HW #5 Solutions - homework 05 FONTENOT, BRIAN Due: Feb 18...

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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 = T-W 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 left-hand cable had tension T 2 and makes an angle of 2 with the ceiling. The right-hand cable had tension 440 N and makes an angle of 36 with the ceiling. The right-hand 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 left-hand 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 free-body diagram below. F 2 F 1 1 2 W g Note: The sum of the x- and y-components 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 2-W 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 left-hand cable makes with respect to the ceiling?...
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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.

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HW #5 Solutions - homework 05 FONTENOT, BRIAN Due: Feb 18...

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