# phys1hw4 - homework 04 – ANDERSON ZACH – Due 4:00 am 1...

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Unformatted text preview: homework 04 – ANDERSON, ZACH – Due: Sep 22 2007, 4:00 am 1 Question 1, chap 5, sect 4. part 1 of 1 10 points An object placed on an equal arm balance requires 12 kg to balance it. When placed on a spring scale, the scale reads 12 kg. Now everything (balance, set of masses, scale, and object) is transported to the moon, where the force of gravity is 1 6 that on earth. The new readings on the balance and spring scale, respectively, are 1. 12 kg, 72 kg 2. 12 kg, 2 kg correct 3. 12 kg, 12 kg 4. 2 kg, 2 kg 5. 2 kg, 12 kg Explanation: The equal arm balance measures mass , whereas the spring scale, properly speaking, measures weight . Since an object’s mass is an inherent property, it will not change when we go to the moon. The apparent weight, how- ever, will be less, since the gravitational force exerted by the moon on the object is smaller. Question 2, chap 5, sect 5. part 1 of 2 10 points Consider the 603 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 360 N and makes an angle of 42 ◦ with the ceiling. The right-hand cable makes an angle of 42 ◦ with the ceiling and has a tension of 360 N . 603 N T 2 3 6 N 4 2 ◦ 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? Correct answer: 450 . 221 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 = 603 N , F 1 = 360 N , θ 1 = 42 ◦ , and θ 2 = 90 ◦- θ . Basic Concept: 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) homework 04 – ANDERSON, ZACH – Due: Sep 22 2007, 4:00 am 2 = (360 N) cos42 ◦ = 267 . 532 N , and F y 2 = F 3- F 1 sin θ 1 (2) = 603 N- (360 N) sin42 ◦ = 603 N- 240 . 887 N = 362 . 113 N , so F 2 = radicalBig ( F x 2 ) 2 + ( F y 2 ) 2 = radicalBig (267 . 532 N) 2 + (362 . 113 N) 2 = 450 . 221 N . Question 3, 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: 53 . 5427 ◦ (tolerance ± 1 %). Explanation: Using Eq. 2, we have θ 2 = arctan parenleftbigg F y 1 F x 1 parenrightbigg = arctan parenleftbigg 240 . 887 N 267 . 532 N parenrightbigg = 53 . 5427 ◦ ....
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## This homework help was uploaded on 04/10/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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phys1hw4 - homework 04 – ANDERSON ZACH – Due 4:00 am 1...

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