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Homework Help: Suspended hotel walkway (Young's modulus problem)

  1. Jan 15, 2008 #1
    Here's a (hopefully) quick question for you guys... I'm completely stumped on this problem; any help would be appreciated.

    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    Young's modulus: [tex]Y = \frac{F\ast L}{A\ast\Delta L}[/tex]

    Y value for steel: [tex]20 \times 10^{10} N/m^2[/tex]
    Y value for aluminum: [tex]7.0 \times 10^{10} N/m^2[/tex]

    3. 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 [tex]0.000254m^2[/tex].

    The wire is solid, so its cross-sectional area is the same as that of any circle; I worked that out to be [tex]0.000126m^2[/tex].

    Taking the Y-values for steel and aluminum, I get the following two equations:

    [tex]\frac{F_1*L_1}{A_1* \Delta L_1} = 7.0 \times 10^{10} N/m^2[/tex]

    [tex]\frac{F_2*L_2}{A_2* \Delta L_2} = 20 \times 10^{10} N/m^2[/tex]

    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 [tex]\Delta L_1 = \Delta L_2 = \Delta L[/tex].

    Rearranging and combining my two equations, I then come up with...

    [tex]\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}[/tex]

    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.
    Last edited: Jan 15, 2008
  2. jcsd
  3. Jan 15, 2008 #2


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    Gold Member

    Your work is very good. Just carry through with your equation, plugging in the values for the respective L's, A's and Y's, and you will come up with F1 as a certain percentage of F2! (and note that F1 + F2 = the total applied load).
  4. Jan 15, 2008 #3
    Thanks! I actually had an epiphany about this right after I woke up this morning... I realized that, of course, [tex]F_1 + F_2[/tex] equals the total applied force (8500N), so rearranging the equations in terms of F and then adding them together allows one to solve the problem very easily.

    Hopefully that'll help someone in the future...
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