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physics hw 4 - homework 04 HILL WILLIAM Due 1:00 am...

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homework 04 – HILL, WILLIAM – Due: Sep 28 2007, 1:00 am 1 Question 1, chap 5, sect 1. part 1 of 2 10 points Four forces act on a hot-air balloon, as shown from the side. 713 N 346 N 1178 N 568 N Note: Figure is not drawn to scale Draw the vectors to scale on a graph to determine the answer. a) Find the magnitude of the resultant force on the balloon. Correct answer: 711 . 891 N (tolerance ± 5 %). Explanation: 713 N 346 N 1178 N 568 N 711 . 891 N Scale: 100 N 58 . 9672 Basic Concepts: F x,net = F x, 1 + F x, 2 F y,net = F y, 1 + F y, 2 F net = radicalBig ( F x,net ) 2 + ( F y,net ) 2 Given: F x, 1 = 713 N F x, 2 = - 346 N F y, 1 = 1178 N F y, 2 = - 568 N Solution: Consider the horizontal forces: F x,net = 713 N + ( - 346 N) = 367 N Consider the vertical forces: F y,net = 1178 N + ( - 568 N) = 610 N
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homework 04 – HILL, WILLIAM – Due: Sep 28 2007, 1:00 am 2 Thus the net force is F net = radicalBig (367 N) 2 + (610 N) 2 = 711 . 891 N . Question 2, chap 5, sect 1. part 2 of 2 10 points b) Find the direction of the resultant force (in relation to the 713 N force, with up being positive). Correct answer: 58 . 9672 (tolerance ± 5 %). Explanation: Basic Concept: tan θ = F y,net F x,net Solution: θ = tan 1 parenleftbigg F y,net F x,net parenrightbigg = tan 1 parenleftbigg 610 N 367 N parenrightbigg = 58 . 9672 . Question 3, chap 5, sect 1. part 1 of 1 10 points Two men decide to use their cars to pull a truck stuck in mud. They attach ropes and one pulls with a force of 837 N at an angle of 31 with respect to the direction in which the truck is headed, while the other car pulls with a force of 1320 N at an angle of 22 with respect to the same direction. 837 N 31 1320 N 22 What is the net forward force exerted on the truck in the direction it is headed? Correct answer: 1941 . 33 N (tolerance ± 1 %). Explanation: Given : F 1 = 837 N , F 2 = 1320 N , θ 1 = 31 , and θ 2 = 22 . For the first vehicle, the forward component is F 1 f = F 1 cos θ 1 = (837 N) cos 31 = 717 . 449 N . For the second vehicle, similarly, F 2 f = F 2 cos θ 2 = (1320 N) cos 22 = 1223 . 88 N . Thus the net forward force on the truck is F f = F 1 f + F 2 f = 717 . 449 N + 1223 . 88 N = 1941 . 33 N . Question 4, chap 5, sect 2. part 1 of 1 10 points You place a box weighing 341 . 4 N on an inclined plane that makes a 36 . 5 angle with the horizontal. Compute the component of the gravita- tional force acting down the inclined plane. Correct answer: 203 . 073 N (tolerance ± 1 %). Explanation: Basic concepts θ W θ θ W W 1
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homework 04 – HILL, WILLIAM – Due: Sep 28 2007, 1:00 am 3 The component of the gravitational force act- ing down the plane is the side opposite the angle θ , so W 1 = W sin θ Question 5, chap 5, sect 3. part 1 of 5 10 points In the parallel spring system, the springs are positioned so that the 49 N weight stretches each spring equally. The spring con- stant for the left-hand spring is 1 . 4 N / cm and the spring constant for the righ-hand spring is 4 . 4 N / cm . 1 . 4 N / cm 4 . 4 N / cm 49 N How far down will the 49 N weight stretch the springs?
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