# An object with a mass of 4 kg is on a plane with an incline of  - pi/8 . If it takes 18 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?

Jan 11, 2016

I found $0.91 \mathmr{and} 0.55$ (but check my maths!).

#### Explanation:

Have a look at the diagram:

I assumed that the negative angle represents the downward inclination of the surface and transferred it as $\theta$ as in figure (I may be wrong though...).
With this convention for the angle and Newton's law:

$N = {W}_{y} = W \cos \left(\theta\right) = m g \cos \left(\theta\right) = 4 \cdot 9.8 \cdot \cos \left(\frac{\pi}{8}\right) = 36.2 N$

Also: ${W}_{x} = W \sin \left(\theta\right) = 4 \cdot 9.8 \cdot \sin \left(\frac{\pi}{8}\right) = 15 N$

Now:

1) Static Friction:
I consider the object still at rest but almost moving. Acceleration will be zero and we will be at maximum static friction where it will exactly be: ${f}_{s} = {\mu}_{s} N$

So, we use Newton's law along $x$ with $a = 0$ to get:

$F - {f}_{s} + {W}_{x} = 0$
$\textcolor{red}{18} - \left({\mu}_{s} \cdot 36.2\right) + 15 = 0$
${\mu}_{s} = 0.91$

2) Kinetic Friction:
From our graph of friction we can see that Kinetic friction will be almost constant and so:
${f}_{k} = {\mu}_{k} N$

Here we can assume that the object will move with constant velocity down the incline giving again zero acceleration. As before we get, using Newton's law along $x$ with $a = 0$:

$F - {f}_{k} + {W}_{x} = 0$
$\textcolor{red}{5} - \left({\mu}_{k} \cdot 36.2\right) + 15 = 0$
${\mu}_{k} = 0.55$