An object with a mass of #2 kg# is on a plane with an incline of # - pi/12 #. If it takes #1 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?

Question

An object with a mass of #2   kg# is on a plane with an incline of # - pi/12  #.  If it takes #1  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?

Leo Abeyta · Accepted Answer

If angle of inclination is #(5pi)/12#:

#mu_s = 3.93#

#mu_k = 4.72#

(see explanation regarding angle)

We're asked to find the coefficient of static friction #mu_s# and the coefficient of kinetic friction #mu_k#, with some given information.

We'll call the positive #x#-direction up the incline (the direction of #f# in the image), and the positive #y# direction perpendicular to the incline plane (in the direction of #N#).

There is no net vertical force, so we'll look at the horizontal forces (we WILL use the normal force magnitude #n#, which is denoted #N# in the above image).

We're given that the object's mass is #2# #"kg"#, and the incline is #-(pi)/12#.

Since the angle is #-(pi)/12# (#-15^"o"#), this would be the angle going down the incline (the topmost angle in the image above). Therefore, the actual angle of inclination is

#pi/2 - (pi)/12 = ul((5pi)/12#

The formula for the coefficient of static friction #mu_s# is

#f_s >= mu_sn#

Since the object in this problem "breaks loose" and the static friction eventually gives way, this equation is simply

#f_s = mu_sn#

Since the two vertical quantities #n# and #mgcostheta# are equal,

#n = mgcostheta = (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos((5pi)/12) = 5.08# #"N"#

Since #1# #"N"# is the "breaking point" force that causes it to move, it is this value plus #mgsintheta# that equals the upward static friction force #f_s#:

#f_s = mgsintheta + 1# #"N"#

#= (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin((5pi)/12) + 1color(white)(l)"N" = 20.0# #"N"#

The coefficient of static friction is thus

#mu_s = (f_s)/n = (20.0cancel("N"))/(5.08cancel("N")) = color(red)(ul(3.93#

The coefficient of kinetic friction #mu_k# is given by

#f_k = mu_kn#

It takes #5# #"N"# of applied downward force (on top of weight) to keep the object accelerating constantly downward, then we have

#f_k = mgsintheta + 5# #"N"#

#= 19.0color(white)(l)"N" + 5# #"N"# #= 24.0# #"N"#

The coefficient of kinetic friction is thus

#mu_k = (f_k)/n = (24.0cancel("N"))/(5.08cancel("N")) = color(blue)(ul(4.72#

These values are fairly high, and it's also interesting to see that the kinetic friction force is greater than the static friction force...but this is merely theoretical!

Nathan Ashcroft · Answer

undefined The coefficient of static friction, $ \mu_s $, can be calculated using the formula:

$$ \mu_s = \tan(\theta) $$

Where:
$ \theta = -\frac{\pi}{12} $

Substituting the given value, we get:

$$ \mu_s = \tan\left(-\frac{\pi}{12}\right) $$

$$ \mu_s ≈ -0.2679 $$

The coefficient of kinetic friction, $ \mu_k $, can be calculated using the formula:

$$ \mu_k = \frac{F_k}{N} $$

Where:
$ F_k = 5 \, \text{N} $ (force required to keep pushing the object)
$ N = mg \cos(\theta) $ (normal force)
$ m = 2 \, \text{kg} $ (mass of the object)
$ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity)
$ \theta = -\frac{\pi}{12} $

Substituting the given values, we first find the normal force:

$$ N = mg \cos\left(-\frac{\pi}{12}\right) $$

$$ N = 2 \times 9.8 \times \cos\left(-\frac{\pi}{12}\right) $$

$$ N ≈ 19.349 \, \text{N} $$

Then, we can calculate $ \mu_k $:

$$ \mu_k = \frac{5}{19.349} $$

$$ \mu_k ≈ 0.2582 $$

Therefore, the coefficient of static friction is approximately -0.2679, and the coefficient of kinetic friction is approximately 0.2582.