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π΄π΅ is a uniform rod of length 140 centimeters and weight 45 kilogram weight.
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Its end π΄ is fixed to a vertical wall by a hinge.
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It is held horizontally in equilibrium by means of a string of length 70 centimeters connected to point πΆ on the rod, which is 56 centimeters away from π΄, and fixed to point π· on the vertical wall vertically above π΄.
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Calculate the tension in the string.
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To answer this question, weβre going to begin by drawing a free-body diagram.
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This is a really simple diagram of the scenario that shows all the important forces weβre interested in.
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Here is the uniform rod of length 140 centimeters attached to a vertical wall at point π΄ by a hinge.
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Now, since the rod itself is uniform, this means we can model the downwards force of its weight to act at a point exactly halfway along the rod.
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So this 45-kilogram-weight force acts at a point 70 centimeters away from π΄.
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Next, we know that the rod itself is attached by a piece of string at point πΆ to a point on the wall π·.
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The distance between π΄ and πΆ is 56 centimeters.
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And the distance between π· and πΆ, the length of the string, is 70 centimeters.
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Weβre being asked to calculate the value of the tension in the string.
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So we model that acting at point πΆ, away from πΆ, and along the length of the string, and weβll call that π.
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So now we have all the relevant forces and dimensions, how are we going to calculate the value of π?
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Well, the key here is this word equilibrium.
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For the object to be in equilibrium, the sum of its forces must be equal to zero.
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But also the sum of the moments about any point must also be equal to zero.
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This means we can find the value of π by finding the sum of the moments about some point on the rod.
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Now, weβre going to take moments about point π΄, and weβre going to take the counterclockwise direction to be positive.
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We also know, of course, that the moment is the product of a force and its perpendicular distance from the pivot.
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So, thinking about the weight force, we know that this moment is going to be negative because itβs trying to rotate the object in a clockwise direction.
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Itβs a force of 45 kilogram weight, and itβs acting at a point 70 centimeters away from π΄.
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So its moment in kilogram weight centimeters is negative 45 times 70.
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But what do we do with our tensional force?
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Well, we need to work out the component of this force thatβs perpendicular to the line segment π΄π΅.
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So weβre going to begin by calculating the value of π.
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Thatβs the included angle in triangle π΄πΆπ·.
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We can use the dimensions given to calculate this value.
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We know, of course, the side adjacent to our included angle and the hypotenuse.
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So we use the cosine ratio.
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cos of π is 56 over 70.
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We could then take the inverse cos of both sides of this equation.
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But in fact we can find some exact values by also calculating the length of the other side.
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Weβre dealing with a right triangle, so weβre going to use the Pythagorean theorem.
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Defining the missing length in this triangle to be π₯, 70 squared equals 56 squared plus π₯ squared.
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Subtracting 56 squared from both sides, and we get π₯ squared equals 1764.
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And the square root of this is 42, so π₯ is equal to 42.
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And this is really useful because it allows us to calculate exact values for the sin, cos, and tan of π.
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For instance, sin of π is the opposite side divided by the hypotenuse, so it would be 42 over 70.
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And tan is opposite over adjacent, so tan of π here is 42 over 56.
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Weβre now ready to calculate the component of the tensional force that acts in a direction perpendicular to the rod.
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Itβs going to be π sin π, since we have the hypotenuse π and weβre trying to find the component opposite to the included angle.
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Since the tensional force is trying to rotate object π΄π΅ in a counterclockwise direction, the moment of this tensional force is going to be positive.
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Itβs π sin π times 56.
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And now we have the sum of all of the moments, and this is equal to zero.
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Negative 45 times 70 is negative 3150.
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And remember, we calculated sin of π to be 42 over 70.
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So π sin π times 56 becomes π times 42 over 70 times 56.
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42 over 70 times 56 simplifies to 168 over five.
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Then, we can add 3150 to both sides of our equation.
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We clear some space and we see that we can calculate the value of π by simply dividing through by 168 over five, giving us 93.75.
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So the tension in the string is 93.75 kilogram weight.