# External and internal forces (Dynamics of system of particles)

This lesson is about External and internal forces (Dynamics of system of particles) . This chapter comes under subject Mechanics and is for B.Sc. Physics students.

While studying Newton’s laws of motion most of the time we have considered objects (for example planets, projectiles etc..) to be point particles rather then extended bodies. We did this because while dealing with problems like projectile motion or planets we do not have to deal the interactions that is there between many internal particles that make up these extended bodies.
In this chapter we will try solving problems involving several interacting particles. While studying dynamics of system of particles we generally consider a system that consists of $n$ number of particles where $n$ could be any integer number. Here you can always treat individual particles as you have always treated before , that is you can define co-ordinate system for each and every particle in the system.You can even use Newton’s second law of motion and even find $n$ second order coupled ODE’s.
A system of $n$ particles may be volume of gas molecules or it may be a rigid body (where constituent particles are restrained from moving related to one another). Here we will mostly focus on overall motion of the system and will also study about rigid body dynamics.

## External and internal forces

For studying mechanics of system of particles we shall first learn to distinguish between external forces acting on particle due to agents outside the system and internal forces on some particle say $i$ due to all other particles of the system. We can say that internal forces in the system is exerted by the particles of the system on one another.
Here we are assuming all the particle to be point particles where each one of them has a specified position in space. Now let us consider a system that consists of $n$. You can name the particles from $1,2,3,…..n$ and $m_{1},m_{2}, m_{3}, ……m_{n}$ be the masses of each one of these particles. If $r_{i}$ is the position and ${\vec F_i}$ is the net force on i’th particle then ${{\vec p}_i} = {m_i}{{\vec r}_i}$ would be the momentum of the i’th particle.So, the equation of motion for the i’th particle can be written as
${{\vec F}_i} = \sum\limits_j {{{\vec F}_{ji}} + } {{\vec F}_{i(ext)}} = \frac{{d{{\vec p}_i}}}{{dt}}$           (1)
Here the force ${\vec F_i}$ on the i’th particle has been splited into two terms.
Firest term:- ${{\vec F}_{ji}}$ which is the first term in above equation (1) is the internal force on the i’th particle and $\sum\limits_j {{{\vec F}_{ji}}}$ denotes sum of forces acting on i’th particle due to all other particles in the system.
Second term:- ${{\vec F}_{i(ext)}}$ is the second term which is the external force acting on the particle i . By external force we mean that the force that is being applied on the particle is due to sources outside the system.
Now here the internal force ${{\vec F}_{ji}}$ will obey the Newton’s Third law of motion according to which

The force exerted by two particles on each other are equal and opposite

Obviously this statement does not hold for all type of forces.
Now we sum the equation (1) over all the particles of the system. So now we have
$\frac{{{d^2}}}{{d{t^2}}}\sum\limits_i {{m_i}{{\vec r}_i} = \frac{d}{{dt}}\sum\limits_i {{{\vec p}_i}} = \sum\limits_i {{{\vec F}_{i(ext)}} + \sum\limits_{j,i\atopi \ne j} {{{\vec F}_{ji}}} } }$
(2)
where $\vec{p}_i=m_{i}\dot{\vec{r}}_i$ is the momentum of the i’th particle.
In the above equation term $\sum\limits_i {{{\vec F}_{i(ext)}}}$ is the sum of all external forces acting on the system i.e., ${{\vec F}_{ext}}$. So,
${{\vec F}_{ext}} = \sum\limits_i {{{\vec F}_{i(ext)}}}$
Again the sum $\sum\limits_{j,i\atopi \ne j} {{{\vec F}_{ji}}}$ is nothing but the sum of all internal forces acting on all the particles.
Newton’s third law of motion plays an important part in dynamics of system of particles due to the internal force acting between the particles of the system.

The forces acting between any two particles of the system on each other are equal in magnitude and opposie in direcction

So here from Newton’s third Law of motion
${{\vec F}_{ji}} = – {{\vec F}_{ij}}$
Due to this equal and opposite force the term $\sum\limits_{j,i\atopi \ne j} {{{\vec F}_{ji}}}$ vanishes
i.e.,
${{\vec F}_{ji}}+{{\vec F}_{ij}}=0$
this implies that , $\sum\limits_{j,i\atopi \ne j} {{{\vec F}_{ji}}} = 0$
This simplifies equation 2 to
${{\vec F}_{ext}} = \sum\limits_i {\frac{{d{{\vec p}_i}}}{{dt}}}$               (3)
Now right hand side of above equation (3) can also be written as
$\sum\limits_i {\frac{{d{{\vec p}_i}}}{{dt}}} = \frac{d}{{dt}}\left( {\sum {{{\vec p}_i}} } \right)$
here $\sum {{{\vec p}_i}}$ is the total momentum of the system and can be denoted by ${\vec p}$.
This implies that $\vec p = \sum {{{\vec p}_i}}$
Putting this in equation (3) we get
${{\vec F}_{ext}} = \frac{{d\vec p}}{{dt}}$
From above equation we can conclude that total external force applied to a system is equal to the rate of change of momentum of the system. This statement is also true even if there are bunch of forces acting on the body.

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