# What are point objects?

I can't seem to get the idea of point mass into my head. Why are equations of physics applicable on only point masses and should be altered while dealing with object that has a collection of points? We don't see point objects in the real world, so why consider using it in a physical science?

• physics.stackexchange.com/q/140885 – BowlOfRed Jun 11 '18 at 5:41
• Because it's often a useful model: it vastly simplifies the calculations, while still yielding reasonably accurate answers. – PM 2Ring Jun 11 '18 at 6:17
• This is an approximation tested and found good for some problems. A next approximation is to consider a rigid 3D body and there are equations for such a model too. Also there are equations for deformable bodies. Each approach has its own field of applicability outlined with inequalities like $D\ll R$ etc. – Vladimir Kalitvianski Jun 11 '18 at 6:56
• – Stéphane Rollandin Jun 11 '18 at 8:21

In many situation the shape and size of an object are not relevant. The properties of such an object can be lumped in a point particle. Even objects that are clearly extended such as stars and planets are often well approximated by point masses. In the case of subatomic particles, such as electrons, we are physically unable to determine the size, we only know that it is very small, so small that it does not affect measurement.

Imagine that I ask you to calculate how long it takes for you to drive 2 km to the bakery with an average speed of 50 km/hr. Do you need to know if it is a Ferrari or an Opel?

Nope.

On your sketch and in any discussions, you can just think of the car as was it no more than a simple point. You can model it as a point. That point would move in the same way anyways, so it doesn't matter. And isn't it smart to remove unnecessary info and make an as-simple-as-possible scenario when solving issues and problems?

• And, as they say: all models are wrong, some models are useful. – Emilio Pisanty Jun 11 '18 at 9:51

Suppose that you were asked to find the acceleration of a block of mass $M$ resting on a frictionless surface when a horizontal force $F$ was applied to the block.
It can be shown that in such an example the block is equivalent to a point mass $M$ located at the centre of mass of the block $C$ and so the free body diagram can be drawn as per the middle image.

Newton's second law can then be applied $F=Ma$ where $a$ is the acceleration of the block.

Using this simple model you could then predict the maximum value of the force $F$ which could be applied to the block before it toppled when the line of action of the force $F$ does not pass through the centre of mass of the block $C$.
The right hand image shows the forces acting on the block with the line of action of the normal reaction $N$ not passing through the centre of mass of the block $C$.
The onset of toppling would occur when the normal reaction $N$ was acting at the left hand end of the block.

Of course all this is very much a simplification but you have probably done lots of experiments in Mechanics (and other branches of Physics) to verify this and that with the assumption that an extended mass can be thought of as a point mass.

A step forward which is used by engineers and physicist is to split the block into a very large number of very small masses and compute what happens numerically.
This is called finite element analysis and it will potentially yield more accurate predictions but of course this it will take a lot longer than the point mass method mentioned above.