An object with a mass of #9# #kg# is on a plane with an incline of #- pi/4 #. If it takes #15# #N# to start pushing the object down the plane and #7# #N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Apr 25, 2016

Answer:

The coefficient of static friction is #mu_s=1.24# and the coefficient of kinetic friction is #mu_k=1.11#.

Explanation:

The negative sign on the angle is not relevant in this case: the angle of the inclined plane is #pi/4# #radians# (#45^o#). The negative sign just shows that the slope increases right-to-left rather than left-to-right.

The force perpendicular to the plane, the normal force, will be given by:

#F_N=mgsinTheta=9*9.8*sin(pi/4)=62.4# #N#

Based on this, the frictional force will be:

#F_"frict"=muF_N#

Rearranging,

#mu=(F_"frict")/(F_N)#

We will find two different values, the static friction, #mu_s#, and the kinetic friction, #mu_k#.

The force parallel to the plane due to gravity will be given by:

#F=mgcosTheta=9*9.8*cos(pi/4)=62.4# #N#

Overcoming the force of static friction requires this parallel force plus #15# #N#, a total frictional force of #77.4# #N#.

#mu_s=(F_"frict")/(F_N)=77.4/62.4=1.24#

Once it is moving, keeping it moving only requires #7# #N# in addition to the parallel force, a total of #69.4# #N#.

#mu_k=(F_"frict")/(F_N)=69.4/62.4=1.11#

Coefficients of friction do not have units.