# Why are Newton's laws valid only for rigid bodies?

As far as I know, Newton originally wrote the first and second laws for a particle. Under the condition that a collection of particles is rigid, we can write Newton's second law for a body as $$F_{net}=Ma_{cm}$$ where $$a_{cm}$$ is the acceleration of the center of mass.

This is confirmed by Wikipedia (look under overview section). But I don't see where the condition for the body being rigid is used in obtaining the above equation from Newton's laws for a particle.

Let there be a rigid body where each particle in the body experiences a net force of $$dF$$. According to Newton's laws, $$dF=dM\ a\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $$dM$$ is a differential mass.

Now, $$\int_{}^{}dF=\int_{}^{}dM\ a$$ where the integral is for all masses. By definition of center of mass, $$r_{cm,anypoint}=\frac{m_{1}r_{1}+m_{2}r_{2}+m_{3}r_{3}+.....m_{n}r_{n}}{m_{1}+m_{2}+m_{3}+m_{4}+......m_{n}}$$ double differentiating with respect to time on bot sides, we get $$a_{cm,anypoint}=\frac{m_{1}a_{1}+m_{2}a_{2}+m_{3}a_{3}+.....m_{n}a_{n}}{m_{1}+m_{2}+m_{3}+m_{4}+......m_{n}}$$ for a collection of differential masses, this can be written as $$(\int_{}^{}dM\ a)=(\int_{}^{}dM)\ a_{cm}$$ this means that $$(1)$$ cna now be written as $$F_{net}=(\int_{}^{}dM)\ a_{cm}$$ and so $$F_{net}=M\ a_{cm}$$

The definition of center of mass is applicable for all bodies, rigid and nonrigid. Where have I used the condition that the body has to be rigid?

• Newton's laws are valid for all bodies. Where does it say that the body has to be rigid? The complications from applying them to fluids are completely covered in continuums mechanics, we simply can't teach that in a beginners class or in highs school science, the math is too nasty. Commented Dec 27, 2015 at 3:33
• I've been taught that the equation $f=Ma$ is valid only for rigid bodies. Do you mean that even if the body is not rigid, the equation would hold irrespective of the complexities of other math? And I've tried to apply the equation for relatively simple non rigid bodies such as pulley systems but it doesn't work Commented Dec 27, 2015 at 3:42
• Would your teacher consider Earth, the sun or any of the gas planets as solid? If not, why limit the applicability of Newton's laws to solids? They work just fine for liquids and gases, one simply has to take a lot more care with forces/pressures/shear etc. on volume elements, if the acting forces are not homogenous. You don't have to invent any of this yourself, by the way. You can look up the entire theory for continuous matter in a good text about continuums (or fluid) mechanics. Commented Dec 27, 2015 at 3:58
• Then can I apply $F_{net}=Ma_{cm}$ for systems with objects moving at non zero relative velocites? Commented Dec 27, 2015 at 5:38
• You have to apply individual forces to individual volume elements, which requires that you replace them with a stress tensor: en.wikipedia.org/wiki/Cauchy_stress_tensor. Like I said, all of this is well understood, it's just not a simple and neat mathematical formalism. One has to work pretty hard to derive it correctly for moving fluids and the resulting equations of motion are very hard to solve. Commented Dec 27, 2015 at 5:56

The body does not have to be rigid. Newton's laws work for all bodies. If you are interested only in the motion of the centre of mass of some object, then it is the case that

$$\vec{F}_\text{net} = M\vec{a}_\text{cm}$$

Where $$\vec{F}_\text{net}$$ is the net applied force and $$\vec{a}_\text{cm}$$ is the acceleration of the centre of mass.

The problem is that for non-rigid bodies this is only a tiny part of what is interesting about the system: you are almost certainly interested in what happens to the shape of the thing or the distribution of mass in it or any number of other interesting things. Imagine, for instance, applying some impulse to a ball (with a bat, say). What we can know immediately is that

$$m\vec{\Delta v}_{cm} = \int_0^T \vec{F}(t)\,dt$$

assuming the impulse is zero outside $$[0,T]$$. This tells us how fast the centre of mass of the ball is moving at the end of the process. But it doesn't tell us anything at all about what shape the ball is during and after the process, and whether, for instance, it falls apart if you hit it hard enough. Those things are interesting to people designing balls, for instance.

To answer those questions you need to treat it in a much more sophisticated way, using continuum mechanics. This, of course, is also using Newton's laws of motion, but it is applying them to the ball considered as a continuous mass distribution, not as a rigid body.