Phys-chapter 9

Phys-chapter 9 - PHYSICS 111 SPRING 2008 HOMEWORK#14...

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Unformatted text preview: PHYSICS 111 SPRING 2008 HOMEWORK #14 SOLUTION Chap 9: Q10,18; P6,30,40,58 Chapter 9, Question 10 . The drawing shows a wine rack for a single bottle of wine that seems to defy common sense as it balances on a table top. Treat the rack and wine bottle as a rigid body and draw the external forces that keep it in equilibrium. In particular, where must the center of gravity of the rigid body be located? Solution . Like the center of gravity of any object in stable equilibrium, the center of gravity (i.e., center of mass) of the rigid body must be located above the base of the rack. Then, if the bottle is nudged to the right or the left, the center of gravity will rise, and cause the bottle and rack to return to the equilibrium postion. To see this, consider the net torque on the rack + bottle. The net torque is the sum of the torque produced by gravity at the center of gravity, plus the torque produced by the normal force of the table at its point of contact with the rack. If we tilt the rack to the right and calculate the net torque relative to the right edge of the base, then the normal force will not contribute a torque because its lever arm will be zero. The center of gravity will contribute the only nonzero torque to cause a counterclockwise rotation back to the horizontal position. Similarly, if the bottle is tilted to the left, the net torque relative to the left edge of the base will cause a clockwise rotation back to the horizontal position....
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This note was uploaded on 03/27/2008 for the course PHYS 111 taught by Professor Shinar during the Spring '08 term at Iowa State.

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Phys-chapter 9 - PHYSICS 111 SPRING 2008 HOMEWORK#14...

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