Chapter 12 Student Solutions Manual

Chapter 12 Student Solutions Manual - Chapter 12 Student...

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Chapter 12 – Student Solutions Manual 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 h away from the wall), and the force of gravity mg orizontally G (acting downward). Since the sphere is in equilibrium they sum to zero. Le θ be the angle between the rope and the vertical. Then Newton’s second law gives t 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 to obtain =+ LL r / 22 222 (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 + = = = ++ 7. We take the force of the left pedestal to be F 1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F 2 and denote its position as x = d . W is the weight of the diver, located at x = L . The following two equations result from setting the sum of forces equal to zero (with upwards positive), and the sum of torques (about x 2 ) equal to zero: 12 1 0 () FFW Fd W L d 0 + −= + (a) The second equation gives 1 3.0m (580 N)= 1160 N 1.5m Ld FW d ⎛⎞ =− ⎜⎟ ⎝⎠ which should be rounded off to . Thus, || 3 1 1.2 10 N F × 3 1 1 . 2 1 0 N . F (b) Since F 1 is negative, indicating that this force is downward.
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(c) The first equation gives 21 580 N+1160 N=1740 N FWF =−= which should be rounded off to . Thus, 3 2 1.7 10 N F 3 2 || 1 . 7 1 0 N F . (d) The result is positive, indicating that this force is upward. (e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched. (f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed. 11. The x axis is along the meter stick, with the origin at the zero position on the scale. The forces acting on it are shown on the diagram below. The nickels are at x = x 1 = 0.120 m, and m is their total mass. The knife edge is at x = x 2 = 0.455 m and exerts force G . The mass of the meter stick is M , and the force of gravity acts at the center of the stick, x = x F 3 = 0.500 m. Since the meter stick is in equilibrium, the sum of the torques about x 2 must vanish: Mg ( x 3 x 2 ) – mg ( x 2 x 1 ) = 0. Thus, 32 0.455m 0.120m (10.0g)=74.4 g. 0.500m 0.455m xx Mm ⎛⎞ −− == ⎜⎟ ⎝⎠ 21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the
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This homework help was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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Chapter 12 Student Solutions Manual - Chapter 12 Student...

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