How can the unit of Newtons measure both gravity and electromagnetism?

Question

For context, I was thinking about forces. Force is the mass of the particle multiplied by the acceleration it undergoes. There are different kinds of forces, for example, gravity and electromagnetism. While gravity uses mass as it's quality, electromagnetism uses charge. Given how charge is not in the force equation, I wondered how newtons could measure electrical forces. The only conclusion I could come up with is that mass and charge are related on a fundamental level, or possibly even the same thing. Am I correct, or is there something else that I'm not understanding?

Answer 1

Force is what a force gauge measures. So, attach a charged object to a force gauge, bring another charged object close to it, and measure a Coulomb force. Or, attach a mass to it, place it in a gravitational field, and measure a gravitational force. Or, attach a mass, accelerate it using the attachment, see $F=ma$. If the gauge is calibrated in newtons, that's what you measure. What produces the force doesn't matter to the gauge.

Answer 2

There's no reason that the force exerted on an object should have any functional relationship to its mass, so I suspect your insistence that it does is the root cause of the misunderstanding. Electromagnetic forces, buoyant forces, and macroscopic contact forces don't depend on mass. Essentially the only force which does is gravity, and the reason for this remained mysterious until it was explained in the context of General Relativity.

It's true that force and mass both appear in Newton's 2nd law, but you should view Newton's 2nd law as the statement that two quantities which have independent definitions and existence (in this case, net force and acceleration) are proportional to one another.

Answer 3

When discussing forces we are usually discussing its applications to massive objects. For most practical purposes, we can just use Newton's second law in the form of $$\mathbf F =m\mathbf a,\label{newt}\tag{1}$$ where $\mathbf F$ is the force (or sum of forces) acting on an object with mass $m$ and acceleration $\mathbf a$. So any force acting on the object affects its motion (acceleration) in proportion to its mass, no matter the nature of $\mathbf F$. The force $\mathbf F$ could be the gravitational force, the force of a spring, electromagnetic forces and so on. In a sense the idea that all forces can be measured in newtons (mass times acceleration, kg m/$s^2$) it is telling you more about its relation to the motion being proportional to mass than about the nature of the force itself.

Examples

Let us compare the action of Earth's gravity $\mathbf g$ on a ball vs the action of an electric field $\mathbf E$ on the same ball. The graviational force close to the surface of the Earth is

$$\mathbf F_g=m\mathbf g\label{eq1}\tag{2}$$ where $\mathbf g$ is the acceleration due to gravity, that points to the center of the Earth.

If the ball is charged with electric charge $q$, the force due to the applied field $\mathbf E$ is given by

$$\mathbf F_e=q\mathbf E.\tag{3}\label{eq2}$$

Both \eqref{eq1} and\eqref{eq2} are different in nature, \eqref{eq1} takes into account the mass of the ball and \eqref{eq2} takes into account only its charge $q$.

But how much would the ball accelerate depends on the mass because the mass appears in the right hand side of \eqref{newt} ! For \eqref{eq1}, we have the special case that both the force and Newton's law depend on mass, so any body under this force will fall at the same rate $\mathbf a=\mathbf g$ independently of the mass. In case \eqref{eq2}, how much the body accelerates is given by $\mathbf a = \frac{q}{m}\mathbf E$, so the larger the mass the less it can be accelerated. In both cases mass is playing a role (different roles if you like) but it is because it appears on the right hanside of \eqref{newt} not on the left hand side.

Note I could go on with any other force. For a spring the force is $\mathbf F_k=k \mathbf r$, where $k$ is the spring constant and $\mathbf r$ is the displacement from equilibrium. In that case the acceleration is $\mathbf a =\frac{k}{m}\mathbf r$ (the higher the mass the less it will accelerate due to $\mathbf F_k$).