The detailed microscopics of friction are complicated. However, by focusing on practical details we can arrive at useful models to treat friction.

Some observations:

- Heavier objects have more friction than lighter ones
- On surfaces with friction it usually takes more force to get an object moving than keep it moving
- It is harder to move objects on rough surfaces than smooth ones

We can thus suppose that friction should be proportional to the force the surface exerts on an object, which is of course the normal force. As an equation this means that the **maximum magnitude** of the frictional force can be expressed as

$\large F_{fr}=\mu N$

and we can also suppose we will need different constants for a stationary object compared to a moving one. i.e we have different coefficients of static friction ($\mu_{s}$) and kinetic friction ($\mu_{k}$).

Some things to note about the equation

$\large F_{fr}=\mu N$

1. The frictional force does not depend on the contact area.

This can be rationalized by the idea of an effective area that depends on the normal force. The harder an object is pressed down the higher the effective area.

2. The frictional force does not depend on the velocity of motion, we can surmise from this that the interaction between the surfaces is not substantially modified once an object is moving.

You can find more details on these points here.

To answer this question we need to compare the applied force to the **maximum** force that static friction can provide, which is $F_{fr}=\mu_{s} N$. We should note that this force only is present when a force is applied and up to the point where the component of the applied force in the direction of motion exceeds the maximum possible static friction force the static friction force will be exactly equal and opposite to the component of the applied force in the direction of motion.

If you are either told that an object is moving or if you have evaluated that the static friction force has been overcome, then you can consider the kinetic frictional force $F_{fr}=\mu_{k} N$ as one of the net forces which determines an objects acceleration.

We will talk in more detail about the role of friction in rolling later in this course, but for now we can also define a coefficient of rolling friction $\mu_{r}$ (usually small) to describe the slowing down of a rolling object.

In all cases consider down the plane to be positive!

$\Sigma F_{\perp}=0$

$\vec{N}=m\vec{g}\cos\theta$

$\Sigma F_{\parallel}=mg\sin\theta-\mu mg\cos\theta-F_{A}$

or

$\Sigma F_{\parallel}=mg\sin\theta+\mu mg\cos\theta-F_{A}$

depending on whether the block is moving up or down the slope. The frictional force always opposes the current velocity.

Drag forces are forces an object experiences opposing their motion in a fluid. Air is considered a fluid and so air resistance is normally considered as drag force.

Small objects moving at slow speeds can be treated as having a drag force

$\large \vec{F}_{D}=-b\vec{v}$

This equation is known as Stoke's Law and the parameter $b$ depends on the nature of the fluid and the dimensions of the object. At these velocities the flow of the fluid through which the moving object passes is smooth and is called laminar flow. Flow at higher velocities frequently becomes turbulent and so at higher velocities and for larger objects the magnitude of the drag force is given by

$\large F_{D}=-\frac{1}{2}C \rho A v^2$

$\rho$ is the density of the fluid, $C$ is a parameter that depends on the shape of the object and $A$ is the area of the object, which we call the drag coefficient.

The dependence of the drag force on area is utilized by skydivers.

As an object falling under the Earth's gravitational field gets faster the drag force will eventually grow to equal the gravitational force at which point it will stop accelerating and reach it's terminal velocity.

For the laminar drag force this is when

$ma=mg-bv=0$

$mg=bv$

$v_{terminal}=\frac{mg}{b}$

For the turbulent drag force this is when

$ma=mg-\frac{1}{2}\rho v^2 C_{d}A=0$

$mg=\frac{1}{2}\rho v^2 C_{d}A$

$v_{terminal}=\sqrt{\frac{2mg}{\rho C_{d}A}}$

To deal with velocity dependent forces we need to use integral calculus. For the laminar flow drag force

$ma=mg-bv$

$\frac{dv}{dt}=g-\frac{b}{m}v$

$\frac{dv}{g-\frac{b}{m}v}=dt$ or better $\frac{dv}{v-\frac{gm}{b}}=-\frac{b}{m}dt$

