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Unformatted text preview: homework 04 – VENNES, ROSS – 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 hotair balloon, as shown from the side. 252 N 118 N 264N 115N 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: 200 . 392 N (tolerance ± 5 %). Explanation: 252 N 118 N 264 N 115 N 2 . 3 9 2 N Scale: 100 N 48 . 034 ◦ 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 = 252 N F x, 2 = 118 N F y, 1 = 264 N F y, 2 = 115 N Solution: Consider the horizontal forces: F x,net = 252 N + ( 118 N) = 134 N Consider the vertical forces: F y,net = 264 N + ( 115 N) = 149 N Thus the net force is F net = radicalBig (134 N) 2 + (149 N) 2 = 200 . 392 N . homework 04 – VENNES, ROSS – Due: Sep 28 2007, 1:00 am 2 Question 2, chap 5, sect 1. part 2 of 2 10 points b) Find the direction of the resultant force (in relation to the 252 N force, with up being positive). Correct answer: 48 . 034 ◦ (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 149 N 134 N parenrightbigg = 48 . 034 ◦ . 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 550 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 929 N at an angle of 23 ◦ with respect to the same direction. 5 5 N 3 ◦ 9 2 9 N 2 3 ◦ What is the net forward force exerted on the truck in the direction it is headed? Correct answer: 1331 . 46 N (tolerance ± 1 %). Explanation: Given : F 1 = 550 N , F 2 = 929 N , θ 1 = 30 ◦ , and θ 2 = 23 ◦ . For the first vehicle, the forward component is F 1 f = F 1 cos θ 1 = (550 N) cos30 ◦ = 476 . 314 N . For the second vehicle, similarly, F 2 f = F 2 cos θ 2 = (929 N) cos23 ◦ = 855 . 149 N . Thus the net forward force on the truck is F f = F 1 f + F 2 f = 476 . 314 N + 855 . 149 N = 1331 . 46 N . Question 4, chap 5, sect 2. part 1 of 1 10 points You place a box weighing 394 . 3 N on an inclined plane that makes a 41 . 4 ◦ angle with the horizontal. Compute the component of the gravita tional force acting down the inclined plane. Correct answer: 260 . 755 N (tolerance ± 1 %). Explanation: Basic concepts θ W θ θ W W 1 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....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Force, Work

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