An object with a mass of $8 k g$ $8 kg$ is on a plane with an incline of $- \frac{π}{3}$ $- pi/3$ . If it takes $9 N$ $9 N$ to start pushing the object down the plane and $7 N$ $7 N$ to keep pushing it, what are the coefficients of static and kinetic friction?

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An object with a mass of $8 k g$ $8 kg$ is on a plane with an incline of $- \frac{π}{3}$ $- pi/3$ . If it takes $9 N$ $9 N$ to start pushing the object down the plane and $7 N$ $7 N$ to keep pushing it, what are the coefficients of static and kinetic friction?

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Answer 1 · 2017-08-07T16:32:09

If angle of inclination is $\frac{π}{6}$ $pi/6$ :

$μ_{s} = 0.710$ $mu_s = 0.710$

$μ_{k} = 0.680$ $mu_k = 0.680$

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

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 incline 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 $8$ $8$ $kg$ $"kg"$ , and the incline is $- \frac{π}{3}$ $-(pi)/3$ .

Since the angle is $- \frac{π}{3}$ $-(pi)/3$ , 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 $9$ $"N"$ is the "breaking point" force that causes it to move, it is this value plus $mgsintheta$ that equals the upward static friction force $f_s$ :

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

$= (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/6) + 9color(white)(l)"N" = color(green)(48.2color(white)(l)"N"$

The coefficient of static friction is thus

$mu_s = (f_s)/n = (color(green)(48.2)cancel(color(green)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 0.710" ")|$

The coefficient of kinetic friction $mu_k$ is given by

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

It takes $7$ $"N"$ of applied downward force (on top of the object's weight) to keep the object accelerating constantly downward, then we have

$color(purple)(f_k) = mgsintheta + 7$ $"N"$

$= 39.2color(white)(l)"N" + 7$ $"N"$ $= color(purple)(46.2color(white)(l)"N"$

The coefficient of kinetic friction is thus

$mu_k = (f_k)/n = (color(purple)(46.2)cancel(color(purple)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.680" ")|$

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An object with a mass of $8 k g$ $8 kg$ is on a plane with an incline of $- \frac{π}{3}$ $- pi/3$ . If it takes $9 N$ $9 N$ to start pushing the object down the plane and $7 N$ $7 N$ 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 $8 k g$ $8 kg$ is on a plane with an incline of $- \frac{π}{3}$ $- pi/3$ . If it takes $9 N$ $9 N$ to start pushing the object down the plane and $7 N$ $7 N$ to keep pushing it, what are the coefficients of static and kinetic friction?