Physics Solution Manual for 1100 and 2101

2 0628 mj 318

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Unformatted text preview: ts of F31 and F32: Chapter 4 Problems F3 = 2 F cos θ = 2 GMm3 r2 181 cos θ The acceleration of sphere 3 is given by Newton's second law: a3 = F3 m3 =2 (6.67 ×10 GM cos θ = 2 2 r −11 N ⋅ m 2 /kg 2 ) ( 2.80 kg ) (1.20 m )2 cos 30.0° = 2.25 × 10 –10 m/s 2 ____________________________________________________________________________________________ 33. SSM WWW REASONING AND SOLUTION There are two forces that act on the balloon; they are, the combined weight of the balloon and its load, Mg, and the upward buoyant force FB . If we take upward as the positive direction, then, initially when the balloon is motionless, Newton's second law gives FB − Mg = 0 . If an amount of mass m is dropped overboard so that the balloon has an upward acceleration, Newton's second law for this situation is FB − ( M − m) g = ( M − m)a But FB = mg , so that Mg – ( M − m ) g = mg = ( M – m ) a Solving for the mass m that should be dropped overboard, we obtain Ma (310 kg )(0.15 m/s 2 ) = = 4.7 kg g + a 9.80 m/s 2 + 0.15 m/s 2 ______________________________________________________________________________ m= 34. REASONING AND SOLUTION acceleration, it follows that Since both motions are characterized by constant yJ yE = 1 a t2 2 JJ 1 a t2 2 EE where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both objects fall the same distance (yJ = yE), the above ratio is equal to unity. Solving for the ratio of the times yields tJ tE = 2 aE GM E / RE RJ = = 2 aJ RE GM J / RJ ME 1 = (11.2 ) = 0.628 MJ 318 ____________________________________________________________________________________________ 182 FORCES AND NEWTON'S LAWS OF MOTION 35. REASONING The gravitational force that the sun exerts on a person standing on the earth 2 is given by Equation 4.3 as Fsun = GM sun m / rsun-earth , where Msun is the mass of the sun, m is the mass of the person, and rsun-earth is the distance from the sun to the earth. Likewise, the gravitational force that the moon exerts on a person standing on the earth is given by 2 Fmoon = GM moon m / rmoon-earth , where Mmoon is the mass of the moon and rmoon-earth is the distance from the moon to the earth. These relations will allow us to determine whether the sun or the moon exerts the greater gravitational force on the person. SOLUTION Taking the ratio of Fsun to Fmoon, and using the mass and distance data from the inside of the text’s front cover, we find GM sun m Fsun Fmoon 2 M rsun-earth = = sun M GM moon m moon 2 rmoon-earth rmoon-earth rsun-earth 1.99 × 1030 kg 3.85 × 108 m = 7.35 × 1022 kg 1.50 × 1011 m 2 2 = 178 Therefore, the sun exerts the greater gravitational force. ____________________________________________________________________________________________ 36. REASONING The gravitational attraction between the planet and the moon is governed by Newton’s law of gravitation F = GMm/r2 (Equation 4.3), where M is the planet’s mass and m is the moon’s mass. Because the magnitude of this attractive force varies inversely with the square of the distance r between the center of the moon and the center of the planet, the maximum force Fmax occurs at the minimum distance rmin, and the minimum force Fmin at the maximum distance rmax. The problem text states that Fmax exceeds Fmin by 11%, or Fmax = 1.11 Fmin. This expression can be rearranged to give the ratio of the forces: Fmax/ Fmin = 1.11. We will use Equation 4.3 to compute the desired distance ratio in terms of this force ratio. SOLUTION From Equation 4.3, the ratio of the maximum gravitational force to the minimum gravitational force is Fmax Fmin = GMm 2 rmin GMm 2 rmax 1 = 2 rmin 1 2 rmax = 2 rmax 2 rmin Chapter 4 Problems 183 Taking the square root of both sides of this expression and substituting the ratio of forces Fmax/ Fmin = 1.11 yields the ratio of distances: rmax Fmax = rmin Fmin = 1.11 = 1.05 The moon’s maximum distance from the center of the planet is therefore about 5% larger than its minimum distance. 37. REASONING We place the third particle (mass = m3) as shown in the following drawing: L D m3 m 2m The magnitude of the gravitational force that one particle exerts on another is given by Newton’s law of gravitation as F = Gm1m2/r2. Before the third particle is in place, this law indicates that the force on each particle has a magnitude Fbefore = Gm2m/L2. After the third particle is in place, each of the first two particles experiences a greater net force, because the third particle also exerts a gravitational force on them. SOLUTION For the particle of mass m, we have Gmm3 Fafter Fbefore = Gm 2m 2 D L2 = L m3 + 1 Gm 2m 2mD 2 2 L 2 + For the particle of mass 2m, we have G 2mm3 Fafter Fbefore = ( L – D) 2 + Gm2m L2 Gm2m L2 = L2 m3 m ( L – D) 2 +1 Since Fafter/Fbefore = 2 for both particles, we have L2 m3 2mD 2 +1 = L2 m3 m ( L – D) 2 +1 or 2D2 = ( L – D ) 2 184 FORCES AND NEWTON'S LAWS OF MOTION Expanding and rearranging this result gives D 2 + 2 LD − L2 = 0 , which can be solved for D using the quadratic formula: D= –2 L ± ( 2 L )2 – 4 (1) ( – L2 ) = 2 (1)...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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