Kinematics
We introduce some basic kinematics examples of motion.
Free Fall
Finding the trajectory of a free falling body has puzzled scientist for a long time. Only in the 17th century Galilei discovered the law of motion of free fall. A free fall is a motion in a vacuum with only the force of gravity working on the object.
Galilei discovered that the distance travelled in a serie of equal units of time is proportional to the odd numbers and is independent of the mass of the body. Let c be the distance travelled in the first interval then the distance in the subsequent intervals of time develops as: 3c,5c,7c,9c,.... From this serie follows that the total distance travelled is a serie of squares: c,4c,9c,16c,25c,..... or in a mathematical formula, where x is the total distance travelled:
\[
x(t)=ct^2
\]
From mathematics we know that the average velocity of a very small interval of time around a point approaches the instantenous velocity at this point. We have therefore for the velocity the following relation:
\[
v(t)=2ct
\]
The speed increase with 2c. The increase of speed is called the acceleration
\[
a(t)=2c
\]
The constant change in velocity is called the gravitational constant, denoted with $g$, and is about $g=2c=9.8 ms^{2}=32 fts^{2}$ around the surface of the earth.
We learned from Galilei:
The gravitational force causes an object to accelerate uniformly with g. As a result the change in distance in the vertical direction is a quadratic function in relation to time and independent of the mass of the body.
Although the calculations are mathematical in nature, the law itself is an observational result, and does not follow from mathematics. Mathematics is only the tool to describe this law.
We can use differential calculus for the case of uniform accelaration to establish the relationships of distance, velocity and acceleration as function of time as follows:
\begin{align*}
a(t)&=g \to \\
a(t)dt&=gdt \to \\
\int{a(t)dt}&=\int{gdt} \to \\
v(t)&=gt+v_0
\end{align*}
\begin{align*}
v(t)dt&=(gt+v_0)dt \to \\
\int{v(t)dt}&=\int{(gt+v_0)dt} \to \\
x(t)&=\tfrac{1}{2}gt^2 + v_0t+x_0
\end{align*}
Motion takes place in a three dimensionsal space: length, breadth and height. The mathematical object of a 3 dimensional vector is usefull in physics to represent position, velocity and acceleration in these 3 dimensions. One of the remarkable things of mechanics is the insight that Galilea brought us too:
Motion in each of the three dimensions of space are mutually independent.
The gravitional force acts only in one dimension, height, and does not impact the other dimensions. If no other forces acts on the body the velocity in these other dimensions remain unchanged during the free fall motion.
We will now turn to a variant of free falling motion, where there is an initial velocity in the other dimensions.
Projectile motion
A stone dropped from an airplane will accelerate in the vertical direction due to gravity and move with constant speed in the horizontal direction. A passenger in the plane will see a straight line motion, whereas an observer on the ground will see a curved motion. This type of curved motion (motion in more than one dimension), with a constant acceleration, is called projectile motion. It is the idealized motion of the flight of a ball.

The motion of a projectile is a two dimensional motion. We choose a coordinate system with the positive yaxis vertically upward and the origin at the point where the projectile begins its flight. Furhter we assume that the projectiles height is limited, so the same gravitational force can be applied during its flight, and also that the reach of the projectile is limited so the earth can be taken as flat for this motion. We can use a constant force vector for this type of motion. The projectile motion starts with a force that launches the projectile with an initial velocity $\vec{v_0}$. After launch only the gravitational force acts on the projectile. The gravitational force is given by:
\[
\vec{F}_g=mg\hat{\vec{ y}}
\]
If we apply Newtons second law $\vec{F}=m\vec{a}$ we get:
\begin{align*}
mg\hat{\vec{y}}=m\vec{a} \\
\vec{a}=g\hat{\vec{y}}
\end{align*}
This shows clearly that the mass of the projectile plays no role in the projectile motion. We can write down the equations for the vector components for acceleration and velocity:
\begin{align*}
a_x&=0 \\
a_y&=g \\
v_{x0}&=v_0\cos\alpha \\
v_{y0}&=v_0\sin\alpha
\end{align*}
From these equations we see that there is no change in the velocity in the horizontal direction. The only force is the gravittional force and this force works only in the vertical direction. Notice also that a body in a projectile motion and free fall fall the same distance in the same time.

We can use this information to write down the equations of motion with constant acceleration for this special case:
\begin{align*}
v_x&=v_0\cos\alpha \\
v_y&=v_0\sin\alpha  gt \\
x &= (v_0\cos\alpha)t \\
y &= (v_0\sin\alpha)t \tfrac{1}{2}gt^2
\end{align*}
We can derive also the equation which gives the relation between the x and y coordinates by combining the last two equations and eliminating $t$. We get:
\begin{align*}
y&=\tan\alpha  \frac{g}{2(v_0\cos\alpha)^2}x^2
\end{align*}
This shows us that the projectile motion is a parabolic motion.
 Angle of sight

A plane is flying with a constant horizontal velocity of 500km/h at a height of 5km toward a target point on the ground. At what angle of sight $\theta$ should a projectile be released to hit the target on the ground ?
 Solution

