An object with a mass of #10 kg# is on a plane with an incline of # - pi/4 #. If it takes #12 N# to start pushing the object down the plane and #8 N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Oct 1, 2017

Answer:

# mu_k ~~ 0.113#
# mu_s ~~ 0.169 #

Explanation:

The minimum force to start pushing will help to overcome the static friction, so we can calculate the static friction from there.

Self created image, ( using MS paint )

The Gravitational force acting on the block :
# F_g = m*g = 10 * 10= 100 N#

Taking components along the incline and perpendicular to the incline, # F_g cos θ = F_g cos 45° = 50sqrt(2) #
and, # F_g sin θ = F_g sin 45° = 50sqrt(2) #

Since there is no motion perpendicular to the incline, the net force in that direction will be zero.

# F_g cos θ = F_n = 50sqrt(2) # ( where #F_n# is normal force. )

Static friction, #f_s = mu_s * F_n #

According to the question, the value of static friction force happens to be 12N.

# implies 12 = mu_s * ( 50 sqrt(2)) #

# implies mu_s ~~ 0.169 #

Now the kinetic friction will be applicable when the body starts to move.

Kinetic friction, #f_k = mu_k * F_n #

According to the question, the value of static friction force happens to be 8N.

# implies 8 = mu_k * ( 50 sqrt(2)) #

# implies mu_k ~~ 0.113 #