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Lenny Conundrum
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Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every... 顯示更多 Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm. If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? 更新: Hurry!!!
最佳解答:
這條鋼條最長的長度是951.5 米 Weight of the bottom spherical weight = volume x density x acceleration = 4/3 x π x r^3 x 7.8 x 9.81 = 4/3 x π x (25)^3 x 7.8 x 9.81 = 510.508806 kg x 9.81 = 5008.091389 N Weight of one smaller spherical weight = volume x density x acceleration = 4/3 x π x r^3 x 7.8 x 9.81 = 4/3 x π x (15)^3 x 7.8 x 9.81 = 110.269902 kg x 9.81 = 1081.747740 N Weight of a section of steel bar that is 1cm long = Area of pentagon section x 1 x density x accleration = (6 x 0.5 x 2 x 2cos 30) x 1 x 7.8 x 9.81 = 81.059978 g x 9.81 = 0.081059978 kg x 9.81 = 0.795198384 N The Highest weight the steel that can withstand = 400 MPa x area of cross section = 400 x 10^6 N/m^2 x (6 x 0.5 x 2 x 2cos 30) = 400 x 100 N/cm^2 x 12cos30 = 415692.1938 N Now we treat one smaller stell ball and its upper 270 cm of petagon stell bar as one unit, and suppose there are Y such units. the weight of this unit = 1081.747740 + 270 x 0.795198384 = 1296.451304 N The total weight of the steel bar = 5008.091389 + (300 - 15)x0.795198384 + Y x 1296.451304 = 5234.72292844 + 1296.451304 Y Largest weight the steel bar could withstand = 415692.1938 5234.72292844 + 1296.451304 Y = 415692.1938 Y = 316.6007621 Note that weight of smaller steel ball / weight of one defined unit = 1081.747740 / 1296.451304 = 83.44 % Thus we know that the highest point of this steel bar is a nearly steel sphere with a little part removed from the top. Since the steel bar is long enough that we could tolerate a error of 15 cm , we assume the upper part is a hemisphere of the smaller steel ball. ∴Maximum possible length = height of the lowest steel ball + 1 lowest 285 cm long steel bar + height of 316 defined unit + the height of the uppest hemisphere = 50 cm + 285 cm + 316 x 300 cm + 15 cm = 951.5 M
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