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PirateFan308
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Homework Statement
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?
Homework 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]
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!