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Unformatted text preview: oldmidterm 04 FRENNEA, KYLE Due: Apr 28 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Given: G = 6 . 6726 10 11 N m 2 / kg 2 Three 3 kg masses are located at points in the xy plane as shown in the figure. 36 cm 55 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 5 . 04111 10 9 N (tolerance 1 %). Explanation: Let : m = 3 kg x = 36 cm , and y = 55 cm . Basic Concepts: Newtons Law of Grav itation: F g = G m 1 m 2 r 2 We calculate the forces one by one, and then add them using the superposition prin ciple. The force from the mass on the right is pointing in the x direction, and has magni tude f 1 = G mm x 2 = Gm 2 x 2 = (6 . 6726 10 11 N m 2 / kg 2 )(3 kg) 2 (0 . 36 m) 2 = 4 . 63375 10 9 N . The other force is pointing in the y direction and has magnitude f 2 = G mm y 2 = Gm 2 y 2 = (6 . 6726 10 11 N m 2 / kg 2 )(3 kg) 2 (0 . 55 m) 2 = 1 . 98524 10 9 N . f 2 f 1 F Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythago ras theorem as well: F = radicalBig f 2 1 + f 2 2 = bracketleftBig (4 . 63375 10 9 N) 2 + (1 . 98524 10 9 N) 2 bracketrightBig 1 2 = 5 . 04111 10 9 N . As you see, this force is very small. If the masses in the picture are standing on a table (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. Note: The angle shown in the figure is = arctan bracketleftbigg f 2 f 1 bracketrightbigg = arctan bracketleftbigg (1 . 98524 10 9 N) (4 . 63375 10 9 N) bracketrightbigg = 23 . 1917 . Question 2 part 1 of 1 10 points oldmidterm 04 FRENNEA, KYLE Due: Apr 28 2007, 4:00 am 2 Planet X has five times the diameter and three times the mass of the earth. What is the ratio of gravitational accelera tion at the surface of planet X to the gravita tional acceleration at the surface of the Earth, g x g e ? 1. g x g e = 2 9 2. g x g e = 3 25 correct 3. g x g e = 5 36 4. g x g e = 4 25 5. g x g e = 5 9 6. g x g e = 6 49 7. g x g e = 9 49 8. g x g e = 1 7 9. g x g e = 2 49 10. g x g e = 3 16 Explanation: Let : M X = 3 M e , R X = 5 R e . Because mg e = GM e m R 2 e , where M e is the mass of the earth and R e is the radius. So that g e = GM e R 2 e . Similar equation applies to the planet X g x = GM x R 2 x . The ratio between the surface gravitational accelerations is g x g e = M x R 2 e M e R 2 x = 3 M e R 2 e M e (5 R e ) 2 = 3 25 . Question 3 part 1 of 1 10 points An astronaut weighs 147 N on the Moons surface. He is in a circular orbit about the Moon at an altitude of twoninth the moons radius....
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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 at Austin.
 Spring '08
 Turner
 Physics, Mass

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