Question

(20%) Problem 5: A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) @theexpertta.com - tracking id: 4N89-A2-C4-45-B07A-17568. In accordance with Expert TA's Terms of Service, copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account D 05 A 50% Part (a) At takeoff, an aircraft travels at 52 m/s, so that the air speed relative to the bottom of the wing is 52 m/s. Given the sea level density of air to be 1.29 kg/mº, how fast must it move over the upper surface to create the ideal lift? Grade Summary Vy= Deductions Potential 100% sino COS tan > 7 8 9 HOME Submissions Attempts remaining: 20 cotan asino acoso E 4 5 6 - (15 per attempt) atan acotano) 1 2 3 - detailed view cosho) tanho cotanh O . END Degrees Radians om CLEAR | sinh) Submit Hint Feedback I give up! Hints: 15 deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. H 5 0% Part (b) How fast must air move over the upper surface at a cruising speed of 243 m/s and at an altitude where air density is one-fourth that at sea level?

Answer 1

lift per square meter - P=1000 /M² solution) P = 1000 N of (a) V = 52 m/s 8 = 1.29 kg / m3 V2 = ? P = 2 8 (v2 - v,2) 1000 = (2) (129) (V2 - 52²) (1000x2) = ² – 52² = 1550.38 = 2 - 52' 1.29 V2 ² = 1550.38 + 5 2 ² = 4254.38 V2 = 54254.38 = 65.2 [R = 6512) mis (6) v = 243 mis p= ) = (122) = 0.32 kg love 0.32 P = 2 (8) ( 12² - ,2) 1000 = (₂) (0.32) ( V₂² - 243²) (2x1000) = 1 - 59049 6250 = 2 ² - 51049 = 6250 +59049 2 = 65°299 ) V₂ = 565299 [12= 25553) mis