solution_pdf hw 6

solution_pdf hw 6 - Patel(ppp285 HW06 TSOI(58160 This...

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Patel (ppp285) – HW06 – TSOI – (58160) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points Three 6 kg masses are located at points in the xy plane as shown. 37 cm 53 cm What is the magnitude oF the resultant Force (caused by the other two masses) on the mass at the origin? The universal gravita- tional constant is 6 . 6726 × 10 11 N · m 2 / kg 2 . Correct answer: 1 . 95196 × 10 8 N. Explanation: Let : m = 6 kg , x = 37 cm = 0 . 37 m , y = 53 cm = 0 . 53 m , and G = 6 . 6726 × 10 11 N · m 2 / kg 2 . The Force From the mass on the right points in the x direction and has magnitude F 1 = G mm x 2 = Gm 2 x 2 = (6 . 6726 × 10 11 N · m 2 / kg 2 ) (6 kg) 2 (0 . 37 m) 2 = 1 . 75466 × 10 8 N . The other Force points in the y direction and has magnitude F 2 = (6 . 6726 × 10 11 N · m 2 / kg 2 ) (6 kg) 2 (0 . 53 m) 2 = 8 . 55157 × 10 9 N . F 2 F 1 F θ The magnitude oF the resultant Force is F = r F 2 1 + F 2 2 = b (1 . 75466 × 10 8 N) 2 + (8 . 55157 × 10 9 N) 2 B 1 / 2 = 1 . 95196 × 10 8 N . 002 (part 2 oF 2) 10.0 points At what angle From the positive x -axis will the resultant Force point? Let counterclockwise be positive, within the limits 180 to 180 . Correct answer: 25 . 9828 . Explanation: The angle θ shown is θ = arctan p f 2 f 1 P = arctan p 8 . 55157 × 10 9 N 1 . 75466 × 10 8 N P = 25 . 9828 . 003 (part 1 oF 3) 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 A uniForm solid sphere oF mass m 2 = 289 kg and radius R 2 = 1 . 22 m is inside and concen- tric with a spherical shell oF mass m 1 = 207 kg and radius R 1 = 2 . 38 m (see the fgure).
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Patel (ppp285) – HW06 – TSOI – (58160) 2 R R a b c m m 1 1 2 2 Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 2 kg located at a dis- tance 0 . 52 m from the center of the sphere and spherical shell. Correct answer: 1 . 10445 × 10 8 N. Explanation: In this case the distance a = 0 . 52 m from the particle of mass m = 2 kg to the center of the sphere (whose mass is m 2 = 289 kg) is smaller than the radius R 2 = 1 . 22 m of the sphere. Thus, the particle is inside the sphere and the gravitational force F a exerted on it increases linearly with the distance from the particle to the center of the sphere F a = G m 2 m R 3 2 a = (289 kg) (2 kg) (1 . 22 m) 3 (0 . 52 m) × 6 . 67259 × 10 11 N m 2 / kg 2 = 1 . 10445 × 10 8 N . 004 (part 2 of 3) 10.0 points Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 2 kg located at a distance 1 . 84 m from the center of the sphere and spherical shell. Correct answer: 1 . 13917 × 10 8 N. Explanation: In this case the distance b = 1 . 84 m from the particle of mass 2 kg to the center of the sphere is greater than the radius R 2 = 1 . 22 m of the sphere, but less than the radius R 1 = 2 . 38 m of the spherical shell. First, the shell does not a±ect the particle and, second, we can regard the sphere as a point mass and apply Newton’s law of gravity in its simplest version F b = G m 2 m b 2 = (289 kg) (2 kg) (1 . 84 m) 2 × 6 . 67259 × 10 11 N m 2 / kg 2 = 1 . 13917 × 10 8 N .
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solution_pdf hw 6 - Patel(ppp285 HW06 TSOI(58160 This...

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