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In this problem on the topic of equilibrium and elasticity, we are shown a non -uniform bar being suspended at rest in a horizontal position by two massless cords.
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The first chord makes an angle of 36 .9 degrees with the vertical, and the second chord makes the angle of 53 .1 degrees with the vertical.
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If we are given the length of the bar l to be 6 .1 meters, we want to calculate the distance x from the left end of the bar to the bar to the bar's center.
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Of mass.
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Now the bar is in equilibrium, so the forces and the torques acting on it each sum to zero.
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So if we let tl be the tension force of the left -hand cord and tr be the tension force of the right -hand cord and m the mass of the bar, we can write the equations of equilibrium as follows.
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Firstly, for the vertical force components, the equilibrium equation is tl cosine theta plus tr cosine phi minus the weight of the bar mg must equal to 0.
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Next for the horizontal force components, we have minus tl sine theta plus tr sine theta or sine phi rather, is equal to zero.
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Lastly, if we equate the talks, we get mg times m.
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X minus tr times l times cosine phi must equal to 0.
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Now the origin here was chosen to be at the left end of the bar, so to allow for the calculating of the talk...