An object with a mass of $6 k g$ $6 kg$ is on a plane with an incline of $- \frac{3 π}{8}$ $-(3 pi)/8$ . If it takes $15 N$ $15 N$ to start pushing the object down the plane and $5 N$ $5 N$ to keep pushing it, what are the coefficients of static and kinetic friction?

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An object with a mass of $6 k g$ $6 kg$ is on a plane with an incline of $- \frac{3 π}{8}$ $-(3 pi)/8$ . If it takes $15 N$ $15 N$ to start pushing the object down the plane and $5 N$ $5 N$ to keep pushing it, what are the coefficients of static and kinetic friction?

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Answer 1 · 2017-09-02T14:01:13

If angle of inclination is $\frac{π}{8}$ $pi/8$ :

$μ_{s} = 0.690$ $mu_s = 0.690$

$μ_{k} = 0.605$ $mu_k = 0.605$

See explanation regarding angle.

Explanation:

We're asked to find the coefficient of static friction $μ_{s}$ $mu_s$ and the coefficient of kinetic friction $μ_{k}$ $mu_k$ , with some given information.

![upload.wikimedia.org]( useruploads.socratic.org )

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

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

We're given that the object's mass is $14$ $14$ $kg$ $"kg"$ , and the incline is $- \frac{3 π}{8}$ $-(3pi)/8$ .

Since the angle is $- \frac{3 π}{8}$ $-(3pi)/8$ , this would be the angle going down the incline (the topmost angle in the image above). Therefore, the actual angle of inclination is

$pi/2 - (3pi)/8 = ul((pi)/8$

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,

$n = mgcostheta = (6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/8) = color(orange)(ul(54.4color(white)(l)"N"$

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 $f_s$ :

$color(green)(f_s) = mgsintheta + 15$ $"N"$

$= (6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) + 15color(white)(l)"N" = color(green)(37.5color(white)(l)"N"$

The coefficient of static friction is thus

$mu_s = (f_s)/n = (color(green)(37.5)cancel(color(green)("N")))/(color(orange)(54.4)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 0.690" ")|$

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

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

$= 22.5color(white)(l)"N" + 5$ $"N"$ $= color(purple)(27.5color(white)(l)"N"$

The coefficient of kinetic friction is thus

$mu_k = (f_k)/n = (color(purple)(27.5)cancel(color(purple)("N")))/(color(orange)(119)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.506" ")|$

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An object with a mass of $6 k g$ $6 kg$ is on a plane with an incline of $- \frac{3 π}{8}$ $-(3 pi)/8$ . If it takes $15 N$ $15 N$ to start pushing the object down the plane and $5 N$ $5 N$ to keep pushing it, what are the coefficients of static and kinetic friction?

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Explanation:

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1 Answer

Explanation:

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An object with a mass of $6 k g$ $6 kg$ is on a plane with an incline of $- \frac{3 π}{8}$ $-(3 pi)/8$ . If it takes $15 N$ $15 N$ to start pushing the object down the plane and $5 N$ $5 N$ to keep pushing it, what are the coefficients of static and kinetic friction?