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

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1. (a) The center of mass is given by x com = [0 + 0 + 0 + ( m )(2.00) + ( m )(2.00) + ( m )(2.00)]/6.00 m = 1.00 m. (b) Similarly, y com = [0 + ( m )(2.00) + ( m )(4.00) + ( m )(4.00) + ( m )(2.00) + 0]/6 m = 2.00 m. (c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have x cog = x 1 m 1 g 1 + x 2 m 2 g 2 + x 3 m 3 g 3 + x 4 m 4 g 4 + x 5 m 5 g 5 + x 6 m 6 g 6 m 1 g 1 + m 2 g 2 + m 3 g 3 + m 4 g 4 + m 5 g 5 + m 6 g 6 = 0.987 m. (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. From & & & 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.
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 & 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 § · = = ° ¨ ¸ © ¹ Therefore, T = F /(2sin θ )= 7.92 × 10 3 N.

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4. 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 & 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.
5. 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 θ be the angle between the rope and the vertical. Then, the vertical component of Newton’s second law is T cos θ mg = 0. The horizontal component is F N T sin θ = 0. (a) We solve the first equation for the tension: T = mg / cos θ . We substitute cos θ = + L L r / 2 2 to obtain 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: sin N F T θ = . Using sin θ = + r L r / 2 2 , we obtain 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 + = = = = = + +

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6. Let " 1 15 = . m and 2 (5.0 1.5) m 3.5 m = = " . We denote tension in the cable closer to the window as F 1 and that in the other cable as F 2 . The force of gravity on the scaffold itself (of magnitude m s g ) is at its midpoint, " 3 2 5 = . m from either end.
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