Let $x$ be the distance on the ground between the position of the plane and the target and $h$ the height of the plane. We have
\[
x=h\tan\theta
\]
To hit the target the projectile should travel this distance during its fall. The distance travelled depends on the time to fall. So we first calculate the time $t$ to fall. The initial velocity of the projectile is $v_0$, but there is no in intial velocity in the vertical direction, the angle of projection is $0^\circ$. We have:
\[
y=\tfrac{1}{2}gt^2
\]
So the time to fall a distance h:
\[
t=\sqrt{\frac{2h}{g}}
\]
The horizontal distance travelled during this time is given by:
\[
x=v_0\sqrt{\frac{2h}{g}}
\]
We must equate these two distances and get
\[
\tan\theta=\frac{v_0}{h}\sqrt{\frac{2h}{g}}
\]
We convert the given values to SI units and plug into this equation and find
\[
\theta=42^\circ
\]
 Motion of a thrown ball

A ball is thrown at an angle $\theta$ and initial velocity $v_0$. Calculate:
 The time when the ball reaches the highest point
 The maximum height
 The time of flight
 The range of the ball
 The velocity of the ball when it hits the ground
 Solution

The time when the ball reaches the highest point
When the ball reaches the highest point the velocity is zero, before this point the velocity is positive and after it is negative. For $t$ we find:
\[
t=\frac{v_0\sin\alphav_y}{g}
\]
So the maximum, $v_y=0$, follows from:
\[
t=\frac{v_0\sin\alpha}{g}
\]
The maximum height
Substitute the relation for $t$ into the equation for $y$ and we find
\[
y_{\text{max}}=\frac{1}{2}\frac{(v_0\sin\alpha)^2}{g}
\]
The time of flight
As the projectile motion is a parabolic motion, which is symmetrical around the axis through its maximum, we know that the time for the ball to go up is equal to the time for the ball to go down. So the total time of the flight of the ball is:
\[
t_{\text{tot}}=2\frac{v_0\sin\alpha}{g}
\]
The range of the ball
The range of the ball is the distance travelled in the horizontal direction during its flight. We susbtitute the time of the flight into the equation for the horizontal distance and get:
\[
x_{\text{range}}=\frac{v_0^2}{g}2\sin\alpha\cos\alpha
\]
Notice that when $\alpha=45^\circ$ we get the maximum range:
\[
x_{\text{max}}=\frac{v_0^2}{g}
\]
Velocity of the ball when it hits the ground
From the symmetry of the path follows:
\[
v_{t_\text{tot}}=v_0
\]
\[
\theta_{t_\text{tot}}=\theta_0
\]
 Projectile trajectory

Graph the trajectory of a ball launched at an angle of $37^\circ$ and with an inital speed of $15m/s$.
 Solution

We modify the program. First our initial velocity vector is no longer zero. We use basic geometric facts to calculate the $x$ and $y$ component of the velocity factor by using Pythagoras square angle law:
\[
v_{0x}=v_0\cos\alpha
\]
\[
v_{0y}=v_0\sin\alpha
\]
Our equations for force, acceleration, velocity and position remain the same, but instead of using constants for our motion we start using parameters. This makes the program more flexible in use. Furthermore we realize that the outcomes for time $t$ is fully determined by the outcomes of time $t\Delta t$. So it is more efficient to calculate the path step by step, where each next step uses the outcomes of the previous step. In the old program we calculated a number of time points, but each calculation started from scratch. Next we build in a constraint as the vertical $>0$. Last we convert the output list of vectors to a tuple so we can immediately graph the path with gnuplot.
 New definition
 path
path::Vector > Vector > Double > Integer > [Vector] > [Vector]
path v0 x0 dt 0 xs = xs
path v0 x0 dt n xs = let
v1 = v0 `vadd` (dt `vscalar` (at 0))
 x0 +v0*dt +0.5*a*dt^2
x1 = (head xs) `vadd` ((dt `vscalar` v0) `vadd` ((0.5*dt*dt) `vscalar` (at 0)))
xs'= x1:xs
in path v1 x1 dt (n1) xs'
 trajectory x,y projectile motion
projectile_xy::Double > Double > Vector > Double > Double > [(Double,Double)]
projectile_xy speed alpha x0 dt n = let
v0 = [speed*cos (alpha*degtrad),speed*sin (alpha*degtrad)]
steps = round(n/dt)
p = path v0 x0 dt steps (x0:[])
in map (\[x,y] > (x,y)) (takeWhile (\x > (last x)>=0) (reverse p))
With this modified program we can easily plot the trajectory as follows:
>plotPath [] (projectile_xy 15 37 [0,0] 0.01 2)
 Gun fired at monkey

A famous physics lecture demonstration is a gun sighted at an elavated target which is released in free fall as the buller leaves the muzzle. The question is does the bullet hit the target ?
 Solution

