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Stress, strain and modulus - log hung from 2 wires

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    There is a 104kg uniform log hung on two steel wires, A and B, both of radius 1.00mm. Initially, wire A was 2.70m long and 2.00mm shorter than wire B. The log is now horizontal. Note that the centre of gravity (com) is not necessarily centered between the wires and the modulus of steel is 20x1010Pa.
    a) What is the magnitude of the force on the log from wire B?
    b) What is the ratio of distance from wire A to com/distance from wire B to com?


    2. Relevant equations

    [tex]modulus = \frac{stress}{strain}[/tex]
    [tex]modulus_B = \frac{F_B~/~m^{2}}{Δl~/~l_{o}}[/tex]
    [tex]F_B = \frac{(Δl)(area)(modulus)}{l_{o}}[/tex]

    3. The attempt at a solution

    I'm assuming the left wire is wire A
    length of A = 2.70 m
    length of B = 2.702 m
    area of wire = [itex]\pi d^2 ÷ 4[/itex]
    let x = change in length of B
    so that x+0.002 = change in length of A
    [itex]F_A[/itex] = force of wire A
    [itex]F_B[/itex] = force of wire B
    total weight of block = (104kg)(9.81) = 1020.24N
    [itex]F_A + F_B[/itex] = 1020.24 N
    distance from wire A = d_A
    distance from wire B = d_B

    a)
    [tex]F_B = \frac{(Δl)(area)(modulus)}{l_{o}}[/tex]
    [tex]F_B = \frac{(x)(7.85398*10^{-7}m^{2})(20*10^{10}Pa}{2.702m}[/tex]
    [tex]F_B = 58134.579x[/tex]

    [tex]F_A = \frac{(Δl)(area)(modulus)}{l_{o}}[/tex]
    [tex]F_A = \frac{(x+0.002)(7.85398*10^{-7}m^{2})(20*10^{10}Pa}{2.70m}[/tex]
    [tex]F_A = 58177.64x + 116.355[/tex]

    [tex]F_A + F_B = 1020.24 N[/tex]
    [tex](58177.64x)+(58130x)+(116.355)=1020.24 N[/tex]
    [tex]x(58177.64+58130)=1020.24-116.355[/tex]
    [tex]x=7.77*10^{-3} m[/tex]

    [tex]F_{B}=(area)(modulus)(strain)[/tex]
    [tex]F_{B}=(7.85398*10^{-7})(20*10^{10})(X ÷ 2.702) = 4.51775*10^{2} N[/tex]
    [tex]F_{A}=5.6846*10^{2} N[/tex]


    b)
    [tex]\sum\tau=0[/tex]
    [tex](d_A)(F_A)-(d_B)(F_B)=0[/tex]
    [tex]\frac{d_A}{d_B} = \frac{F_B}{F_A}[/tex]
    [tex]\frac{d_A}{d_B} = \frac{4.51775*10^2}{5.6846*10^2}[/tex]
    [tex]\frac{d_A}{d_B} = 1.25829 = 1.26[/tex]

    Unfortunately, the website said that my answers for a) and b) are wrong. I know that since a) is wrong, that automatically makes b) wrong. Any help would be appreciated!
     
  2. jcsd
  3. Jan 13, 2012 #2

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    In part a, your approach is excellent. I didn't check all your math, but I notice you have calculated the area incorrectly. If the wire has a 1mm radius, it's area in x-section is ∏r^2 = 3.14E-06 m^2. Looks like you used d =1 mm instead of r = 1mm.:frown: Redo the math and you should get the correct answer.

    In part b, aside from making the necessary correction in numbers, it's da/db, but your result calculates db/da.

    Otherwise, you've done yourself proud:approve:
     
  4. Jan 13, 2012 #3
    Wow, what a simple mistake! Thanks for catching that - I would have been trying to figure it out for hours!
     
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