HW4 - homework 04 GAUTHIER, JOSEPH Due: Sep 28 2007, 1:00...

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Unformatted text preview: homework 04 GAUTHIER, JOSEPH 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. 1020 N 480 N 1285N 569N 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: 896 . 803 N (tolerance 5 %). Explanation: 1020 N 480 N 1285 N 569 N 8 9 6 . 8 3 N Scale: 100 N 52 . 9768 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 = 1020 N F x, 2 =- 480 N F y, 1 = 1285 N F y, 2 =- 569 N Solution: Consider the horizontal forces: F x,net = 1020 N + (- 480 N) = 540 N Consider the vertical forces: F y,net = 1285 N + (- 569 N) = 716 N homework 04 GAUTHIER, JOSEPH Due: Sep 28 2007, 1:00 am 2 Thus the net force is F net = radicalBig (540 N) 2 + (716 N) 2 = 896 . 803 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 1020 N force, with up being positive). Correct answer: 52 . 9768 (tolerance 5 %). Explanation: Basic Concept: tan = F y,net F x,net Solution: = tan 1 parenleftbigg F y,net F x,net parenrightbigg = tan 1 parenleftbigg 716 N 540 N parenrightbigg = 52 . 9768 . 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 941 N at an angle of 30 with respect to the direction in which the truck is headed, while the other car pulls with a force of 1366 N at an angle of 23 with respect to the same direction. 9 4 1 N 3 1 3 6 6 N 2 3 What is the net forward force exerted on the truck in the direction it is headed? Correct answer: 2072 . 34 N (tolerance 1 %). Explanation: Given : F 1 = 941 N , F 2 = 1366 N , 1 = 30 , and 2 = 23 . For the first vehicle, the forward component is F 1 f = F 1 cos 1 = (941 N) cos30 = 814 . 93 N . For the second vehicle, similarly, F 2 f = F 2 cos 2 = (1366 N) cos23 = 1257 . 41 N . Thus the net forward force on the truck is F f = F 1 f + F 2 f = 814 . 93 N + 1257 . 41 N = 2072 . 34 N . Question 4, chap 5, sect 2. part 1 of 1 10 points You place a box weighing 242 . 8 N on an inclined plane that makes a 43 . 1 angle with the horizontal. Compute the component of the gravita- tional force acting down the inclined plane. Correct answer: 165 . 899 N (tolerance 1 %). Explanation: Basic concepts W W W 1 homework 04 GAUTHIER, JOSEPH 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 , so W 1 = W sin Question 5, chap 5, sect 3....
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This note was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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HW4 - homework 04 GAUTHIER, JOSEPH Due: Sep 28 2007, 1:00...

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