretch by \(\Delta L -2~\rm mm \) so that the log is horizontal in the end.

Answer 5

oh, right :
\[\Delta L_B = \Delta L_A -0.002\tag{2}\]

Answer 6

two extra unknowns hmm ..

Answer 7

From stress strain relation :

\[\dfrac{T_A}{\pi(1.2/1000)^2} = 200*10^9*\dfrac{\Delta L_A}{2.5-0.002}\tag{3}\]

\[\dfrac{T_B}{\pi(1.2/1000)^2} = 200*10^9*\dfrac{\Delta L_B}{2.5}\tag{4}\]

Answer 8

four equations and four unknowns : \(T_A, T_B, \Delta L_A, \Delta L_B\)

i hope they are independent equations... let me feed them to wolfram :)

Answer 9

thats it! matches with the answer XD
http://www.wolframalpha.com/input/?i=solve+103*9.8+%3D+a%2Bb%2C+y%3Dx-0.002%2C+%5Cdfrac%7Ba%7D%7B%5Cpi%281.2%2F1000%29%5E2%7D+%3D+200*10%5E9*%5Cdfrac%7Bx%7D%7B2.5-0.002%7D%2C+%5Cdfrac%7Bb%7D%7B%5Cpi%281.2%2F1000%29%5E2%7D+%3D+200*10%5E9*%5Cdfrac%7By%7D%7B2.5%7D

Answer 10

any idea on how to work part c ?

Answer 11

Net torque = 0?

Answer 12

\(-d_AT_A +d_BT_B=0 \implies \dfrac{d_A}{d_B} = \dfrac{T_B}{T_A}\)

perfect! thanks

Answer 13

Omg so smart, I will never be that good in math Q.q mind blown.