# $F=ma$ proof, undesired result

I believe we all know the famous equation $$F=ma$$

My attempt to prove this:

Newton's second law states that the rate of change of momentum of a body is directly proportional to the resultant force acting on the body, and this takes place in the same direction as the resultant force.

Hence, I can get $\frac{d\bar{p}}{dt}\varpropto\sum\bar{F}$. From here, we have $\sum\bar{F}=k\frac{d\bar{p}}{dt}$, where $k$ is a constant. We find that $k=1$ by comparing S.I. units. Now we have $$\sum\bar{F}=\frac{d\bar{p}}{dt}$$

By Newton's second law, $p=mv$, $$\sum\bar{F}=\frac{d(m\bar{v})}{dt}$$

By product rule, $$\sum\bar{F}=m\frac{d\bar{v}}{dt}+\bar{v}\frac{dm}{dt}$$

How do I get $F=ma$ from here? I know that the left term is $ma$, meaning that the left term $\bar{v}\frac{dm}{dt}$ is $0$. Why is this so?

• Since $m$ has no dependence of $t$, the change in mass over time ($\frac{dm}{dt}$) will be 0 Jan 28, 2018 at 10:05
• I think the formula is more like $\sum F=\sum \frac{dp}{dt}$, it seems to be the same as the rocket equation, since that one splits the terms depending on what velocity which mass has.
– Emil
Jan 28, 2018 at 14:33
• $m dv/dt + dm/dt v + ndu/dt + dn/dt u$ with m rocket without the exhaust, n the exhaust mass, u the exhaust velocity and v the rocket without exhaust velocity. It looks the same to me at least :-)
– Emil
Jan 28, 2018 at 14:47
• Also, I think atoms tend to not change mass except maybe in chemical reactions.
– Emil
Jan 28, 2018 at 14:51
• Any reason for downvote? Jan 29, 2018 at 11:48

The famous equation is only true when the mass remains constant, that is, when the derivative of mass is zero. Thus what you have derived is in fact the correct law of dynamics.

The second law only applies to constant-mass system, so: $$\frac{dm}{dt}=0\;\;\Longrightarrow\;\; F=ma$$

Of course the second law still applies to any system, but in such a case of variable mass you should apply it to the system composed by the body itself and the mass which is being added or removed.

When you us the equation $F=ma$ the assumption made but often not stated is that the mass is constant.

You question highlights the fact that when mass flows the application of Newton's second law must be done with care as illustrated in these examples.

You should be knowing that the derivative of a constant is 0.

That's why $\vec{v} \dfrac{dm}{dt} = 0$ as mass is constant.

Hence, it comes down to $\vec{F} = m \vec{a}$.

Your equation would hold true when the system has a variable mass, like in rocket motion.

• "Your equation would hold true when the system has a variable mass, like in rocket motion." Is that really true? Shouldn't that be $F+v_r\frac{dm}{dt}=m\frac{dv}{dt}$, being $v_r$ the velocity of the added/subtracted mass? Jan 28, 2018 at 10:17