Question 190318
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The amount of weight a rectangular cross-section beam can support is directly proportional to the square of the horizontal dimension and to the cube of the vertical dimension, i.e.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  W = kH^2V^3]


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  630 = k(3)^2(5)^3 = 1125k]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k = \frac{630}{1125} = 0.56]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  W = 0.56\,\cdot\,H^2V^3]


If you turn the beam on its side (note the switch of values between H and V):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  W = 0.56\,\cdot\,(5)^2(3)^3 = (0.56)(25)(27) = 378]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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