If there was no gravity the bullet, $B$, would travel in a straight line in the direction of its initial velocity vector, $\vec{v}_{0B}$, towards the target, $T$, and hit the target at time $t^*$. This time $t^*$ is such that
\[
\vec{r}_{0T}=\vec{v}_{0B}t^*
\]
where $\vec{r}_{0T}$ is the position vector of $T$ at $t=0$. Now the consequence of gravity is that the bullet will not follow a straight line, but will fall due to gravity. We descibe the motion of the bullet during $t^*$ without gravity, so it arrives at the postion of the target at $t=0$ then we correct the vertical distance due to the fall of gravity which is to lower the position with $0.5gt^*$. However we must also adjust the position of the target with the same amount. So bullet and target are at the same position at $t^*$ independent of the horizontal speed.
Uniform circular motion
Another type of motion is uniform circular motion. In this type of motion the magnitude of the velocity remains te same during a revolutuin, but changes only in direction. The direction is always tangent to the circle. Because the velocity changes, there is an acceleration. This acceleration is directed towards the center of the circle and the following relation holds between the magnitudes of the acceleration $a$,velocity $v$ and radius $r$ of the circle:
\[
a=\frac{v^2}{r}
\]
 Orbit satellite around the earth

Calculate the speed of an earth satellite assume that it is travelling at an altitude of 225km above the surface of the earth where $g=9.1m/s^2$. What is the reason that the satellite does not fall down to the earth ? Calculate the path of the satellite.
 Solution

We derive a formula for $v$ based on the formula for the acceleration:
\[
g=\frac{v^2}{(r)}
\]
In this case $r$ is the sum of the altitude $h=225km$ and the radius of the earth $R=6440km$.
\[
v=sqrt{g(R+h)}=7788m/s=7.8km/s=28080km/h
\]
The orbit of the satellite is about $42*10^3km$. So one orbit takes about 90m.
Next we turn to the question why the satellite does not fall down to the earth. The satellite is like a projectile and it also falls due to gravity towards the earth. If there was no gravity involved the satellite would move away from the earth in a straight line with uniform speed tangent to the earth. The difference between this tangent straight line and the actual path taken is the amount the satellite has fallen towards the earth. The difference between a satellite and a thrown ball is that the satellite will not fall into the earth. The earth has a spherical form and the surface curves. An object falls in his first second of motion about $0.5*9.8m\approx 5m$. If an object moves in this one second a horizontal distance $X$ such that the earth curves over this distance away at least 5m then the object would never fall into the earth. The earth would curve away from the object and the object would fall around the earth. There is a theorem from plane geometry that defines the relation between the horizontal distance $X$, the vertical distance curved away $Y$ and the radius of the circle.

\[
X=\sqrt{2rY}=\sqrt{2*6440*1000*5}\approx 8000m
\]
So, to escape from falling into the earth the object must have a horizontal speed of at least $8000m/s$. If the speed is exactly $8000m/s$ the path will be a circular, if the speed is greater than $8000m/s$ the path will be elleptical. The $8000 m/s$ figure used in the above discussion applies to satellites launched from heights just above Earth's surface. Since gravitational influences decrease with the height above the Earth, the orbital speed required for a circular orbit is less than $8000m/s$ at significantly greater heights above Earth's surface. There is an upper limit on the orbital speed of a satellite. If launched with too great of a speed, a projectile will escape Earth's gravitational influences and continue in motion without actually orbiting the Earth. Such a projectile will continue in motion until influenced by the gravitational influences of other celestial bodies.
A satellite is continuously in free fall. Astronauts in an orbit around the earth experience this by the situation of weightlessness.

Suppose we now want to calculate the path of the orbit of the satelite. We assume the orbit is circular. So we know the radius is constant during the motion. We have $\vec{F}=mg\vec{r}$, where $\vec{r}$ is the position vector of the point on the circle. The length of this position vector remains constant, but the direction changes continuously. Now we derive the second order differential equations for this motion.
We start with Newton's universal force law:
\begin{align*}
\vec{F}&=G\frac{m_em_s}{r^2}\hat{\vec{r}} \\
&=m_sg\hat{\vec{r}}
\end{align*}
This replacement with the constant $g$ is allowed because the distance $r$ of the satellite is constant during its motion. The direction of the force is opposite to the direction of the position vector $r$ so we can derive
\begin{align*}
\frac{F_x}{F}&=\frac{x}{r} \\
\frac{F_y}{F}&=\frac{y}{r} \\
\end{align*}
from which follows:
\begin{align*}
F_x&=\frac{mgx}{r} \\
F_y&=\frac{mgy}{r} \\
\end{align*}
now applying Newton's second law:
\begin{align*}
\frac{mgx}{r}&=ma_x \\
\frac{mgy}{r}&=ma_y \\
\end{align*}
which result in following second order differential equations:
\begin{align*}
a_x&=\frac{gx}{r}\\
a_y&=\frac{gy}{r}\\
\end{align*}
 Orbit moon

The moon revolves around the earth in 27.3 days. Assume that the orbit is circular with a radius of 385000 km, with the center at the middle of the earth. Find the acceleration of the moon towards the earth.
 Solution

First we convert everything to SI units.
\[
P=27.3d=27.3d*86400*s/d=2.36*10^6s
\]
\[
r=385000km=385000km*1000*m/km=3.85*10^8m
\]
Next we find the velocity by
\[
v=\frac{2\pi r}{P}=1025m/s
\]
Now we can plug the veocity and radius in the formula for the acceleration:
\[
a=\frac{v^2}{r}=0.00273 m/s^2
\]
Notice that this acceleration is about 3600 times smaller than $g=9.8m/^2$. The radius of the earth is about 6440km. This is about 1/60 of the radius of the orbiting moon. This fact brought Newton to discover his law of gravitation where the force is inversely proportional to the square of the radius.
Planetary motion
In this section we explore the planetary motions. Analyzing motion starts with finding the acceleration. Finding the acceleration starts with finding the applicable forces with their force law. For planetary motion this force law is Newtons universal gravitation law. This law implies a force between two particles along the line of joining them. For the celestial bodies we assum they are spherical and therefore we can treat these bodies as pointlike particles.
 Kepler's law of periods

