practice problem 1
A 100 N sign is to be suspended by two cables. Determine the tension in each cable for the given angles … a. 90°, 90°| b. 180°, 45°| c. 120°, 30°|
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For all solutions, let T1 be the cable on the left and T2 be the cable on the right. The sign always has weight (W), which points down. The sign isn't going anywhere (it's not accelerating), therefore the three forces are in equilibrium. Describe this state using the language of physics — equations; in particular, component analysis equations. As always, make a nice drawing to show what's going on. Use a ruler and a protractor if you wish. a.
[magnify] The two upward components should equal one another. Together they should equal the weight, which means each one is carrying half the load.T1 = T2 = ½W = ½(100 N) = 50 N|
[magnify] Weight points down (270°) and T1 points to the left (180°). These are both good vectors — good in the sense that they are easy to deal with. T1 is the troublemaker. Break it up into components and state the conditions for equilibrium in the vertical and horizontal directions. I like to put negative vectors on the left side of the equals sign and positive vectors on the right side. I also suggest working through the vertical equation first. | vertical| | | | horizontal| |
∑ F−y = | ∑ F+y| | | | ∑ F−x = | ∑ F+x| | | | | | | | | |
W = | T2 sin θ2| | | | T1 = | T2 cos θ2| | | | | | | | | |
T2 = | W| = | 100 N| | T1 = | 141 N cos 45°| | | sin θ2| | sin 45°| | | | |
T2 = | 141 N| | | | T1 = | 100 N| |
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[magnify] Weight is the only force with a convenient direction. Resolve the tensions into their components. State the equilibrium condition along both axes. I suggest working with the horizontal equation first. | horizontal| | | vertical| |
∑ F−x = | ∑ F+x| | ...