An object with a mass of #8 kg# is on a plane with an incline of # - pi/3 #. If it takes #15 N# to start pushing the object down the plane and #12 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

If angle is #pi/6#:

#mu_s = 0.798#

#mu_k = 0.754#

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

Since the angle is #-(pi)/3#, 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)/3 = ul((pi)/6#

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 = (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/6) = color(orange)(ul(68.0color(white)(l)"N"#

Since #15# #"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 + 15# #"N"#

#= (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/6) + 15color(white)(l)"N" = color(green)(54.2color(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(" ")(" " 0.798" ")|#

The coefficient of kinetic friction #mu_k# is given by

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

It takes #8# #"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 + 8# #"N"#

#= 39.2color(white)(l)"N" + 12# #"N"# #= color(purple)(51.2color(white)(l)"N"#

The coefficient of kinetic friction is thus

#mu_k = (f_k)/n = (color(purple)(51.2)cancel(color(purple)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.754" ")|#

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Answer 2

To calculate the coefficients of static and kinetic friction, we need to use the equations involving force and friction on an inclined plane. The force required to start pushing the object down the plane is equal to the product of the coefficient of static friction and the normal force exerted on the object. Similarly, the force required to keep pushing the object once it's in motion is equal to the product of the coefficient of kinetic friction and the normal force.

Given that it takes 15 N to start pushing the object down the plane and 12 N to keep pushing it, and the mass of the object is 8 kg, we can calculate the normal force using the formula:

Normal force (N) = mass (kg) * gravitational acceleration (m/s^2)

Then, we can use the force equations involving friction to solve for the coefficients of static and kinetic friction.

First, we solve for the normal force:

Normal force = 8 kg * 9.8 m/s^2 = 78.4 N

To find the coefficient of static friction:

15 N = coefficient of static friction * 78.4 N

Coefficient of static friction = 15 N / 78.4 N

To find the coefficient of kinetic friction:

12 N = coefficient of kinetic friction * 78.4 N

Coefficient of kinetic friction = 12 N / 78.4 N

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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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