The third law of Kepler states that for any object orbitting the sun:
\[
\frac{T^2}{S^3}=C
\]
where the $T$ the period of one revolution about the sun and $S$ the average distance to the sum is, C is a constant.
Calculate $C$.
Show based on the universal law of Newton that this ratio is a constant.
 Solution

We take the earth and mars data form NASA planetary fact sheet and get:
\begin{align*}
T_e&=365.2 days \\
S_e&=149.6*10^6km \\
T_m&=687 days \\
S_m&=227.9*10^6km
\end{align*}
then if we put these numbers in the ratio we get for the earth
\[
\frac{T_e^2}{S_e^3}=3.984\times 10^{20}days^2/km^3
\]
and for mars
\[
\frac{T_m^2}{S_m^3}=3.987\times 10^{20} days^2/km^3
\]
both ratio's are almost the same. The difference is due to inaccuracies. Let us convert this amount into SI units with following conversion factor:
\[
\frac{days^2}{km^3}=\frac{(24*3600s)^s}{1000m^3}=7.46s^2/m^3
\]
\[
\frac{T^2}{S^3}=2.97\times10^{19}s^2/m^3
\]
 Orbit earth

Calculate the orbit of the earth around the sun.
 Solution

Newton's Laws
Newton's laws are demonstrated by a serie of examples.
Trajectory
Newton's second law states in essence that each motion is determined by a second order differential equation with two initial conditions for each degree of freedom:
\[
\ddot{x}(t_0)=\frac{1}{m}F(x(t_0),\dot{x}(t_0),t_0)
\]
We describe a numerical method to find the trajectory $x(t)$. We know that a smooth functioncan be extrapolated with a Taylor expansion if we use small timesteps $\Delta t$:
\[
x(t_0+\Delta t) = x(t_0)+\Delta t \dot{x}(t_0)+...
\]
and of course also
\[
\dot{x}(t_0+\Delta t)=\dot{x}(t_0)+\Delta t \ddot{x}(t_0)+...
\]
but we have:
\[
\ddot{x}(t_0)=\frac{1}{m}F(x(t_0),\dot{x}(t_0),t_0)
\]
Repeat this process $n$ times to compute $x(t+n\Delta t)$.
If we want to have an accurate result we cannot just plug in the $\dot{x}(t_0+\Delta t)$ to calculate $x(t_0+\Delta t)$ because during the time interval the velocity has not been constant. Therefore we assume that during the small step of time, where the acceleration is more or less constant, we calculate the velocity and then the change in distance as follows:
\[
\Delta x_n = v_{avg}\Delta t
\]
with:
\[
v_{avg}=\tfrac{1}{2}((a_n\Delta t+v_{n1}) + v_{n1})=\tfrac{1}{2}a_n\Delta t + v_{n1}
\]
So
\[
\Delta x_n = \tfrac{1}{2}a_n(\Delta t)^2 + v_{n1}\Delta t
\]
To have the cumulative numbers we just add up these small intervals:
\[
x_n=x_0+\sum_{i=1}^{n}{\Delta x_i}=x_0+\sum_{i=1}^{n}\tfrac{1}{2}a_n(\Delta t)^2 + v_{n1}\Delta t
\]
\[
v_n=v_0+\sum_{i=1}^{n}{a_i}\Delta t
\]
To avoid the calculation of the average of the velocities, we can follow another approach to calculate the
velocity at the time $t+0.5\Delta t$ instead of at time $t+\Delta t$. To start we calculate $v(0.5\Delta t)=v(0)+0.5\Delta t a(0)$ and then use following equations for each time step:
\begin{align*}
x(t+\Delta t)&=x(t)+v(t+0.5\Delta t)\Delta t \\
v(t+0.5\Delta t)&=v(t0.5\Delta t)+a(t)\Delta t \\
a(t)&=F(t)/m
\end{align*}
This approach clearly shows that the trajectory is determined by two initial conditions per degree of freedom. A degree of freedom is a coordinate in which an object is free to move. The fundamental law of mechanics is equivalent to a second order differential equation and its solution is $x(t)$ with two integration constants $x(t_0),\dot{x}(t_0)$.
These relationships can be concisely programmed in a language as Haskell. Haskell is a functional programming language and very usefull for mathematic type of programming. We simply defined the vector as a list and defined simple operations for adding and multiplication with a scalar.
{
Newton's second law
eulerStep: calculates a new state t+dt of the trajectory by using the previous state t of the position vector r
and velocity vector v and a supplied force law. The force law uses the previous state and calculates the
acceleration at t+dt. The acceleration is used to calculate the velocity at t+dt.
We assume the acceleration is constant during dt and therefore the change in velocity is dt.a(t+dt).
Next we calculate the position by assuming that the speed during dt is the average of v(t) and v(t+dt).
To calculate a trajectory first define the force law f and then call the eulerStep:
> take n (iterate (eulerStep dt f) s0)
}
type Time = Double
type TimeStep = Double
type Position = Vector
type Velocity = Vector
type Acceleration = Vector
type Force = State > Acceleration
type State = (Time, (Position,Velocity,Acceleration))
eulerStep::TimeStep > Force > State > State
eulerStep dt f (t,s0) = (t+dt,s1)
where
(r0,v0,a0) = s0
a1 = f ((t+dt),s0)
v1 = v0 `vadd` (dt `vscalar` a1)
r1 = r0 `vadd` (dt `vscalar` v0) `vadd` ((0.5*dt*dt) `vscalar` a1)
s1 = (r1, v1, a1)
 Force laws
freefall_earth::Force  used for freefall and projectile motion
freefall_earth s = [0,g_earth]
orbit_earth::Force  used for orbital motion around earth
orbit_earth (t,s) = a
where
(r,_,_) = s
a = (gm_earth / (vnorm r)**3) `vscalar` r
orbit_sun::Force  used for orbital motion around sun
orbit_sun (t,s) = a
where
(r,_,_) = s
Conservation linear momentum
Bullet and gun
Newton's third law states that if two bodies $A$ and $B$ interact then the forces between them are equal and opposite.
\[
\vec{F}_{AB}=\vec{F}_{BA}
\]
the notation $\vec{F}_{AB}$ means the force on $A$ caused by $B$. Now its second law states that a force causes an acceleration inversely proportional to the mass of an object. An acceleration is just a change of velocity. So we have:
\begin{align*}
\Delta \vec{v}_A &= \frac{\vec{F}_{AB}}{m_A} \\
\Delta \vec{v}_B &= \frac{\vec{F}_{BA}}{m_A} \\
\end{align*}
but form this follows:
\[
m_A\Delta v_A=m_B\Delta v_B
\]
or
\[
m_A\Delta v_A+m_B\Delta v_B=0
\]
we clearly see that total of the quantity $mv$, which is called momentum, is not changed by the interaction of bodies, but remains constant (of course not neccessarily zero). The change in velocity of one body,resulting from the force, is compensated by the other body. This ratio of their velocities is inversely proportional to the ratio of their masses:
\[
\frac{\Delta v_A}{\Delta v_B}=\frac{m_B}{m_A}
\]
Let us assume that $m_A>m_B$ and let us change our notation $M=m_A$ and $m=m_B$ it is immediately clear that
\[
\Delta v_A \less \Delta v_B
\]
Let us from now on denote $v=v_A$ and $V=v_B$ to remind us of this fact. If we reformulate our earlier result:
\[
\frac{\Delta V}{\Delta v}=\frac{M}{m}
\]
but as speed determines distance we have the same relationship for the distance travelled by $m$ which we denote by $D$ and $M$ which we denote by $d$:
\[
\frac{\Delta D}{\Delta d}=\frac{M}{m}
\]
the distance $\Delta D$ is $\frac{M}{m}$ times bigger than the distance $\Delta d$. The ratios of their masses determines therefore the ratio of their changed velocities and distances. We could define a point $C$ on the line between $M$ and $m$ such that:
\[
\frac{Cm}{CM}=\frac{D}{d}=\frac{M}{m}
\]
during uniform motion this point would be at rest. This point is called the center of gravity between two masses.

