Confusion over units in force equation?

Question

While discussing Newton's laws, our book says

Force is proportional to rate of change of momentum

so they say

F is proportional to mass * acceleration if mass is constant

So $F=kma$ where $k$ is a constant.

They then say we choose a unit of force such that it produces acceleration of $1\ \mathrm{m/s}^2$ in $1\ \mathrm{kg}$ mass so $1\ \mathrm{N}=k\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{m/s}^2$. Then they say $k=1$.

How is $k=1$? It should be $1\,\mathrm{N}/(1\,\mathrm{kg\, m/s}^2)$, which is different than just $1$. Force is always written as $F=ma$ not $F=kma$ which seems false.

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This question is different as it asks the actual concept of dimensions rather than other number the question asker of other question was confused about the choice of number not of dimension.

Answer 1

The unit of force in the International System is Newton (N), which is equal to kg m/s$^2$. Newton is a derived unit, all units can be expressed as a product of the seven base units of the SI. Therefore, the $k$ in the formula is dimensionless.

The book explains that you can choose $k=1$ so that Force can be defined as being equal to rate of change of momentum, instead of proportional to rate of change of momentum. If early physicist had defined Force as half the rate of change of momentum, the correct formula would have been $F'=\frac{1}{2}ma$, where $F'$ is the newly defined force. $F$ and $F'$ have the same dimensions (and units), but they are different magnitudes.

You can compare this to the concepts of radius and diameter: you can state that the radius of a circumference is proportional to the length of its perimeter by a factor of $\frac{1}{2\pi}$. You can also say that the diameter is proportional to the length of its perimeter by a factor of $\frac{1}{\pi}$. The radius and the diameter are both lengths (measured in $m$), but they are defined differently.

Answer 2

If you have a proportionality $F\propto m\,a$ (or $F \propto \frac{dp}{dt}$) then using a constant $k$ produces an equality $F=k\,m\,a$ or ($F= k\, \frac{dp}{dt}$).

In terms of dimensions $[F]=[k]\,[m]\,[a]\Rightarrow [F]=[k]\rm \,M\,LT^{-2}$.

The BIMP booklet on SI units (page 118) states that the name of the derived unit of force is called the newton (symbol $\rm N$) and expressed in SI base units as $\rm kg\, m \, s^{-2}$ means that $[F]=\rm M\,LT^{-2}$ and so $[k]=1$ ie a dimensionless constant.

Answer 3

I'm breaking your question into parts to identify what confuses you:

so $F=kma$ where $k$ is a constant

OK until there.

They say we choose unit of force such that it produces acceleration of 1ms-2 in 1 kg mass so 1N=k*1kg*1ms-2

So they "define" $1N$ such that $F=1 [N] = k \times 1[kg] \times 1[m.s^{-2}]$. If $k$ where equal to $2[K]$ ([K] = Kelvin), then we would have $[1N] = 2 [K.m.s^{-2}]$ as a definition of $1N$. But that's not the case, because that's not how $1N$ is defined (see @zdimension's comment below for example). The issue is that the definition in your book, from what you report, is not a definition (except if the author defined "force" before), as my example above has just proven.

Answer 4

I remember also being confused when first starting to learn physics.

Force isn't just "proportional" to the rate of change in momentum. A force IS the rate of change in momentum.

A force of one newton means you're changing the momentum of the object you're applying the force to by $1kg(M/s)^2$. Hence if the object weighs one kilogram, it'll accelerate at 1 meter per second squared. If it weighs 2 kilograms, it'll accelerate at half a meter per second squared.

That "unit" you're saying $k$ should be, $\frac{1N}{1Kg(M/s)^2}$ is, by the definition of a Newton, just equal to $1$.

Why?

Because a Newton is $1kg(M/s)^2$!!!!!!!!!!

Answer 5

As already said, the simplest way out of this dilemma is to realise that force is the rate of change of momentum. But when we write this as an equation we will have to introduce a constant of proportionality if the units on the two sides of the equation are not the same.

For example, if we measure force in dynes, mass in kilograms and acceleration in metres per hour squared then to get correct numerical results we must have

$F = \frac{1}{129.6}ma$

But this would be crazy. So (in SI units) we measure mass in kilograms, acceleration in metres per second squared and force in kilogram metres per second squared. If we work in these units life is simpler because we just have

$F=ma$

And to save having to write "kilogram metres per second squared" over and over again, we just call these units "Newtons" for short.

Note that the SI system is not perfect. Working in SI units still involves various constants of proportionality such as the speed of light $c$ and the gravitational constant $G$. so we have

$E=mc^2\\F=\frac{Gm_1m_2}{r^2}$

Physicists sometimes work in an alterative system of units called Planck units, where these constants (and others) become $1$ too. In Planck units we have

$E=m\\F=\frac{m_1m_2}{r^2}$

Answer 6

There's a tricky issue behind this question: is force another name for rate of change of momentum ($d\vec p/dt$), or is force something that may give rise to rate of change of momentum?

I have to admit that I'd go for the 'weak' second alternative: force as agency often associated with $d\vec p/dt$. When we talk of 'applying a force' do we have the $d\vec p/dt$ of some body in mind, or an agency such as a hand pulling? Perhaps more cogently: when we apply two forces to a body at the same time, neither of the individual forces is usually equal to the body's $d\vec p/dt.$

Nonetheless, we don't have to insist that this distinction between force and $d\vec p/dt$ is reflected in the unit of force. We can call the strength of a force in newton the $d\vec p/dt$ in kg m s$^{-2}$ of a body to which (only) that force is applied. That is 1 N = 1 kg m s$^{-2}.$

[There's a partial analogy with work and energy. Energy isn't the same thing as work but, arguably, the energy of a system may be defined as the amount of work it can do. To use the same unit for energy that we use for work seems the obvious thing to do. Maybe it's not quite so clearcut in the case of force and $d\vec p/dt,$ but I did say partial analogy!]