s63 - [IMAGE] Consider the sketch on the right; half the...

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Solution to Problem 63 Congratulations to this week’s winners Nathan Pauli, Ray Kremer, Tracy Thatcher Further correct solutions were also received from Daniel Statman, Al Zimmermann, William Webb. A large number of incorrect solutions were submitted. Al Zimmermann solved the problem in an unusual way. Quoting from his submission: "I ordered a bar of steel 800 2/3 feet long. I also placed an order for a couple of tanks that could be used to push the ends together until they were only 800 feet apart. And I picked up a tape measure at the local hardware store so that I could measure how high above the ground the bar buckled. But then I started to worry: would a bar compressed in this way really form a circular arc? I suspected not. This was the catch. Defeated I did it the hard way, with a pencil and paper." One only hopes Al was able to make some other use of the tanks.
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Unformatted text preview: [IMAGE] Consider the sketch on the right; half the arc is 400 1/3 foot and half the chord is 400 feet long, R is the radius of the circle containing the arc and Q is the central angle cutting off half the arc. The length of half the arc is given by 400 1/3 = R sin(Q) while the length of half the chord is given by 400 = R Q . Solving for R and setting the equations equal gives sin(Q) / Q = (400 1/3) / 400, which one can solve numerically to get Q = .0706900643447 (approximately). The height of the curved bar above the straight bar is R- x , where x is the distance from the center of the circle to the middle of the straight bar, so x = 400 / tan(Q). Doing the arithmetic gives a deflection of approximately 14.1439032171 feet or about 14 feet 2 inches. this page....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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s63 - [IMAGE] Consider the sketch on the right; half the...

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