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Unformatted text preview: louey (cal2859) Ch5HW1 florin (56930) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 6) 10.0 points A load of 900 kg is suspended as shown in the following figure. 62 38 1 2 3 Calculate the tension in all three wires (that is, the magnitude of the tension force exerted by each of these wires.) Begin with wire 1. Use g = 9 . 8 m / s 2 . Correct answer: 8820 N. Explanation: Wire 1 must exactly balance the downward pull of gravity, so the tension must be F 1 = (900 kg)(9 . 8 m / s 2 ) = 8820 N . 002 (part 2 of 6) 10.0 points Now calculate the tension in wire 2. Correct answer: 4204 . 62 N. Explanation: This part may be trickier than it seems at first. You might assume that each of the upper ropes contributes half of the tension in the lower rope, but we dont really know this. We know two things: 1) since this situation is an example of static equilibrium, the x components of the tension of the upper ropes must be equal and opposite, and 2) their y components of tension must add to 8820 N. We start by writing down an equation for the x components: F 2 ,x = F 3 ,x F 2 cos 2 = F 3 cos 3 . We can also write an equation for the y components: F 2 ,y + F 3 ,y = F 1 F 2 sin 2 + F 3 sin 3 = F 1 . Combining these two equations, we can eliminate F 3 from the system to find that F 2 = F 1 sin 38 + cos 38 cos 62 sin62 = 4204 . 62 N . 003 (part 3 of 6) 10.0 points Now calculate the tension in wire 3. Correct answer: 7057 . 47 N. Explanation: From the equations we deduced in part 2, we know that F 3 = cos38 cos62 F 2 , so we have F 3 = cos 38 cos 62 (4204 . 62 N) = 7057 . 47 N . 004 (part 4 of 6) 10.0 points These wires are made of a material whose value for Youngs modulus is 1 . 2 10 11 N / m 2 ....
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This note was uploaded on 03/20/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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