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Answer 1 · 2016-12-04T02:23:38

$mu_s(7.5xx9.8-18sin38^@)N$

$F-> "Applied force "=18N$

$alpha-> "angle of applied force with the horizontal "=38^@$

$F_v->"vertical component of applied force"=Fsinalpha$

$F_h->"horizontal component of applied force"=Fcosalpha$

$m->"mass of the object"=7.5kg$

$N->"Normal reaction"=mg-Fsinalpha$

$F_"fric"->"Limiting value of friction" =mu_sxxN$

If horizontal component of the applied force represents the limiting value of the frictional force then

$F_"fric"=F_h$

$=>mu_sxxN=Fcosalpha$

$=>mu_s=(Fcosalpha)/N=(Fcosalpha)/(mg-Fsinalpha)$

$=(18xxcos38)/(7.5xx9.8-18xxsin38)~~0.23$