cos60cos20sin 60sin 20TTTWT′°=°′+°horizontal forcesvertical forces.(a) We solve the above simultaneous equations and find 15N.tan 60 cos20sin 20WT==°°−°(b) Also, we obtain T´= Tcos 20º / cos 60º = 29 N. 68. We denote the tension in the upper left string (
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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.