N=Mgcosα:NormalReactionFR=μMgcosα:Frictionfroce
Writing equation of motion.......
Ma=Mgsinα−μMgcosα Where "a" is the acceleration of the block which is zero since block is moving with constant velocity. Hence......
μ=tanα=tan23°μ=0.424..........(a)Ans And Hence
FR=.424×6.5×9.8×cos23°FR=24.861N...........(b)Ans Now part (c)
μ→43μ Now again writing equation of motion,
Ma′=Mgsinα−43μMgcosαa′=gsinα−43μgcosαa′=9.8×sin23°−43×0.424×9.8×cos23°a′=3.829−2.869=0.96meter/sec we see that if board is lubricated then block starts accelerating.
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