Question

Question #a8cb9

Answer 1

`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`