221sin FTmgmaTmaθ−−==which can be added to obtain F– m2gsin = (m1+ m2)a. This yields the acceleration 2212 N (1.0 kg)(9.8 m/s )sin 371.53 m/s .1.0 kg 3.0 kga−°==+Thus, the tension is T= m1a= (3.0 kg)(1.53 m/s2) = 4.6 N. 82. We take +xuphill for the m2= 1.0 kg box and +xrightward for the m1= 3.0 kg box (so the accelerations of the two boxes have the same magnitude and the same sign). The
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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.