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This is problem 111 of chapter 7 .4 of vector mechanics for engineers, statics, and dynamics.
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In this problem, there is a bridge suspended by cables.
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The load each cable can support is 11 .1 kips per foot in the horizontal direction.
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The length is 4 ,150 feet, and the sag is 464 feet.
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So our task for this problem in a is a, is a.
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Determine the max tension in each cable, and for b, the length of the cable.
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So the first thing we're going to do is determine the weight of half the cable.
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It'll be w times half the length.
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So we'll do it 11 .1 times the length over 2, which is 4 ,150 divided by 2.
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This equals 23 ,032 .5 kips.
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Now we're going to draw for you about a diagram from b to a, which is half the cable.
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So we're going to take the moment around b and set that equal to zero.
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And this is going to be the tension of a times 464, because that is, there's 464 feet, in the vertical direction between a and b.
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We're going to subtract the weight, which is 23 ,000 and 32 .5 times 1 ,0003 ,000.
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37 .5 because there's 1 ,037 .5 feet in the horizontal direction between w and b...