An object with a mass of #4 kg# is on a plane with an incline of # - pi/8 #. If it takes #18 N# to start pushing the object down the plane and #5 N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Jan 11, 2016

Answer:

I found #0.91 and 0.55# (but check my maths!).

Explanation:

Have a look at the diagram:
enter image source here
I assumed that the negative angle represents the downward inclination of the surface and transferred it as #theta# as in figure (I may be wrong though...).
With this convention for the angle and Newton's law:

#N=W_y=Wcos(theta)=mgcos(theta)=4*9.8*cos(pi/8)=36.2N#

Also: #W_x=Wsin(theta)=4*9.8*sin(pi/8)=15N#

Now:

1) Static Friction:
I consider the object still at rest but almost moving. Acceleration will be zero and we will be at maximum static friction where it will exactly be: #f_s=mu_sN#

So, we use Newton's law along #x# with #a=0# to get:

#F-f_s+W_x=0#
#color(red)(18)-(mu_s*36.2)+15=0#
#mu_s=0.91#

2) Kinetic Friction:
From our graph of friction we can see that Kinetic friction will be almost constant and so:
#f_k=mu_kN#

Here we can assume that the object will move with constant velocity down the incline giving again zero acceleration. As before we get, using Newton's law along #x# with #a=0#:

#F-f_k+W_x=0#
#color(red)(5)-(mu_k*36.2)+15=0#
#mu_k=0.55#