Suppose a bullet has a mass of 1 ounce (= 1/16 pound) and a muzzle velocity of 1600 feet per second. Its forward momentum is then 1 / 16 x 1600 = 100 pounds feet per second. The gun must therefore have a momentum of 100 pounds feet per second. If the gun’s mass is 12.5 pounds, its recoil velocity would be 100 pounds feet per second divided by 12.5 pounds =  8 feet per second. In other words, the gun’s recoil velocity is 8 feet per second in the direction opposite that of the bullet.
Another example is a balloon from which air escapes from the nozzle. The shrinking balloon will careen crazily until it becomes deflated. The balloon is propelled forward by pushing its contents rearward through the nozzle. This same principle is the basis for rocket propulsion, which expels highvelocity exhaust gases from the rocket engine. The rocket motor is thus a selfsufficient means of propulsion. It can therefore operate in “outer
space".
Forces on mass
Force on a mass on a horizontal surface
The most simple situation is a force on a single mass. In analyzing mechanics problems we isolate the different masses and draw all the forces on an individual mass to find the net resultant force acting on this mass. We draw these forces in a free body diagram. In this case there are 3 forces at work on the mass. The gravitational force $F_g$, the normal force $F_N$ and the applied force $F$. The gravitational force and the normal force act in the vertical direction and cancel each other. The acceleration of the mass is in the direction of the applied net force $F$ and is proportional to the applied force and inversely proportional to the mass: $a=\frac{F}{m}$.
Force on a mass on a horizontal surface at an angle
When the force acts at an angle the force is resolved in a vertical and horizontal component. The vertical component is cancelled out by the normal force that the support surface exerts on the mass.
Strings
Consider a block of mass $m$ suspended from a fixed beam by means of a string. We suppose the string has neglectible mass compared to the block, and the string is inextensible.
The beam and the block pull on the string. As reaction the string pulls the beam and block together. This force is called the tension force $\vec{F}_T$ on the beam and the block. The tension force at each end of the string is equal in magnitude and opposite in direction. These forces act so as to oppose the stretching of the string: i.e., the beam experiences a downward force, whereas the block experiences an upward force. When the block is in equilibrium, no acceleration, we have
\[
Tmg=0
\]
Consider now a block mass m is suspended by three strings.
Determine the tension forces $F_T$, $F_{T_1}$ and $F_{T_2}$.
It is easily seen that the two forces on the block must cancel so we have again $F_T=mg$. To find the other tension forces we draw a free body diagram.
The net resultant force in the horizontal and vertical direction are zero. So we resolve the three forces in these directions. We take the direction to the right and up as positive. For the horizontal directions we have:
\begin{align*}
F_{T_1}\cos 60^\circ  F_{T_2}\cos 30^\circ=&0 \\
\frac{F_{T_1}}{2}\frac{\sqrt{3}F_{T_2}}{2}=0 \\
F_{T_1}=\sqrt{3}F_{T_2}
\end{align*}
For the vertical direction we have:
\begin{align*}
F_{T_1}\sin 60^\circ + F_{T_2}\sin 30^\circ mg=&0 \\
\frac{\sqrt{3}F_{T_1}}{2}+\frac{F_{T_2}}{2}mg&=0 \\
\end{align*}
Substitution of $F_{T_1}=\sqrt{3}F_{T_2}$ into the last equation we find:
\begin{align}
F_{T_2}&=\frac{mg}{2} \\
F_{T_1}&=\frac{\sqrt{3}mg}{2}
\end{align}
Inclined motion
With an incline we can demonstrate different forces of motion in action. As well as the usage of vector components in the different directions of motion.
Suppose we have the following situation of a block with mass $m$ on a wedge.
We analyze different types of problems.
 Motion of a block on a fixed wedge without friction

