19. (a) Since the acceleration of the block is zero, the components of the Newton’s second law equation yieldT– mgsin θ= 0 FN–mgcos = 0. Solving the first equation for the tension in the string, we find Tmg==°=sin..859 830422kgm / sN .bgch(b) We solve the second equation in part (a) for the normal force FN:()2cos8.5 kg 9.8 m/scos 3072 N .NFmg°=(c) When the string is cut, it no longer exerts a force on the block and the block
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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.