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This is problem 99 in chapter 7 .4 of vector mechanics for engineers, statics, and dynamics.
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So in this problem, a cable stretches from a to e, and there are three loads applied, one at b, which is two kips.
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One at c, two kips, and one at d also two kips.
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The vertical distance from a to e is 7 .5 feet.
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The horizontal distance from a to b is six feet, from b to c is nine feet.
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C to d is 6 feet, and d to e is 9 feet.
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And the vertical distance from c to e is 15 feet.
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And for a, we have to find the vertical distance from b to e, the distance b, and the vertical distance from d to e, which is the distance d.
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And also for part b, we had to find the max tension in this cable.
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All right, so the first thing we're going to do is draw a free body diagram of the whole entire cable, from a to e.
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And we're going to start off with taking the moment of a and so that equal to zero.
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This will be ey times 30 because there is 30 feet in the horizontal direction between a and e.
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We're subtract e x times 7 .5 because there are 7 .5 feet in the vertical direction between a and e.
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And we're going to subtract 2 times 21 for the load at d.
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And there's time 21 because there's 21 feet between a and d in the horizontal direction, minus 2, which is the load at c times 15 because there's 15 feet in the horizontal direction between a and c, minus 2 times 6.
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And this is the load at b.
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And there is 6 feet between a and b and the whole.
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Horizontal direction.
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So we're going to simplify this.
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This is 30ey minus 7 .5 x.
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And we're going to move all the loads to the other side and it totals 84.
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All right, we're going to go ahead and start isolating x.
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So we're going to move ey to the other side like so.
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And we're going to go ahead and solve for x.
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It's not going to be simplified, but it's how we're going to start solving it.
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And because it's negative, of 30 over negative 7 .5, this will become positive.
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All right.
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So in the next step, we're going to actually have to use this later.
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So we're going to do that.
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All right, so for the next step, we're going to do a free body diagram from c to e.
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And we're going to do the sum of of moments around c, so that equal to zero.
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I'm going to do ey times 15, because there's 15 feet in the horizontal direction between c and e.
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I'm going to subtract x times 15 as well because there is 15 feet in the vertical direction between c and e.
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And we're subtract 2 times 6, which is the load at d.
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It creates a negative moment around c bases in the clockwise direction, and it's times 6 because there's 6 feet in between c and d in the horizontal direction.
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And it equals 0.
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So we're going to go ahead and simplify.
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So 15ey minus 15 ex.
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I'm going to go ahead and move the 2 times 6 over to the other side, so it's going to be positive 12...