# F=ma, what is m actually means?

1. Apr 28, 2006

### hongkongrubbish

My maths teacher taught me that vector cannot divide vector
but the physics states that F and A are vectors
so m=F/A, what m means?
are there any differences about maths vector and phy vectors?:rofl:

2. Apr 28, 2006

### robphy

m can be thought of a scalar ("scaling factor") that gets you the vector F from the vector a.... in particular, it tells you that vector F is parallel to vector a. You need not attempt to divide vector F by vector a in order to mathematically interpret m in that equation.

One difference is that physics-vectors often carry units. So, e.g., one cannot add F and a, since they have different units [since m carries units of mass].

3. Apr 28, 2006

### andrevdh

Physics use mathematics as tools in order to quantatively describe the nature of the universe. That means that all the normal mathematical rules apply when you do the calculations, but some physical meaning is attached to the mathematical quantities also. The vector equation
$$\vec F=m\vec a$$
states that the force vector is obtained by scaling (making longer or shorter by the mass) the acceleration vector with the mass of the object like robphy remarked in his post.

So we physicist decide how to use the maths to our advantage, they are just tools for us (we are not allowed to reformulate their rules though), but it has happened in some cases that the maths actually helped us to discover how the universe actually works! Which is quite fantastic since maths is our creation, that is all because we have done such a great job at designing it logically. For example in Kepler's time it was thought that the planets experienced a force pulling them along their path, that is some force exists that is acting in the direction of the motion of the planets. Using calculus and his belief that there is a force of attraction between any two bodies that is directed towards them (a central force) he proofed Kepler's laws mathematically. Thereby suggesting that the planets are moving along their path due to a central force and not a force that is along their paths. This prompted scientist to investigate if such a central force does exist - and the rest is history.

Of cause you can calculate the mass with
$$m=\frac{F}{a}$$
but these two equations are from two different domains in mathematics and completely different sets of rules apply to them.

Last edited: Apr 29, 2006
4. Apr 29, 2006

### ekinnike

is it just simply m = mass?

5. Apr 29, 2006

### bavenger

yep
force = mass times acceleration
so m = mass

6. Apr 29, 2006

### Chi Meson

A small hair-splitting comment here:

Newton's second law intends to quantify the acceleration that ocurs when an unbalanced force is applied to an object, a =F/m . In this relationship, m represents the quantity of inertia (which is an object's resistance to change in motion). It just so happens that the quantity of mass is equivalent to the quantity of inertia. This little factoid gets more peculiar as you get deeper into physics.

But technically, the m in this relationship refers to "inertial mass" (as opposed to "gravitational mass"--even though both "masses" still appear to be the same thing).

7. Apr 29, 2006

### Andrew Mason

In order to divide a vector by another vector, you would have to define vector division. I am not aware of such a definition.

$$\vec F = m\vec a$$ means

$$|\vec F|\hat F = m|\vec a|\hat a$$

where $\hat F \text{and} \hat a$ are unit vectors in the direction of $\vec F \text{and} \vec a$ respectively.

This means that:

$$m = \frac{|\vec F|}{|\vec a|}$$ and

$$\hat F = \hat a$$

AM

8. Apr 29, 2006

### robphy

This works for a positive m, which is fine for mass... but this is too restrictive in general. It is more correct to say
$$|m| = \frac{|\vec F|}{|\vec a|}$$
Consider $$\vec F=q\vec E$$, where q need not be positive.

A better way that can account for the sign is to write m as a ratio of two dot-products:
$$m = \frac{\vec F\cdot\vec a}{\vec a\cdot\vec a}$$
However, this by itself doesn't tell you that $$\vec F$$ is parallel to $$\vec a$$.