ch06-p033 - 33 The free-body diagrams for the two blocks...

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These expressions are combined (to eliminate F' ) and we arrive at F mg m mM s = + F H G I K J μ 1 which we find to be F = 4.9 × 10 2 N. 33. The free-body diagrams for the two blocks, treated individually, are shown below (first m and then M ). F' is the contact force between the two blocks, and the static friction force G f s is at its maximum value (so Eq. 6-1 leads to f s = f s ,max = s F' where s = 0.38). Treating the two blocks together as a single system (sliding across a frictionless floor), we apply Newton’s second law (with +
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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