md3 - Version 088/ABBCA midterm 03 Turner(58120 This...

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Version 088/ABBCA – midterm 03 – Turner – (58120) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The figure below shows a vertical force applied tangentially to a uniform cylinder of weight 8 N. The coefficient of static friction between the cylinder and all surfaces is 0 . 24. F 8 N Find the maximum force F that can be ap- plied without causing the cylinder to rotate. 1. 4.41881 2. 3.05265 3. 5.57637 4. 1.75679 5. 5.15062 6. 4.56065 7. 2.04638 8. 3.80159 9. 3.94465 10. 7.8907 Correct answer: 1 . 75679 N. Explanation: Let : W = 8 N and μ = 0 . 24 . f 2 N 2 f 1 N 1 R F W When the force F is the largest, the fric- tional forces should be the largest since they act to keep the cylinder from rotating, so f 1 = μ N 1 and f 2 = μ N 2 . summationdisplay F x = f 1 − N 2 = 0 f 1 = N 2 = f 2 μ summationdisplay T = f 1 R + f 2 R F R = 0 F = f 1 + f 2 = f 2 μ + f 2 = 1 + μ f 2 μ f 2 = μ 1 + μ F and f 1 = f 2 μ = F 1 + μ . summationdisplay F y = 0 , so W = F + f 2 + N 1 = F + μ 1 + μ F + f 1 μ = parenleftbigg 1 + 2 μ 1 + μ parenrightbigg F + F μ (1 + μ ) = μ + 2 μ 2 + 1 μ (1 + μ ) F F = μ (1 + μ ) W 2 μ 2 + μ + 1 = 0 . 24 (1 + 0 . 24) 2 (0 . 24) 2 + 0 . 24 + 1 (8 N) = 1 . 75679 N . 002 10.0 points A stainless-steel orthodontic wire is applied to a tooth that is out of line by 25 . The wire has an unstretched length of 2 . 8 cm and a diameter of 0 . 21 mm . 25 25 If the wire is stretched 0 . 09 mm , find the magnitude of the force on the tooth. (Disre- gard the width of the tooth). Young’s modu- lus for stainless steel is 1 . 8 × 10 11 Pa . 1. 28.1372
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Version 088/ABBCA – midterm 03 – Turner – (58120) 2 2. 26.0882 3. 16.9381 4. 17.8972 5. 20.6978 6. 19.4881 7. 26.9872 8. 21.022 9. 23.3455 10. 9.67923 Correct answer: 16 . 9381 N. Explanation: Let : L 0 = 2 . 8 cm = 0 . 028 m , θ = 25 , r = 0 . 105 cm = 0 . 000105 m , Δ L = 0 . 09 mm = 9 × 10 5 m , and Y = 1 . 8 × 10 11 Pa . 25 25 F F We need the tension required to stretch the wire by 0 . 09 mm: Y = F L 0 A Δ L F = Y A Δ L L 0 = Y ( π r 2 ) Δ L L 0 The resultant force exerted on the tooth has components summationdisplay F x = F cos θ F cos θ = 0 and summationdisplay F y = F sin θ F sin θ , so F net = 2 F sin θ = 2 π Y r 2 Δ L sin θ L 0 = 2 π (1 . 8 × 10 11 Pa)(0 . 000105 m) 2 × (9 × 10 5 m) sin 25 0 . 028 m = 16 . 9381 N , which has a magnitude of 16 . 9381 N , and is directed down the page. 003 10.0 points Two wires made of the same material are stretched by equal forces. The second wire is twice as long as the first one and also has twice the diameter of the first wire. If the first wire is stretched by 5 . 6 cm , then by how long is the second wire stretched? 1. 4.3 2. 1.75 3. 2.25 4. 0.9 5. 0.55 6. 2.8 7. 3.65 8. 3.75 9. 1.65 10. 4.35 Correct answer: 2 . 8 cm. Explanation: Let : Δ L 1 = 5 . 6 cm , L 2 = 2 L 1 , and r 2 = 2 r 1 . Young’s modulus is Y = F L A Δ L = F L π r 2 Δ L Y F = L π r 2 Δ L . The forces are the same, so L 1 π r 2 1 Δ L 1 = L 2 π r 2 2 Δ L 2 Δ L 2 = L 2 L 1 parenleftbigg r 1 r 2 parenrightbigg 2 Δ L 1 = 2 parenleftbigg 1 2 parenrightbigg 2 Δ L 1 = 1 2 Δ L 1 = 1 2 (5 . 6 cm) = 2 . 8 cm .
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Version 088/ABBCA – midterm 03 – Turner – (58120) 3 004 10.0 points Masses M 1 and M 2 are supported by wires that have equal lengths when unstretched.
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