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Unformatted text preview: 533 Chapter 12 1. (a) The center of mass is given by com ( )(2.00 m) ( )(2.00 m) ( )(2.00 m) 1.00 m. 6 m m m x m + + + + + = = (b) Similarly, we have com ( )(2.00 m) ( )(4.00 m) ( )(4.00 m) ( )(2.00 m) 2.00 m. 6 m m m m y m + + + + + = = (c) Using Eq. 1214 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, 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. 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 r F is exerted, we find (since the acceleration is zero) 2 T sin θ = F , where θ 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 θ = 30º. Since α = 180º – 2 θ is the angle between the two segments, then we find α = 120º. 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 r F is exerted, we find (since the acceleration is zero) 2 T sin θ = F , where θ 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 θ = = ° CHAPTER 12 534 Therefore, T = F /(2sin θ ) = 7.92 × 10 3 N. 4. From r r r τ = × r F , 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. 5. Three forces act on the sphere: the tension force r T of the rope (acting along the rope), the force of the wall N F r (acting horizontally away from the wall), and the force of gravity mg r (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 θ – mg = 0 horizontal component: F N – T sin θ = 0....
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 Summer '10
 Prof.Yang
 Physics, Center Of Mass, Force, Friction, Mass

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