We assume the incline is fixed to the earth, so it does not move. Determine the acceleration of the block in the horizontal and vertical dimension.
 Solution

In this example we have two forces on the body: the gravitational force $\vec{F}_g$ and the normal force, $\vec{F}_N$. The normal force is the force which the wedge exerts on the block and which is perpendicular to the surface of contact. We draw these forces in a free body diagram and use a rectangular coordinate system with respect to the ground. In the diagram we have resolved the two vectors into components. The gravitational force has been resolved into two components perpendicular, $F_g\cos\alpha$, and parallel, $F_g\sin\alpha$, to the incline of the wedge. The normal force has been resolved into two components one in the vertical $F_N\cos\alpha$ and the other in the horizontal direction $F_N\sin\alpha$. These components play an important role in the solution of problems where an incline is involved.
To find the angle $\alpha$ between the lines drawn in the diagram we use the following rule.
When two intersecting lines are perpendicular to two lines then these lines make the same angle as these other lines. Therefore the angle between the gravitational force (which is perpendicular to the ground) and the component perpendicular to the incline is the same as the angle between the incline and the ground, namely $\alpha$.
There are two forces at work so we must find the net resultant force on the block. The gravitational force works only in the vertical direction therefore we choose to write down the equations of motion with the vectors components in the horizontal and vertical direction.
\begin{align}
mg  F_N\cos\alpha = m a_y \label{eq10} \\
F_N\sin\alpha=m a_x \label{eq11}
\end{align}
We have now 2 equations with three unknowns $F_N,a_x,a_y$. We must add another equation to solve this problem. In fact we can assume in this particular problem that the normal force is cancelled by the gravitational force in the direction perpendicular to de incline of the wedge, $F_g\cos\alpha$. This means that the block has no acceleration in the direction of the normal force and keeps in contact with the incline during its motion.
\begin{equation}
F_N=F_g\cos\alpha=mg\cos\alpha
\end{equation}
Substituting this value for $F_N$ into $\ref{eq10}$ we get:
\begin{align}
mg  mg\cos^2\alpha &= m a_y \to \notag \\
F_y&=mg\sin^2\alpha \\
a_y&=g\sin^2\alpha
\end{align}
Substituting this value for $F_N$ into $\ref{eq11}$ we get:
\begin{align}
mg\cos\alpha\sin\alpha &= m a_x \to \notag \\
F_x&=mg\cos\alpha\sin\alpha \\
a_x&=g\cos\alpha\sin\alpha
\end{align}
If we look at the diagram we see that we can arrive at the same result if we use first the components of $F_g$ resolved in the direction parallell and perpendicular to the incline and then notice that the motion is caused by $F_g\sin\alpha$. If we resolve $F_g\sin\alpha$ in the horizontal and vertical direction we get the same result.
 Motion of a block on a nonfixed wedge without friction

Now suppose that the wedge can move along the ground and has mass $M$. Determine the acceleration of the wedge,$a_w$, and the acceleration of the block relative to the wedge, $a_{wb}$.
 Solution

This situation is almost the same as the previous one except that the motion of the wedge causes also an acceleration in the direction perpendicular to the incline of the wedge. In the previous case there was no acceleration in this direction. The block has an acceleration $a_b$, the wedge an acceleration $a_w$, the relative acceleration of the block versus the wedge $a_{bw}=a_ba_w$. See the figure below.
For the force diagram we have to look now at the two bodies seperatly. On the block we have as before acting the gravitational force $\vec{F_g}_b=mg$ and the normal force $\vec{F_N}_b$. On the wedge we have the gravitational force $\vec{F_g}_w=Mg$, the normal force $\vec{F_N}_w$, and the reaction force $\vec{F_N}_b$.

