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Ch5-H1-solutions

# Ch5-H1-solutions - chauhan(mmc2762 Ch5-H1 turner(56910 This...

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chauhan (mmc2762) – Ch5-H1 – turner – (56910) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of6)10.0points A load of 800 kg is suspended as shown in the following figure. 61 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: 7840 N. Explanation: Wire 1 must exactly balance the downward pull of gravity, so the tension must be F 1 = (800 kg)(9 . 8 m / s 2 ) = 7840 N . 002(part2of6)10.0points Now calculate the tension in wire 2. Correct answer: 3848 . 29 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 don’t 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 7840 N. We start by writing down an equation for the x components: F 2 ,x = F 3 ,x F 2 cos θ 2 = F 3 cos θ 3 . We can also write an equation for the y components: F 2 ,y + F 3 ,y = F 1 F 2 sin θ 2 + F 3 sin θ 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 61 sin 61 = 3848 . 29 N . 003(part3of6)10.0points Now calculate the tension in wire 3. Correct answer: 6255 . 01 N. Explanation: From the equations we deduced in part 2, we know that F 3 = cos 38 cos 61 F 2 , so we have F 3 = cos 38 cos 61 (3848 . 29 N) = 6255 . 01 N . 004(part4of6)10.0points These wires are made of a material whose value for Young’s modulus is 1 . 4 × 10 11 N / m 2 . The diameter of the wires is 1 . 1 mm. What is the strain (fractional stretch) in each wire?

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