# An object with a mass of 5 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 3 N to keep pushing it, what are the coefficients of static and kinetic friction?

Feb 8, 2016

Coefficient of static friction : $\setminus {\mu}_{s} = 0.8188$
Coefficient of kinetic friction : $\setminus {\mu}_{k} = 0.4805$

#### Explanation:

Step1 : Draw a free body diagram showing all these forces.
(1) Gravitational force (weight) : $\setminus \vec{{w}_{}}$, $\setminus q \quad$ (2) Normal force : $\setminus \vec{{N}_{}}$,
(3) Frictional force : $\setminus \vec{{F}_{f}}$, $\setminus q \quad$ (4) Applied Force : $\setminus {\vec{{F}_{}}}_{a p p}$.

Step 2 : Pick a 2D cartesian system with its X axis parallel to the incline and directed downward and its Y axis perpendicular to the plane.

Step 3: Resolve these forces into their X and Y components carefully assigning the signs:
$\setminus \vec{{w}_{}} = \setminus \vec{{w}_{x}} + \setminus \vec{{w}_{y}}$;
$\setminus \vec{{w}_{x}} = \left(+ m g \sin \setminus \theta , 0\right)$; $\setminus \quad$ $\setminus \vec{{w}_{y}} = \left(0 , - m g \cos \setminus \theta\right)$;
$\setminus \vec{{N}_{}} = \left(0 , N\right)$; $\setminus \quad$ $\setminus \vec{{F}_{f}} = \left(- \setminus \mu N , 0\right)$; $\setminus \quad$ $\setminus {\vec{{F}_{}}}_{a p p} = \left(F , 0\right)$

Step 4: Equilibrium condition is that the net force acting on the object must be zero. Apply this condition by adding the vectors component-wise and set each component separately to zero.

Net Force : $\setminus \vec{{F}_{N}} = \setminus \vec{{w}_{}} + \setminus \vec{{N}_{}} + \setminus \vec{{F}_{f}} + \setminus {\vec{{F}_{}}}_{a p p} = \left(0 , 0\right)$

$\setminus \vec{{F}_{N}} = \left(m g \sin \setminus \theta , - m g \cos \setminus \theta\right) + \left(0 , N\right) + \left(- \setminus \mu N , 0\right) + \left(F , 0\right)$
$\setminus q \quad$$= \left(m g \sin \setminus \theta - \setminus \mu N + \setminus F , N - m g \cos \setminus \theta\right) = \left(0 , 0\right)$

Y Component: $N - m g \cos \setminus \theta = 0 \setminus \rightarrow N = m g \cos \setminus \theta$$\setminus \quad$ (Eqn 1)
X Component: $m g \sin \setminus \theta - \setminus \mu N + \setminus F = 0 \setminus \rightarrow \setminus \mu N = m g \sin \setminus \theta - F$
Therfore, $\setminus \mu = \setminus \frac{m g \sin \setminus \theta + F}{N} = \setminus \frac{m g \sin \setminus \theta + F}{m g \cos \setminus \theta} = \tan \setminus \theta + \setminus \frac{F}{m g \cos \setminus \theta}$

Static Equilibrium: When the vector sum of these forces is zero and the object is at rest, the object is in static-equilibrium and the friction in that cases is the static friction. The coefficient of friction is the static friction coefficient ($\setminus \mu \setminus \rightarrow \setminus {\mu}_{s}$)

Dynamics Equilibrium: When the vector sum of these forces is zero and the object slides with a constant velocity, the object is in dynamic equilibrium and the friction in that case is kinetic friction.The coefficient of friction is the coefficient of kinetic friction,
($\setminus \mu \setminus \rightarrow \setminus {\mu}_{k}$)
$\setminus {\mu}_{s} = \tan \setminus \theta + \setminus \frac{F}{m g \cos \setminus \theta} = 0.8188$
$\setminus {\mu}_{k} = \tan \setminus \theta + \setminus \frac{F}{m g \cos \setminus \theta} = 0.4805$

$\setminus \theta = \setminus \frac{\pi}{8} r a d$; $\setminus q \quad$ $m = 5 k g$; $\setminus q \quad$ $g = 9.8 m {s}^{- 2}$

$F = 18 N$ (static) $\setminus q \quad$ $F = 3 N$ (kinetic)