SM_chapter12 - 12 Static Equilibrium and Elasticity CHAPTER...

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12 Static Equilibrium and Elasticity CHAPTER OUTLINE 12.1 The Conditions for Equilibrium 12.2 More on the Center of Gravity 12.3 Examples of Rigid Objects in Static Equilibrium 12.4 Elastic Properties of Solids ANSWERS TO QUESTIONS *Q12.1 The force exerts counterclockwise torque about pivot D. The line of action of the force passes through C, so the torque about this axis is zero. In order of increas- ing negative (clockwise) values come the torques about F, E and B essentially together, and A. The answer is then τ D > τ C > τ F > τ E = τ B > τ A Q12.2 When you bend over, your center of gravity shifts forward. Once your CG is no longer over your feet, gravity contributes to a nonzero net torque on your body and you begin to rotate. Q12.3 Yes, it can. Consider an object on a spring oscillating back and forth. In the center of the motion both the sum of the torques and the sum of the forces acting on the object are (separately) zero. Again, a meteoroid flying freely through interstellar space feels essentially no forces and keeps moving with constant velocity. Q12.4 (a) Consider pushing up with one hand on one side of a steering wheel and pulling down equally hard with the other hand on the other side. A pair of equal-magnitude oppositely- directed forces applied at different points is called a couple. (b) An object in free fall has a non-zero net force acting on it, but a net torque of zero about its center of mass. *Q12.5 Answer (a). Our theory of rotational motion does not contradict our previous theory of transla- tional motion. The center of mass of the object moves as if the object were a particle, with all of the forces applied there. This is true whether the object is starting to rotate or not. Q12.6 A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide at the edge of a swimming pool each have a center of mass that is not within the bulk of the object. Q12.7 Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywood along the string of the plumb bob. Now suspend the plywood with the nail through a different point on the plywood, not along the first line you drew. Again hang the plumb bob from the nail and trace along the string. The center of gravity is located halfway through the thickness of the plywood under the intersection of the two lines you drew. *Q12.8 In cases (a) and (c) the center of gravity is above the base by one-half the height of the can. So (b) is the answer. In this case the center of gravity is above the base by only a bit more than one-quarter of the height of the can. 311 13794_12_ch12_p311-336.indd 311 13794_12_ch12_p311-336.indd 311 1/8/07 8:21:33 PM 1/8/07 8:21:33 PM
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*Q12.9 Answer (b). The skyscraper is about 300 m tall. The gravitational field (acceleration) is weaker at the top by about 900 parts in ten million, by on the order of 10 4 times. The top half of the uniform building is lighter than the bottom half by about (1 2)(10 4 ) times. Relative to the center of mass at the geometric center, this effect moves the center of gravity down, by about (1 2)(10 4 )(150 m) 10 mm.
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