If we consider an object that starts falling from rest

$\int_{0}^{v}\frac{dv}{v-\frac{gm}{b}}=-\frac{b}{m}\int_{0}^{t}dt$

$\ln{(v-\frac{gm}{b})}-\ln{(-\frac{gm}{b})}=-\frac{b}{m}t$

$\ln{(\frac{v-\frac{gm}{b}}{-\frac{gm}{b}})}=-\frac{b}{m}t$

$v-\frac{gm}{b}=-\frac{gm}{b}e^{-\frac{b}{m}t}$

$v=\frac{gm}{b}(1-e^{-\frac{b}{m}t})$

for the higher velocity drag force you can find the derivation on Wikipedia under derivation for the velocity v as a function of time.

The exponential function $e^{x}$ is frequently encountered in physics. This is largely because of it's special property that is it's own derivative, ie.,

$\frac{d}{dx}e^{x}=e^{x}$

This means that frequently the solution to a differential equation will in some way involve exponential functions.

In the equation we just derived for the velocity of a freely falling object subject to a laminar drag force we have an exponential function of the form

$e^{-kt}$

which is characteristic of exponential decay, though in fact our function is of the form $1-e^{-kt}$ so instead of going from 1 and approaching zero it starts at zero and approaches 1.

The constant $k$ defines how quickly the exponential function changes it's value. When $t=1/k$ it takes the value $e^{-1}\approx 0.37$

*Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.*

From Newton's first law we can see that circular motion can only occur when a net force acts on an object.

Unlike in baseball, cricketers may not extend their elbows during the bowling action. (Clearly, flexing the elbow, or “chucking” is “just not cricket”).

See some bowling basics (or even learn about a **better** sport).

Neglecting effects such as spin, you can see that the path of the ball after release is essentially tangential to it's velocity.

In uniform circular motion, the magnitude of the velocity does not change, but it's direction does. The vector for the change of the velocity always points towards the center of the circle. There is thus an acceleration, named the **centripetal acceleration** pointing to the center of the circle. For small changes in the position.

$\Delta v=2v\sin(\frac{\Delta\theta}{2})\approx v \Delta\theta$

$\Delta\theta=\frac{\Delta l}{r}$

$v\Delta\theta=v\frac{\Delta l}{r}$

$a_{R}=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t} = \lim_{\Delta t\to 0} \frac{v}{r}\frac{\Delta l}{\Delta t}=\frac{v^2}{r}$

We can look at the direction of the acceleration of an object moving in circular motions using the accelerometers in a smartphone or tablet.To display the acceleration vector for the x-y plane of your phone or tablet click here for iOs or click here for Android (Your device must have an accelerometer and gyroscope chip for this to work. For reasons I don't begin to comprehend Apple and Google define the accelerometer axes differently which is why there are two versions of this.).

Newton's second law tells us we can only have a centripetal acceleration towards the center of a circular motion path if the sum of the forces in the radial direction

$\Sigma F_{R}=ma_{R}=m\frac{v^{2}}{r}$

This **centripetal** force must be provided in some way for circular motion to occur.

Do not be misled in to thinking that such a thing as **centrifugal** force exists. As we saw earlier if the centripetal force is removed an object continues in a tangential, not perpendicular path to the motion. Of course in situations where you are in circular motion, it can feel as if you are being pushed out from the center, but this is in fact you applying the equal and opposite reaction force required by Newton's 3rd Law. The force on **you** is directed inwards.

$\Sigma F = m_{p} 20g $

$\frac{mv^{2}}{r}=m20g$

$v^{2}=200g$

$v=44.3 \mathrm{ms^{-1}}$

Frequency (revolutions per second) $f$ [ $s^{-1}$ or $\mathrm{Hz}$]

Period (time for one revolution) $T=\frac{1}{f}$ [$\mathrm{s}$]

Velocity $v=\frac{2\pi r}{T}=2\pi r f$ [$ms^{-1}$]

The relationship between frequency and period can be explored here.

Springs extend or contract as a force is exerted on them. The force on a spring can be considered to be directly proportional to it's extension. If a spring is used to provide the centripetal force on an object, the equal and opposite reaction force on it will stretch it out.

