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concentrated force located at the edge of the bottom block (which is the point about which we compute torques, in the following). If (as indicated in our sketch, where G F top has magnitude mg /2) we consider equilibrium of torques on the rightmost brick, we obtain mg b L mg Lb 11 1 22 − F H G I K J =− () which leads to b 1 = 2 L /3. Once we conclude from symmetry that b 2 = L /2 then we also arrive at h = b 2 + b 1 = 7 L /6. 75. We locate the origin of the x axis at the edge of the table and choose rightwards positive. The criterion (in part (a)) is that the center of mass of the block above another must be no further than the edge of the one below; the criterion in part (b) is more subtle and is discussed below. Since the edge of the table corresponds to x = 0 then the total center of mass of the blocks must be zero. (a) We treat this as three items: one on the upper left (composed of two bricks, one directly on top of the other) of mass 2 m whose center is above the left edge of the bottom brick; a single brick at the upper right of mass m which necessarily has its center over the
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