Question

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?

Answer 1

Coefficient of static friction : `\mu_s = 0.8188`
Coefficient of kinetic friction : `\mu_k = 0.4805`

Explanation:

Step1 : Draw a free body diagram showing all these forces.
(1) Gravitational force (weight) : `\vec{w_{}}`, `\qquad` (2) Normal force : `\vec{N_{}}`,
(3) Frictional force : `\vec{F_f}`, `\qquad` (4) Applied Force : `\vec{F_{}}_{app}`.

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:
`\vec{w_{}}=\vec{w_x}+\vec{w_y}`;
`\vec{w_{x}}=(+mgsin\theta,0)`; `\quad` `\vec{w_{y}}=(0, -mgcos\theta)`;
`\vec{N_{}}=(0, N)`; `\quad` `\vec{F_{f}}=(-\muN, 0)`; `\quad` `\vec{F_{}}_{app}=(F, 0)`

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 : `\vec{F_{N}} = \vec{w_{}}+\vec{N_{}}+\vec{F_{f}}+\vec{F_{}}_{app}=(0, 0)`

`\vec{F_{N}}=(mgsin\theta,-mgcos\theta)+(0, N)+(-\mu N, 0)+(F, 0)`
`\qquad` `=(mgsin\theta-\muN+\F, N-mgcos\theta)=(0, 0)`

Y Component: `N-mgcos\theta=0 \rightarrow N = mgcos\theta` `\quad` (Eqn 1)
X Component: `mgsin\theta-\muN+\F=0 \rightarrow \mu N = mgsin\theta-F`
Therfore, `\mu = \frac{mgsin\theta+F}{N}=\frac{mgsin\theta+F}{mgcos\theta}=tan\theta+\frac{F}{mgcos\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 (`\mu\rightarrow\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,
(`\mu\rightarrow\mu_k`)
`\mu_s = tan\theta+\frac{F}{mgcos\theta}=0.8188`
`\mu_k = tan\theta+\frac{F}{mgcos\theta}=0.4805`

`\theta=\pi/8 rad`; `\qquad` `m=5kg`; `\qquad` `g=9.8ms^{-2}`

`F=18N` (static) `\qquad` `F=3N` (kinetic)