ch12-p008 - . 8. Let 1 = 15 m and 2 = (5.0 1.5) m = 3.5 m ....

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22 23 1 12 2 (80kg)(9.8m/s )(3.5m)+(60kg)(9.8m/s )(2.5m) 5.0m 8.4 10 N. ws mg F + == + AA (b) Equilibrium of forces leads to (60kg+80kg)(9.8m/s ) 1.4 10 N sw FF m gm g += + = which (using our result from part (a)) yields F 2 2 53 10 .N . 8. Let A 1 15 = .ma n d 2 (5.0 1.5) m 3.5 m =− = A . We denote tension in the cable closer to the window as
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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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