# Newton's laws of motion

In this page, I am going to discuss Newton's laws of motion as they apply to point masses (or point particles).

## Newton's First Law (Law of Inertia)

In an inertial reference frame, a point mass remains at rest or continues to move uniformly unless acted upon a force or equivalently any particle persists on its own state of rest or uniform motion unless it is compelled to change that state by an external force.

Where uniform motion means that the derivative of the position vector function is a constant vector, with respect our reference frame. Mathematically:

${\displaystyle {\frac {d}{dt}}{\vec {r}}(t)={\dot {\vec {r}}}={\vec {v}}}$

where ${\displaystyle {\vec {v}}}$ is a non time-varying vector.

The first Law of Newton is unintuitive as it does not correspond to our every day experiences. For Instance, we never see an object moving with constant veclocity as this state requires the asbense of forces. This condition is mathematically expressed as:

${\displaystyle \sum _{i=1}^{N}{\vec {F}}_{i}={\vec {F}}_{1}+{\vec {F}}_{2}+...{\vec {F}}_{N}={\vec {0}}}$

where ${\displaystyle {\vec {F}}_{i}}$ is i-th force, and N is the total number of forces acting on the point particle of our interest.

Examples of forces that prevent objects from undergoing uniform motion in our world are:

- Graviatational forces between objects (obeying the law of Universal Gravitation)

- Electrostatic forces between charged particles (obeying Coulomb's Law)

- the Lorentz Force (exerted on a charged particle when moving through a magnetic field)

- Frictional Forces

- Forces by gases when moving through them (example: The atmosphere)

## Newton's Second Law

For any particle of non zero mass m, the vector summation of all the forces acting on the particle is always equal to the mass m times the body's acceleration.

This law is mathematically expressed as:

${\displaystyle \sum _{i=1}^{N}{\vec {F}}_{i}=m{\vec {a}}}$
where ${\displaystyle {\vec {F}}_{i}}$is the i-th force, N the total number of forces acting on the particle and ${\displaystyle {\vec {a}}}$ the vector function of acceleration of the particle.

Many choose to write the previous equation as follows:

${\displaystyle {\vec {a}}={\frac {1}{m}}\sum _{i=1}^{N}{\vec {F}}_{i}\quad ,m>0}$
This form is frequently chosen because it distinguishes the cause (the forces exerting on the particle) and the mechanical effect in the real world (the particle's acceleration).

• Relation with momentum

The First equation can be written as:

${\displaystyle \sum _{i=1}^{N}{\vec {F}}_{i}=m{\vec {a}}\Leftrightarrow \sum _{i=1}^{N}{\vec {F}}_{i}={\frac {d(m{\vec {v}}(t))}{dt}}\Leftrightarrow \sum _{i=1}^{N}{\vec {F}}_{i}={\frac {d{\vec {p}}}{dt}}}$
where ${\displaystyle {\vec {p}}}$is called the momentum of the particle and it is defined to be ${\displaystyle {\vec {p}}=m{\vec {v}}}$

• Newton's second law as a differential equation

Forces in classical mechanics are depended from the position of a particle in space (Example: Gravitational Forces), from the velocity of a particle moving in space (Example: Forces exerted by gases or the Lorentz force) and there can be time-varying forces

(${\displaystyle F_{total}=g(t)}$, where g(t) is some function of time with dimensions ${\displaystyle [g(t)]=[M]\cdot {\frac {[L]}{[T]^{2}}}}$).

When the second law of Newton is written in the form

${\displaystyle \sum _{i=1}^{N}{\vec {F}}_{i}({\vec {r}},{\dot {\vec {r}}},t)=m{\ddot {\vec {r}}}}$
then it is a differential equation that we need to solve for ${\displaystyle {\vec {r}}(t)}$

To illustrate that, consider the following cases:

• Horizontal Motion with Linear Drag:

Consider a particle S moving horizontally in a linearly resistive medium (where at ${\displaystyle t=0,x=0,v=v_{o}}$) and also assume that the force exerted by the medium on the particle is ${\displaystyle F_{x}=-bv_{x}}$. The second law of Newton will completely define the trajectory of the particle. Therefore:

