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# ch12 - Chapter 12 1(a The center of mass is given by xcom 0...

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519 Chapter 12 1. (a) The center of mass is given by com 0 0 0 ( )(2.00 m) ( )(2.00 m) ( )(2.00 m) 1.00 m. 6 m m m x m (b) Similarly, we have com 0 ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 0 2.00 m. 6 m m m m y m (c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have 6 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 cog 6 1 1 2 2 3 3 4 4 5 5 6 6 1 0.987 m. i i i i i i i x m g x m g x m g x m g x m g x m g x m g x m g m g m g m g m g m g m g (d) Similarly, we have 6 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 cog 6 1 1 2 2 3 3 4 4 5 5 6 6 1 0 (2.00)(7.80 ) (4.00)(7.60 ) (4.00)(7.40 ) (2.00)(7.60 ) 0 8.0 7.80 7.60 7.40 7.60 7.80 1.97 m. i i i i i i i y m g y m g y m g y m g y m g y m g y m g y m g m g m g m g m g m g m g m m m m m m m m m m 2. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.35 m is the horizontal distance between the axles; (3.35 1.85) m 1.50 m is the horizontal distance from the rear axle to the center of mass; F 1 is the force exerted on each front wheel; and F 2 is the force exerted on each back wheel. (a) Taking torques about the rear axle, we find 2 3 1 (1360kg)(9.80m/s )(1.50m) 2.98 10 N. 2 2(3.35m) Mg F L

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CHAPTER 12 520 (b) Equilibrium of forces leads to 1 2 2 2 , F F Mg from which we obtain 3 2 3.68 10 N F . 3. Three forces act on the sphere: the tension force T of the rope (acting along the rope), the force of the wall N F (acting horizontally away from the wall), and the force of gravity mg (acting downward). Since the sphere is in equilibrium they sum to zero. Let be the angle between the rope and the vertical. Then Newton’s second law gives vertical component : T cos mg = 0 horizontal component: F N T sin = 0. (a) We solve the first equation for the tension and obtain T = mg / cos . We then substitute cos L L r / 2 2 : 2 2 2 2 2 (0.85 kg)(9.8 m/s ) (0.080 m) (0.042 m) 9.4 N 0.080 m mg L r T L . (b) We solve the second equation for the normal force and obtain sin N F T . Using sin r L r / 2 2 , we have 2 2 2 2 2 2 2 (0.85 kg)(9.8 m/s )(0.042 m) 4.4 N. (0.080 m) N Tr mg L r r mgr F L L L r L r 4. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2 T sin = F , where is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). Setting T = F therefore yields = 30º. Since = 180º – 2 is the angle between the two segments, then we find = 120º.
521 5. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a “kink” similar to that shown for problem 10 (see the figure that accompanies that problem in the text). By analyzing the forces at the “kink” where F is

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