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HW #11-solutions

# HW #11-solutions - wolz(cmw2833 HW#11 Antoniewicz(57420...

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wolz (cmw2833) – HW #11 – Antoniewicz – (57420) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)5.0points A uniform horizontal beam of weight 400 N and length 4 . 49 m has two weights hanging from it. One weight of 300 N is located 0 . 95637 m from the left end; the other weight of 200 N is located 0 . 95637 m from the right end. What must be the magnitude of the one additional force on the beam that will produce equilibrium? Correct answer: 900 N. Explanation: Let : W = 400 N , W 1 = 300 N , and W 2 = 200 N . For the beam to preserve translational equi- librium, the net force acting on the beam has to be zero, so the additional force should be F = W + W 1 + W 2 = 400 N + 300 N + 200 N = 900 N . 002(part2of2)5.0points At what distance from the left end must this force be applied for the beam to be in equilib- rium? Correct answer: 2 . 10182 m. Explanation: The beam also has to preserve its rotational equilibrium; i.e. , the net torque acting on the beam has to be zero, so F x = W L 2 + W 1 L 1 + W 2 L 2 x = W L 2 F + W 1 L 1 F + W 2 L 2 F = (400 N) (4 . 49 m) 2 (900 N) + (300 N) (0 . 95637 m) 900 N + (200 N) (3 . 53363 m) 900 N = 2 . 10182 m . 003(part1of4)3.0points Two weights are suspended as shown in the diagram. On the left, a weight of 79 N is suspended from its upper left by a string of tension T 1 angled at 44 with the horizontal. A string carrying tension T 2 is stretched be- tween them angled at 12 with the horizontal. On the right, the unknown weight W 2 is also suspended from its upper right by a string of tension T 3 angled at 50 with the horizontal. 79 N W 2 12 T 2 T 1 T 3 50 44 What is the tension T 1 ? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 145 . 822 N. Explanation: Let : W 1 = M 1 g = 79 N , θ 1 = 44 , θ 2 = 12 , and θ 3 = 50 . For the 79 N weight, horizontal equilibrium yields T 1 cos θ 1 = T 2 cos θ 2 T 2 = T 1 cos θ 1 cos θ 2

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wolz (cmw2833) – HW #11 – Antoniewicz – (57420) 2 and vertical equilibrium T 1 sin θ 1 = W 1 + T 2 sin θ 2 T 1 sin θ 1 = W 1 + T 1 cos θ 1 cos θ 2 sin θ 2 = W 1 + T 1 cos θ 1 tan θ 2 . T 1 sin θ 1 - T 1 cos θ 1 tan θ 2 = W 1 T 1 = W 1 sin θ 1 - cos θ 1 tan θ 2 = 79 N sin 44 - cos 44 tan 12 = 145 . 822 N . 004(part2of4)3.0points What is the tension T 2 ? Correct answer: 107 . 239 N. Explanation: From the horizontal equation, T 2 = T 1 cos θ 1 cos θ 2 = (145 . 822 N) cos 44 cos 12 = 107 . 239 N . 005(part3of4)2.0points What is the tension T 3 ? Correct answer: 163 . 188 N. Explanation: For the weight W 2 horizontal equilibrium gives T 3 cos θ 3 = T 2 cos θ 2 T 3 = T 2 cos θ 2 cos θ 3 = (107 . 239 N) cos 12 cos 50 = 163 . 188 N . 006(part4of4)2.0points What is the weight W 2 ? Correct answer: 147 . 305 kg. Explanation: For the weight W 2 vertical equilibrium gives W 2 = T 2 sin θ 2 + T 3 sin θ 3 = (107 . 239 N) sin 12 + (163 . 188 N) sin 50 = 147 . 305 N .
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