12_P65InstructorSolution

12_P65InstructorSolution - 12.65. Visualize: Please refer...

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12.65. Visualize: Please refer to Figure P12.65. Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by 2. L For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of the brick 2, the center of gravity of the top two bricks is 11 2 2 12 com 12 3 2 () . 24 L mm L mx mx x L m ⎛⎞ + ⎜⎟ + ⎝⎠ == = + Thus the top two bricks can extend 4 L past the edge of the third brick. The top three bricks have a center of mass 33 123 com 123 35 5 4 . 36 LLL m mx x
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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