Question #a8cb9

1 Answer
Dec 4, 2016

Answer:

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

Explanation:

#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#