This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 lefthand cable had tension T 2 and makes an angle of θ 2 with the ceiling. The righthand cable had tension 360 N and makes an angle of 42 ◦ with the ceiling. The righthand 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 lefthand cable slanted at an angle of θ 2 with respect to the wall? Correct answer: 450 . 221 N (tolerance ± 1 %). Explanation: Observe the freebody diagram below. F 2 F 1 θ 1 θ 2 W g Note: The sum of the x and ycomponents 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 lefthand 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 ◦ ....
View
Full
Document
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.
 Fall '08
 Turner
 Work

Click to edit the document details