ch12-p059 - 59. (a) The center of mass of the top brick...

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x mm L m L com = 30+ / 2 4 = 8 b g b g shows that a 4 = L /8. (e) We find 4 1 25 / 24 i i ha L = == ¦ . 59. (a) The center of mass of the top brick cannot be further (to the right) with respect to the brick below it (brick 2) than L /2; otherwise, its center of gravity is past any point of support and it will fall. So a 1 = L /2 in the maximum case. (b) With brick 1 (the top brick) in the maximum situation, then the combined center of mass of brick 1 and brick 2 is halfway between the middle of brick 2 and its right edge. That point (the combined com) must be supported, so in the maximum case, it is just
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