By swinging a weight on a spring in horizontal circle above my head I can demonstrate that the force required depends on the velocity.

If we now try to move the mass on spring in a vertical circle we can see it doesn't happen. The force required is different at the top and bottom because gravity alternatively assists or hinders us at the top and bottom of the path and as the spring changes it's length according to the tension on it the resulting motion is not circular.

A ball on a string works, providing the velocity is greater than $v=\sqrt{rg}$

A plane in a circular loop is a similar problem. Here we can also consider the force $F_{A}$ that the plane needs to apply as it goes through the loop. Things not attached to the plane will still be subject to gravity.

See 5.6 of this physclips video for a nice example.

How fast does it go?

$F_{T}\cos\theta=mg$ $F_{T}\sin\theta=\frac{mv^{2}}{r}$

$v^2=\frac{rF_{T}\sin\theta}{m}$ $v=\sqrt{\frac{r}{m}(\frac{mg}{\cos\theta})\sin\theta}=\sqrt{\frac{rg\sin\theta}{\cos\theta}}$

Rather than $r$ we would like an expression in $l$ so we sub in $r=l\sin\theta$

$v=\sqrt{\frac{lg\sin^{2}\theta}{\cos\theta}}$

To get the period of rotation, divide the path length $2\pi r$ = $2\pi l\sin\theta$ by $v$

$T=2\pi l \sin\theta \sqrt{\frac{\cos\theta}{lg\sin^{2}\theta}}=2\pi\sqrt{\frac{l\cos\theta}{g}} $

When a car rounds a bend the question of whether it slips or not is a **static** friction problem. We consider the car not to be moving in the direction perpendicular to it's motion. It will remain stationary in this direction if the maximum possible static friction force

$\mu_{s}N=\mu_{s}mg>\frac{mv^{2}}{r}$

or

$v^{2}<\mu_{s} g r$

As you can see, on wet or icy roads you should slow down round bends!

Roads designed for high speed traffic will often used banked turns to increase the maximum speed for which slipping does not occur. A well designed banked turn means the car should not rely on friction. (It also makes the problem easier..)

The relationship between safe speed, radius and the bank angle can be found from considering the forces.

In the absence of friction car's should go round the banked curve at a specific velocity related to the banking angle.

$F_{N}\sin\theta=\frac{mv^{2}}{r}$

$F_{N}\cos\theta=mg$

$mg\frac{\sin{\theta}}{\cos{\theta}}=\frac{mv^{2}}{r}$

$v^{2}=rg\tan{\theta}$

In reality, the frictional force allows cars to round the curve within a range of safe velocities determined by the maximum frictional force that can be provided.

For a car that is going fast the frictional force points down the incline and stops the car from flying out, but this only works up the point where the frictional force is equal to $\mu F_{N}$. So the maximum safe velocity $v_{high}$ is obtained from

$F_{N}\sin\theta+\mu F_{N}\cos\theta=\frac{mv_{high}^{2}}{r}$

$F_{N}\cos\theta-\mu F_{N}\sin\theta=mg$

$\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}=\frac{v_{high}^2}{rg}$

$v_{high}^{2}=rg\frac{\sin\theta+\mu\cos\theta}{\cos\theta-\mu\sin\theta}$

However it also possible to slide down the slope if a car goes too slowly. In this case friction acts up the slope and stops the car sliding down, which again only works up the point where the frictional force is equal to $\mu F_{N}$. So the minimum safe velocity $v_{low}$ is obtained from

$F_{N}\sin\theta-\mu F_{N}\cos\theta=\frac{mv_{low}^{2}}{r}$

$F_{N}\cos\theta+\mu F_{N}\sin\theta=mg$

$\frac{\sin\theta-\mu\cos\theta}{\cos\theta+\mu\sin\theta}=\frac{v_{low}^2}{rg}$

$v_{low}^{2}=rg\frac{\sin\theta-\mu\cos\theta}{\cos\theta+\mu\sin\theta}$