When a body moves on a surface or through a viscous medium (e.g. air, water), there are forces of friction because the body interacts with its surroundings.
Force of static friction, fs : the force that counteracts the applied force and keeps the object from moving .
Force of kinetic friction, fk : the retarding frictional force on the object in motion.
fs increases as the magnitude of the applied force increases, keeping the object in place.
When it is on the verge of moving, fs is at maximum.
When the applied force exceeds (fs)max, the object accelerates .
Once the object is in motion, the retarding frictional force becomes less than (fs)max.
Empirical Laws Of Friction
The direction of the force of static friction between any two surfaces in contact is opposite the direction of any applied force.
$$f_{s} \leq \mu_{s} n$$
,where
μs is coefficient of static friction
The direction of the force of kinetic friction acting on an object is opposite the direction of its motion and is given by:
$$f_{k} = \mu_{k} n$$
, where
μk is coefficient of kinetic friction
Coefficients Of Friction
The values of μs and μk depend on the nature of the surfaces (0.05-1.5), but μk is generally < μs.
The coefficients of friction are nearly independent of the area of contact between the surfaces.
Although μk varies with v, we normally neglect this (cf. stick-and slip motion at low v).
| $\mu_{s}$ | $\mu_{k}$ | |
|---|---|---|
| Steel on steel | 0.74 | 0.57 |
| Aluminum on steel | 0.61 | 0.47 |
| Copper on steel | 0.53 | 0.36 |
| Rubber on concrete (Dry) | 1.0 | 0.8 |
| Rubber on concrete (Wet) | 0.30 | 0.25 |
| Zinc on cast iron | 0.85 | 0.21 |
| Copper on cast iron | 1.05 | 0.29 |
| Glass on glass | 0.94 | 0.40 |
| Copper on Glass | 0.68 | 0.53 |
| Teflon on Teflon | 0.04 | 0.04 |
| Teflon on steel | 0.04 | 0.04 |
| Synovial joints in humans | 0.01 | 0.003 |
Experimental Determination Of $\mu_{s}$ And $\mu_{k}$
Suppose a block is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block starts to move. By measuring the critical angle, $\theta_{c}$ at which this slippage just occurs, we can obtain $\mu_{s}$.
Once the block starts to move, it accelerates down the incline. However, if $\theta$ is reduced to a value less than $\theta_{c}$, it may be possible to find an angle $\theta_{c}^{\prime}$ such that the block moves down the incline with constant speed. We can then obtain $\mu_{k}$
Let’s work out the details:
The horizontal forces acting on the block:
$$\begin{aligned} \sum F_{x} &= 0 \\ mg \sin{\theta}-f_{s} &= 0 \end{aligned}$$
The above equation will give the static frictional force on the block:
$$f_{s} = mg \sin{\theta}$$
The vertical forces acting on the block:
$$\begin{aligned} \sum F_{y} &= 0 \\ n-mg \cos{\theta} &= 0 \end{aligned}$$
The above equation will give the normal force on the block by the ramp:
$$n = mg \cos{\theta}$$
Hence, we have:
$$f_{s} = n \tan{\theta}$$
Since $f_{s} \leq \mu_{s} n$, at max angle $\theta_{c}$, we can calculate $\mu_{s}$:
$$\mu_{s} = \tan{\theta_{c}}$$
If moving at constant $\vec{v}$, from $f_{k} = \mu_{k} n$, we have:
$$\mu_{k} = \tan{\theta_{c}^{\prime}}$$
