This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 05 FONTENOT, BRIAN Due: Feb 18 2008, 4:00 am 1 Question 1, chap 5, sect 5. part 1 of 2 10 points A child holds a sled on a frictionless, snow covered hill, inclined at an angle of 26 . 8 7 N = F 26 If the sled weighs 87 N, find the force ex erted on the rope by the child. Correct answer: 38 . 1383 N (tolerance 1 %). Explanation: Given : W = 87 N , = 26 , and = 0 . Consider the free body diagram for the block m g s i n N = m g c o s N F W Basic Concepts: If we tilt our world, and consider the forces parallel to the hill, F net = summationdisplay F up summationdisplay F down = 0 then the forces perpendicular to the hill, F net = summationdisplay F out summationdisplay F in = 0 Solution: Consider the free body diagram for the sled: The weight of the sled has compo nents W sin acting down the hill and W cos acting straight into the hill. The system is in equilibrium, so for forces parallel to the hill, F net = TW sin = 0 = T = W sin = (87 N) sin26 = 38 . 1383 N . Question 2, chap 5, sect 5. part 2 of 2 10 points What force is exerted on the sled by the hill? Correct answer: 78 . 1951 N (tolerance 1 %). Explanation: For forces perpendicular to the hill, F net = N  W cos = 0 = N = W cos = (87 N) cos26 = 78 . 1951 N . Question 3, chap 5, sect 5. part 1 of 2 10 points Consider the 634 N weight held by two cables shown below. The lefthand cable had tension T 2 and makes an angle of 2 with the ceiling. The righthand cable had tension 440 N and makes an angle of 36 with the ceiling. The righthand cable makes an angle of 36 with the ceiling and has a tension of 440 N . 634 N T 2 4 4 N 3 6 2 a) What is the tension T 2 in the lefthand cable slanted at an angle of 2 with respect to the wall? homework 05 FONTENOT, BRIAN Due: Feb 18 2008, 4:00 am 2 Correct answer: 517 . 319 N (tolerance 1 %). Explanation: Observe the freebody diagram below. F 2 F 1 1 2 W g Note: The sum of the x and ycomponents of F 1 , F 2 , and W g are equal to zero. Given : W g = 634 N , F 1 = 440 N , 1 = 36 , and 2 = 90  . Basic Concept: Vertically and Horizontally, we have F x net = F x 1 F x 2 = 0 = F 1 cos 1 F 2 cos 2 = 0 (1) F y net = F y 1 + F y 2 W g = 0 = F 1 sin 1 + F 2 sin 2W g = 0 (2) Solution: Using Eqs. 1 and 2, we have F x 2 = F 1 cos 1 (1) = (440 N) cos36 = 355 . 967 N , and F y 2 = F 3 F 1 sin 1 (2) = 634 N (440 N) sin36 = 634 N 258 . 626 N = 375 . 374 N , so F 2 = radicalBig ( F x 2 ) 2 + ( F y 2 ) 2 = radicalBig (355 . 967 N) 2 + (375 . 374 N) 2 = 517 . 319 N . Question 4, chap 5, sect 5. part 2 of 2 10 points b) What is the angle 2 which the lefthand cable makes with respect to the ceiling?...
View
Full
Document
This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Friction, Work

Click to edit the document details