The vertical force components on the wedge cancel each other. The horizontal component of the reaction force $\vec{F_N}_b$ causes the acceleration $a_w$. Applying Newton's second law gives the following equation
\begin{equation}
F_{N_b}\sin\alpha = M a_w
\end{equation}
The gravitational and normal force on the block causes the acceleration $a_b$. These net forces can be resolved into a vertical component and a horizontal component in the same way as in the previous example. We equate these forces by using Newton's second law to the accelerations in the vertical direction $a_{b_y}=a_wb \sin\alpha$ and the horizontal direction, $a_{b_x}=a_wb\cos\alpha  a_w$.
\begin{align}
mg  F_{N_b}\cos\alpha &= m a_{wb} \sin \alpha \\
F_{N_b} \sin \alpha &= m (a_{wb}\cos\alpha  a_w) \\
\end{align}
We have now three equations with three unknowns: $F_{N_b},a_{wb},a_w$ which we solve:
\begin{align}
a_{wb}&=\frac{(M+m)g\sin\alpha}{M+m\sin^2\alpha} \\
a_w&=\frac{mg\sin\alpha\cos\alpha}{M+m\sin^2\alpha} \\
F_{n_b}&=\frac{mMg\cos\alpha}{M+m\sin^2\alpha} \\
a_{b_x}&=\frac{Mg\sin\alpha\cos\alpha}{M+m\sin^2\alpha} \\
a_{b_y}&=\frac{(M+m)g\sin^2\alpha}{M+m\sin^2\alpha}
\end{align}
The algebraic derivation can be found here

From the first equation we get:
\[
F_{N_b}=\frac{Ma_w}{\sin\alpha} \label{eq01}
\]
We substitute $\ref{eq01}$ in the second equation:
\begin{align}
mgMa_w\frac{\cos\alpha}{\sin\alpha}&=ma_{wb}\sin\alpha \to \notag \\
Ma_w\frac{\cos\alpha}{\sin\alpha}&=mgma_{wb}\sin\alpha \to \notag \\
a_w&=\frac{m}{M}(ga_{wb}\sin\alpha)\frac{\sin\alpha}{\cos\alpha} \label{eq02}
\end{align}
We substitute $\ref{eq01}$ in the third equation:
\begin{align}
Ma_w&=m(a_{wb}\cos\alphaa_w) \to \notag \\
a_w&=\frac{m}{M}(a_{wb}\cos\alphaa_w) \to \notag\\
a_w+\frac{m}{M}a_w&=\frac{m}{M}a_{wb}\cos\alpha \to \notag\\
a_w(1+\frac{m}{M})&=\frac{m}{M}a_{wb}\cos\alpha \to \notag\\
a_w&=\frac{m}{M+m}a_{wb}\cos\alpha \label{eq03}
\end{align}
We now equate $\ref{eq02}$ and $\ref{eq03}$:
\begin{align}
\frac{m}{M+m}a_{wb}\cos\alpha&=\frac{m}{M}(ga_{wb}\sin\alpha)\frac{\sin\alpha}{\cos\alpha} \to \notag \\
(\frac{m}{M+m}\cos\alpha+\frac{m}{M}\frac{\sin^2\alpha}{\cos\alpha})a_{wb}&=\frac{m}{M}g\frac{\sin\alpha}{\cos\alpha} \to \notag \\
(\frac{mM\cos\alpha+m(M+m)\frac{\sin^2\alpha}{\cos\alpha}}{M(M+m)})a_{wb}&=\frac{m}{M}g\frac{\sin\alpha}{\cos\alpha} \to \notag \\
a_{wb}&=\frac{m}{M}g\frac{\sin\alpha}{\cos\alpha}\frac{M(M+m)}{mM\cos\alpha+m(M+m)\frac{\sin^2\alpha}{\cos\alpha}} \to \notag \\
a_{wb}&=g\frac{\sin\alpha}{\cos\alpha}\frac{(M+m)}{M\cos\alpha+(M+m)\frac{\sin^2\alpha}{\cos\alpha}} \to \notag \\
a_{wb}&=\frac{g\sin\alpha(M+m)}{M\cos^2\alpha+(M+m)\sin^2\alpha} \to \notag \\
a_{wb}&=\frac{g\sin\alpha(M+m)}{M(\cos^2\alpha+\sin^2\alpha)+m\sin^2\alpha} \to \notag \\
a_{wb}&=\frac{g\sin\alpha(M+m)}{M+m\sin^2\alpha} \label{eq04}
\end{align}
Next we substitute $\ref{eq04}$ into $\ref{eq02}$
\begin{align}
a_w&=\frac{m}{M}(g\frac{g\sin\alpha(M+m)}{M+m\sin^2\alpha}\sin\alpha)\frac{\sin\alpha}{\cos\alpha} \to \notag \\
a_w&=\frac{m}{M}\frac{gM+gm\sin^2\alpha gM\sin^2\alpha gm\sin^2\alpha }{M+m\sin^2\alpha}\frac{\sin\alpha}{\cos\alpha} \to \notag \\
a_w&=\frac{m}{M}\frac{gM(1\sin^2\alpha}{M+m\sin^2\alpha}\frac{\sin\alpha}{\cos\alpha} \to \notag \\
a_w&=\frac{gm\cos^2\alpha}{M+m\sin^2\alpha}\frac{\sin\alpha}{\cos\alpha} \to \notag \\
a_w&=\frac{gm\cos\alpha \sin\alpha}{M+m\sin^2\alpha}
\end{align}
$F_{N_b},a_{b_x},a_{b_y}$ are found by substitution.

