Question #4fdc1

1 Answer
Oct 7, 2017

drawn

The situation as described in the problem has been shown in the free body diagram shown above .

Here

  • Mass of the box #m=50kg#
  • Acceleration due to gravity #g=9.8m"/"s^2#
  • Angle of inclination of the inclined plane #theta=30^@#
  • Coefficient of static friction #mu=??#

The frictional force acting on the block when it slides downward is
#f=mumgcostheta#

Considering the equilibrium condition at the position when the block tends to slide downward under gravity, we can write

#mumgcostheta=mgsintheta#

#=>mu=tantheta=tan30=1/sqrt3#