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Halfpixel
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Here's a (hopefully) quick question for you guys... I'm completely stumped on this problem; any help would be appreciated.
A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above and a vertical column underneath. The steel cable is 1.27cm in diameter and is 5.75m long before loading. The aluminum column is a hollow cylinder with an outside diameter of 16.24cm and an inside diameter of 16.14cm, and an unloaded length of 3.25m. When the walkway exerts a load force of 8500N on one of the support points, how much does the point move down?
Young's modulus: Y = \frac{F\ast L}{A\ast\Delta L}
Y value for steel: 20 \times 10^{10} N/m^2
Y value for aluminum: 7.0 \times 10^{10} N/m^2
I started by getting the applicable cross-sectional area of the column and the wire. Since the column is hollow, its area should be:
A = Outer area - inner area
which I worked out to be 0.000254m^2.
The wire is solid, so its cross-sectional area is the same as that of any circle; I worked that out to be 0.000126m^2.
Taking the Y-values for steel and aluminum, I get the following two equations:
\frac{F_1*L_1}{A_1* \Delta L_1} = 7.0 \times 10^{10} N/m^2
\frac{F_2*L_2}{A_2* \Delta L_2} = 20 \times 10^{10} N/m^2
Though the force exerted on the wire is not the same as the force exerted on the column and their modulus values are different, the walkway must be in equilibrium so the amount that the wire stretches must be equal to the amount that the column shrinks.
So \Delta L_1 = \Delta L_2 = \Delta L.
Rearranging and combining my two equations, I then come up with...
\Delta L = \frac{F_1*L_1}{A_1* 7.0 \times 10^{10} N/m^2} = \frac{F_2*L_2}{A_2 * 20 \times 10^{10} N/m^2}
But since I don't know what portion of the force is exerted on the wire and which portion is exerted on the column (I know it's not 50/50), I don't know where to go from here. I know I'm missing something and I'm sure it's fairly obvious, but I'm completely stuck. Any help would be greatly appreciated.
Homework Statement
A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above and a vertical column underneath. The steel cable is 1.27cm in diameter and is 5.75m long before loading. The aluminum column is a hollow cylinder with an outside diameter of 16.24cm and an inside diameter of 16.14cm, and an unloaded length of 3.25m. When the walkway exerts a load force of 8500N on one of the support points, how much does the point move down?
Homework Equations
Young's modulus: Y = \frac{F\ast L}{A\ast\Delta L}
Y value for steel: 20 \times 10^{10} N/m^2
Y value for aluminum: 7.0 \times 10^{10} N/m^2
The Attempt at a Solution
I started by getting the applicable cross-sectional area of the column and the wire. Since the column is hollow, its area should be:
A = Outer area - inner area
which I worked out to be 0.000254m^2.
The wire is solid, so its cross-sectional area is the same as that of any circle; I worked that out to be 0.000126m^2.
Taking the Y-values for steel and aluminum, I get the following two equations:
\frac{F_1*L_1}{A_1* \Delta L_1} = 7.0 \times 10^{10} N/m^2
\frac{F_2*L_2}{A_2* \Delta L_2} = 20 \times 10^{10} N/m^2
Though the force exerted on the wire is not the same as the force exerted on the column and their modulus values are different, the walkway must be in equilibrium so the amount that the wire stretches must be equal to the amount that the column shrinks.
So \Delta L_1 = \Delta L_2 = \Delta L.
Rearranging and combining my two equations, I then come up with...
\Delta L = \frac{F_1*L_1}{A_1* 7.0 \times 10^{10} N/m^2} = \frac{F_2*L_2}{A_2 * 20 \times 10^{10} N/m^2}
But since I don't know what portion of the force is exerted on the wire and which portion is exerted on the column (I know it's not 50/50), I don't know where to go from here. I know I'm missing something and I'm sure it's fairly obvious, but I'm completely stuck. Any help would be greatly appreciated.
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