${\displaystyle F_{x}=m{\ddot {x}}\Leftrightarrow -bv_{x}=m{\ddot {x}}\Leftrightarrow -bv_{x}=m{\dot {v}}_{x}}$
Finally, the differential equation takes the following form:
${\displaystyle {\dot {v}}_{x}=-kv_{x}\quad ,k=b/m}$
which is a first-order differential equation (whose general solution must contain exactly one arbitary constant) The equation above states that the derivative of the velocity is -k times the function of velocity, and the function that satisfies this propery is the exponential function:

${\displaystyle v_{x}(t)=Ae^{-kt}}$
We know that ${\displaystyle v_{x}(0)=v_{o}}$so ${\displaystyle v_{x}(t)}$becomes:
${\displaystyle v_{x}(t)=v_{o}e^{-kt}}$
Graph of the particle's velocity for ${\displaystyle (A,k)=(4,1)}$

Notice that the${\displaystyle \lim _{t\to \infty }v_{x}(t)=0\,{\frac {m}{s}}}$, so after a sufficient amount of time (theoretically infinite) the particle will come to a stop. Based on the particles velocity we can find the position since it is true that: ${\displaystyle x(t)=x(0)+\int \limits _{0}^{t}v_{x}\centerdot dt}$ Therefore, the position function will be:

${\displaystyle x(t)={\frac {v_{o}}{k}}(1-e^{-kt})}$
Graph of the particle's position for ${\displaystyle (A,k)=(4,1)}$

Notice that ${\displaystyle \lim _{t\to \infty }x(t)={\frac {v_{o}}{k}}}$

• Force proportional to the distance (Linear motion)

Suppose we have a particle S ${\displaystyle (t=0,\,x(0)=0,\,{\dot {x}}(0)=v_{xo})}$ and a Force ${\displaystyle F_{x}}$ is exerted on it where ${\displaystyle F_{x}=kx\,,\,k>0}$

Once again, The second law will determine the particle's position function:

${\displaystyle m{\ddot {x}}=kx\Leftrightarrow {\ddot {x}}=a^{2}x\,\,,\,a={\sqrt {\frac {k}{m}}}}$
The solution of this second order differential equation is the function:

${\displaystyle x(t)=C_{1}e^{at}+C_{2}e^{-at}}$
Where C1 and C2 are arbitary constants that are determined by the inital conditions. Therefore:

${\displaystyle {\begin{cases}x(0)=0\Leftrightarrow C_{1}+C_{2}=0\\{\dot {x}}(0)=v_{xo}\Leftrightarrow C_{1}-C_{2}={\frac {v_{xo}}{a}}\end{cases}}}$
The position function becomes:

${\displaystyle x(t)={\frac {v_{xo}}{2a}}e^{at}-{\frac {v_{xo}}{2a}}e^{-at}={\frac {v_{xo}}{a}}({\frac {e^{at}-e^{-at}}{2}})={\frac {v_{xo}}{a}}sinh(at)}$
Graph of the particle's velocity ${\displaystyle (v_{xo},a)=(0.3,1)}$

Graph of position:

• Constant Force

This time, the force exerted on an arbitary particle S is ${\displaystyle {\vec {F}}(t)=\langle A,0\rangle \,\,\,A>0}$ (Assume ${\displaystyle t=0,\,{\vec {r}}(0)={\vec {0}},\,{\dot {\vec {r}}}(0)=\langle v_{xo},v_{yo}\rangle }$). The second law of Newton gives us immediately the second derivative of position.

${\displaystyle {\ddot {\vec {r}}}=\langle {\frac {A}{m}},0\rangle }$
where m is the mass of the particle. Based on the definition of velocity we get:

${\displaystyle {\dot {\vec {r}}}=\langle v_{xo}+{\frac {A}{m}}t,v_{yo}\rangle }$
Integrating the velocity function with respect to time gives us the particle's position:

${\displaystyle {\vec {r}}=\langle v_{xo}t\,+{\frac {A}{2m}}t^{2},v_{yo}t\rangle }$
Depending on the constants ${\displaystyle A,v_{xo},v_{yo}}$ the particle will undergo a parabolic or linear trajectory.