Chapter12 - 1. (a) The center of mass is given by xcom = 0...

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1. (a) The center of mass is given by com 000() ( 2 . 0 0 m )() ( 2 . 0 0 m ( 2 . 0 0 m ) 1.00 m. 6 mmm 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 mmmm 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 1 111 222 333 444 555 666 cog 6 11 22 33 44 55 66 1 0.987 m. iii i ii i xmg xmg xmg x mg mg mg = = = ¦ ¦ (d) Similarly, y cog = [0 + (2.00)( m )(7.80) + (4.00)( m )(7.60) + (4.00)( m )(7.40) + (2.00)( m )(7.60) + 0]/(8.00 m + 7.80 m + 7.60 m + 7.40 m + 7.60 m + 7.80 m ) = 1.97 m.
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2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where G F is exerted, we find (since the acceleration is zero) 2 T sin θ = 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 α = 180º – 2 θ is the angle between the two segments, then we find = 120º.
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3. 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). By analyzing the forces at the “kink” where G F is exerted, we find (since the acceleration is zero) 2 T sin θ = F , where is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are colinear). In this problem, we have 1 0.35m tan 11.5 . 1.72m §· == ° ¨¸ ©¹ Therefore, T = F /(2sin ) = 7.92 × 10 3 N.
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4. From G G τ rF , we note that persons 1 through 4 exert torques pointing out of the page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the page. (a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 2. (b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 7. G
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5. Three forces act on the sphere: the tension force G T of the rope (acting along the rope), the force of the wall N F G (acting horizontally away from the wall), and the force of gravity mg G (acting downward). Since the sphere is in equilibrium they sum to zero. Let θ 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: T = mg / cos . We substitute cos =+ LL r / 22 to obtain 222 (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: sin N FT = . Using sin rL r / , we obtain 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 Lr + = = = ++
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6. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is the horizontal distance between the axles; (3.05 1.78) m 1.27 m =− = A 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.
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Chapter12 - 1. (a) The center of mass is given by xcom = 0...

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