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?

Answer 1

If angle of inclination is #pi/12#:

#mu_s = 0.321#

#mu_k = 0.297#

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 #18# #"kg"#, and the incline is #-(pi)/3#.

Since the angle is #-(5pi)/12#, 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 - (5pi)/12 = ul((pi)/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

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

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

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

Since #9# #"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#:

#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 #mu_k# is given by

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

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

#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" ")|#

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Answer 2

Static friction coefficient: ( \mu_s = \tan\left(\frac{\pi}{2} - \theta\right) )

Kinetic friction coefficient: ( \mu_k = \tan\left(\frac{\pi}{2} - \arctan\left(\frac{f_k}{f_n}\right)\right) )

Where: ( \theta = -\frac{5\pi}{12} ) (angle of the incline) ( f_k = 5 , \text{N} ) (force to keep pushing) ( f_n = mg \cos(\theta) ) (normal force) ( m = 18 , \text{kg} ) (mass of the object) ( g = 9.8 , \text{m/s}^2 ) (acceleration due to gravity)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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