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This is problem 101 of chapter 7 .4 of vector mechanics for engineers, statics, and dynamics.
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So in this problem, there is a cable going from a to d, and there is a mass at b of 70 kilograms, and there's a mass at c of 25 kilograms.
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The horizontal distance from a to b is 4 meters, from b to c is 6 meters, and from c to d is 4 meters.
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The vertical distance from c to d is 3 meters and from a to b is 5 meters.
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And there is a force p in the negative x direction at b.
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And our task for this problem is to find the value of p to maintain equilibrium.
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So we're going to start out with determining the weight of b.
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And to do this, it is the mass of b times gravity.
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So this is 70 times 9 .81, which is 6 .1.
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686 .7 newtons.
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And we also need to calculate the weight of c.
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This is the mass of c times g, which is 25 times 9 .81 for gravity.
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And this equals 245 .25 newtons.
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So now we're going to draw a free body diagram from c to d.
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Now we're going to include the reaction forces at d.
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Dy is in the positive y direction, and dx is in the positive direction.
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So we will be taking the sum of moments of c and setting that equal to zero.
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So it'll be d .y times four because there's four meters in the horizontal direction between d and c minus d x times three because there's three meters in the vertical direction between c and d.
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And i'm just going to go ahead and solve for d .y to simplify, just three -force dx.
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And we're going to need this equation later on.
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Next we are going to draw a free body diagram from b to d.
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We're going to take the sum of moments around b and saying that equal to 0.
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So this will equal d .y times 10.
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These are 10 meters in the horizontal direction between d and b.
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Minus d x times 5.
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These are 5 meters in the vertical direction between d and b.
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Minus the weight of c, which is 245 .25 times 6...