22231122(80kg)(9.8m/s )(3.5m)+(60kg)(9.8m/s )(2.5m)5.0m8.4 10 N.wsmgF+==+=×AA(b) Equilibrium of forces leads to (60kg+80kg)(9.8m/s ) 1.4 10 NswFF mgmg+=+=which (using our result from part (a)) yieldsF2253 10.N.8. Let A115=.mand2(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.