This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 088/ABBCA – midterm 03 – Turner – (58120) 1 This printout should have 19 questions. Multiplechoice 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 + μf 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 = . 24(1 + 0 . 24) 2 (0 . 24) 2 + 0 . 24 + 1 (8 N) = 1 . 75679 N . 002 10.0 points A stainlesssteel 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 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 = 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 A Δ L F = Y A Δ L L = Y ( π r 2 ) Δ L L 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 = − 2 π (1 . 8 × 10 11 Pa)(0 . 000105 m) 2 × (9 × 10 − 5 m) sin25 ◦ . 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....
View
Full
Document
This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Force, Friction

Click to edit the document details