# An object with a mass of #18 kg# is on a plane with an incline of # -(5 pi)/12 #. If it takes #9 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

**If angle of inclination is #pi/12#:**

#mu_s = 0.321#

#mu_k = 0.297#

**See explanation regarding angle.**

#### Explanation:

We're asked to find the coefficient of static friction

We'll call the positive *up* the incline (the direction of

There is no net vertical force, so we'll look at the **horizontal** forces (we WILL use the normal force magnitude

We're given that the object's mass is

Since the angle is

#-(5pi)/12# , this would be the angle goingdownthe incline (the topmost angle in the image above). Therefore, theactualangle ofinclinationis

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

The formula for the coefficient of static friction

#f_s <= mu_sn#

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

#color(green)(ul(f_s = mu_sn#

Since the two vertical quantities

#n = mgcostheta = (18color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/12) = color(orange)(ul(171color(white)(l)"N"#

Since

#color(green)(f_s) = mgsintheta + 9# #"N"#

#= (18color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) + 9color(white)(l)"N" = color(green)(ul(54.7color(white)(l)"N"#

The coefficient of static friction is thus

#mu_s = (f_s)/n = (color(green)(54.2)cancel(color(green)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 0.321" ")|#

The coefficient of kinetic friction

#color(purple)(ul(f_k = mu_kn#

It takes

#color(purple)(f_k) = mgsintheta + 5# #"N"#

#= 45.7color(white)(l)"N" + 5# #"N"# #= color(purple)(50.7color(white)(l)"N"#

The coefficient of kinetic friction is thus

#mu_k = (f_k)/n = (color(purple)(50.7)cancel(color(purple)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.297" ")|#