# Question about inertial mass and gravitational mass

I know that inertial mass $m_i$ is the quantity that appears in Newton's second law: $F=m_ia$ and that gravitational mass $m_{g_1}$ is the quantity that appears in Newton's gravitational law: $F_g=Gm_{g_1}m_{g_2}/r^2$ and that we can measure it using a balance.

This is my problem:

Every text says that we can calculate $m_i$ simply by $m_i=F/a$ but nobody says which unit of measure I should use for $F$. This is a problem for me.

For example, using a dynamometer and so the static definition of force, one could say that $1 N$ is equal to the weight of a $102 g$ gravitational mass. So in this case we use the gravitational mass to calculate the inertial mass. Isn't it strange?

Moreover, if in this way I calculate $m_i$ then just using a simple balance I can verify that $m_i=m_g$. So, in theory, we don't need any difficult experiments to find that $m_i=m_g$.

What is/are my mistake/s?

• Related: physics.stackexchange.com/q/70186/2451 and links therein. – Qmechanic May 28 '16 at 12:03
• Sorry what do you mean? Is the answer to my question in that link? It seems different to me. thank you – Richard May 28 '16 at 12:08
• @Richard Advice: in such situations, extend your question or/and explain in a comment, what is the essential point which makes them different. In theory, if you fix the question after it was closed, then it should be reopened, but in practice, you have much better chance to avoid its close as to reopen it later. – peterh May 28 '16 at 12:16
• thank you for the advice. In my opinion the difference is huge because the linked question is about the statements of Newton's law. My question is about the definitions of mass. – Richard May 28 '16 at 12:19
• @Richard I'm afraid you uses. Otherwise I didn't really understand your point. The spring deformation is associated to $F$. So for your collection of masses you have $F=m_ja_j$. Using this equation and the fact that $F$ is unique (the deformations were chosen to be always the same value) you can eliminate $F$ for any two given pair of particles. – Diracology May 28 '16 at 12:37

## 2 Answers

You do not need a unit for force when measuring inertial mass in Newtonian Mechanics. The only things you really need are the Newton's second law and the concepts of inertial frame and acceleration. The way you shall proceed is the following.

Take a collection $\{m_i\}$ of (unknown) masses and a spring. Use the spring horizontally to accelerate the masses in a horizontal and friction less plane and measure their accelerations with respect to an inertial frame. Maintaining the same spring deformation for all masses you have $F=m_ja_j$. Then you can eliminate $F$ between any pair of masses. Choose one of them to be the standard one, namely $m_0$. Then any other inertial mass will be given in units of $m_0$, $$m_j=\frac{a_0}{a_j}m_0.$$ You can repeat the process using other forms of interaction and you shall obtain the same ratios $a_0/a_j$ which means the mass of a particle is an intrinsic property.

On the other hand, when you use a spring balance or a dynamometer you are actually measuring the strength of gravity instead of the inertia of the particle. In the equilibrium position the right hand side of $\sum F=m_ia$ ($i$ stands for inertial mass) vanish and you lose any information about inertia. To retain information about inertial mass you have to consider dynamic configurations such as a pendulum, a sliding block on a inclined plane or a free fall of a body. In any of these cases you will get equations of motions with gravitational mass on one side and inertial mass on the other. In principle there would be no reason for them to be equal. What we do is to compare the period of the pendulum, or the times taken during the sliding or the free fall with the observed results and conclude that $m_i/m_g=1$.

Let's look at:

$$F_g=Gm_{g_1}m_{g_2}/r^2$$

For an object with mass $m_{g_1}$, on the surface of the earth, then:

$$F_g=Gm_{g_1}M_E/R_E^2$$

Where $M_E$ is the mass of the Earth and $R_E$ the radius of the Earth. You can now verify that:

$$GM_E/R_E^2=g=9.81\:\mathrm{m/s^2}$$

So we could have written the second expression as:

$$F_g=m_{g_1}g$$

An object has ONE mass (what you call $m_{g_1}$), regardless of whether it is subject to acceleration, a central gravitational field or not even subject to any forces at all. It's an intrinsic property of the object. It's pointless to talk of "gravitational mass" or "inertial mass", it's a distinction without a difference.

As regards units of measurement, in science we use the SI system of units: $\mathrm{N}$ (Newton) for force and $\mathrm{kg}$ for mass. But it's possible to use other units of measurements (like Imperial units), as long one uses one system or the other consistently the obtained results can always be converted from one system to the other.