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HW #5 Solutions - homework 05 FONTENOT BRIAN Due 4:00 am...

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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 . 87 N μ = 0 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 , θ = 26 , and μ = 0 . Consider the free body diagram for the block m g sin θ N = m g cos θ μ 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 θ acting down the hill and W cos θ acting straight into the hill. The system is in equilibrium, so for forces parallel to the hill, F net = T - W sin θ = 0 = T = W sin θ = (87 N) sin 26 = 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 θ = 0 = ⇒ N = W cos θ = (87 N) cos 26 = 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 θ 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 2 440 N 36 θ 2 a) What is the tension T 2 in the left-hand cable slanted at an angle of θ 2 with respect to the wall?
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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 θ 1 θ 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 , θ 1 = 36 , and θ 2 = 90 - θ . BasicConcept: Vertically and Horizontally, we have F x net = F x 1 - F x 2 = 0 = F 1 cos θ 1 - F 2 cos θ 2 = 0 (1) F y net = F y 1 + F y 2 - W g = 0 = F 1 sin θ 1 + F 2 sin θ 2 - W g = 0 (2) Solution: Using Eqs. 1 and 2, we have F x 2 = F 1 cos θ 1 (1) = (440 N) cos 36 = 355 . 967 N , and F y 2 = F 3 - F 1 sin θ 1 (2) = 634 N - (440 N) sin 36 = 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 θ 2 which the left-hand cable makes with respect to the ceiling? Correct answer: 46 . 5201 (tolerance ± 1 %). Explanation: Using Eq. 2, we have θ 2 = arctan parenleftbigg F y 1 F x 1 parenrightbigg = arctan parenleftbigg 258 . 626 N 355 . 967 N parenrightbigg = 46 . 5201 . Question 5, chap 5, sect 6. part 1 of 3 10 points A block of mass 1 . 84572 kg lies on a friction- less table, pulled by another mass 3 . 62536 kg under the influence of Earth’s gravity.
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