Motion of a block on a nonfixed wedge without friction

Suppose now we apply a force to the wedge such that the block remains stationary on the wedge. Determine the force $F$ and the acceleration $a$ of the wedge.
 Solution

To keep the block stationary on the wegde both the wedge and the block must have the same horizontal acceleration $a$. So, the force $F$ acts only in the horizontal direction. The net resultant force acting on the horizontal direction on the wedge is given by:
\begin{equation}
FF_N\sin\alpha=Ma
\end{equation}
The horizontal force on the block results from the normal force in the horizontal direction:
\begin{equation}
F_N\sin\alpha=ma \label{eq06}
\end{equation}
The net resultant force in the vertical direction on the block must be cancelled. This results from the force $F$ which is transmitted via the normal force on the block.
\begin{align}
mgF_N\cos\alpha&=0 \to \\
F_N&=\frac{mg}{\cos\alpha}
\end{align}
We can substitute this into $\ref{eq06}$:
\begin{align}
ma&=\frac{mg}{\cos\alpha}\sin\alpha \to \notag \\
a&=g\frac{\sin\alpha}{\cos\alpha}=g\tan\alpha
\end{align}
Then the force $F$ results from:
\begin{align}
F&=F_N\sin\alpha+Ma \to \notag \\
F&=(m+M)a \to \notag \\
F&=(m+M)g\tan\alpha
\end{align}
 Motion along a wedge with friction

Find the statical friction coefficient, $\mu_s$.
 Solution

In this example we have three forces on the body: the gravitational force $\vec{F}_g$, the normal force, $\vec{F}_N$, and the statical friction force, $\vec{F}_{fs}$ which can achieve a maximum value of $\max \vec{F}_{fs}=\mu_s\vec{F}_N$. We draw these forces in a free body diagram and find the components of the gravitational force in the direction of the incline and the normal force. We choose these directions as the orientation of our reference frame. See the figure below.
We find:
\[
F_{gx}=F_g\sin\alpha=mg\sin\alpha
\]
\[
F_{gy}=F_g\cos\alpha=mg\cos\alpha
\]
$F_{gx}$ achieves a maximum when $\alpha=90^\circ$. As long as the block is not sliding we must have $F_{gx}\leq F_{fs}$ and at the moment the block starts sliding $F_{gx}>F_{fs}$. We approximate the value of $\alpha$ where $F_{gx}=F_{fs}$. This gives:
\[
mg\sin\alpha^*=\mu_s F_N
\]
As the object does not move in the $y$ direction, the normal force and the force of gravity in this direction cancel each other, $F_{gy}=F_N=mg\cos\alpha$. We substitute this value for $F_N$ in the above equation and get:
\[
\mu_s=\tan \alpha^*
\]
 Universal law of gravitation

Derive Newton's universal law of gravitation for circular motion.
 Solution

Netwon had to unify the explanation for the falling apple, (free fall), the motion of a canonball, (projectile motion) and the motion of planets, (circular or eliptic motion), all in the same law. From kinematics we know exactly the relationships of motion between acceleration $a$, velocity $v$ and position $r$. Newton introduced the concept of force as the cause for an acceleration as: $a=\vec{F}/m_A$. We observe in free fall motion that gravitation gives the same acceleration to each body independent of its mass, it is immediately clear that we should put in $F$ the mass of the body, since that is the only way to cancel out its effect on the acceleration. However something should cause the motion between bodies so we take another body as the most logical explanation of the motion, $m_E$, of the motion of $m_A$. Next we can observe the motion of almost circular heavenly bodies, like the moon. From calculus we have for a circular motion found:
\begin{align*}
a&=\frac{v^2}{r} \to \\
v^2 &= ar
\end{align*}
If we compare the velocities of a number of these planetary motions with respect to the earth, $M_E$, we find the following relationship:
\[
\frac{v^2_A}{v^2_B}=\frac{M_E/R_A}{M_E/R_B}
\]
So the square of the speed inversely depends on the distance of the body measured by its radius. Then simply must follow (we must introduce a constant because it has been cancelled out in the previous step):
\begin{align*}
C\frac{M_E}{R_A}&=aR_A \to \\
a&=C\frac{M_E}{R_A^2}
\end{align*}
Now we must only go from $a$ back to $F=m_Aa$ as follows:
\[
F=C\frac{M_Em_A}{R_A^2}
\]
Here we have derived Newton's universal law of gravitation for circular motion.
Classical Mechanics Examples
THE GYROSCOPE
Torque, Moment of Inertia, Angular momentum