# dimensional homogeneity

• May 4th 2011, 08:18 PM
hazeleyes
dimensional homogeneity
Hi, How to show that Newton's second law $F=\frac{d\left(mv\right) }{dt}$ for dimensional homogeneity.

thanks
• May 4th 2011, 09:14 PM
TheEmptySet
Quote:

Originally Posted by hazeleyes
Hi, How to show that Newton's second law $F=\frac{d\left(mv\right) }{dt}$ for dimensional homogeneity.

thanks

You need to show that the units match on both sides of the equation.

Force has units

$F=\frac{(\text{kg})(\text{m})}{\text{s}^2}$

$m=\text{kg}$

$v=\frac{\text{m}}{\text{s}}$

Now we need to take the derivative on the right hand side

$\frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)$

So now analyze the units

$\frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}$

So the units on both sides are the same
• May 4th 2011, 11:06 PM
bugatti79
Quote:

Originally Posted by TheEmptySet
You need to show that the units match on both sides of the equation.

Force has units

$F=\frac{(\text{kg})(\text{m})}{\text{s}^2}$

$m=\text{kg}$

$v=\frac{\text{m}}{\text{s}}$

Now we need to take the derivative on the right hand side

$\frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)$

So now analyze the units

$\frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}$

So the units on both sides are the same

May I ask how is mass a function of time?
• May 4th 2011, 11:14 PM
bugatti79
Quote:

Originally Posted by bugatti79
May I ask how is mass a function of time?

OK, it could be the a mass flow rate like density by the cross sectional area by velocity
• May 5th 2011, 01:56 AM
Ackbeet
Quote:

Originally Posted by bugatti79
May I ask how is mass a function of time?

With rockets, you are ejecting mass to propel yourself forward. Mass is most definitely a function of time then. The most general form of Newton's Second Law is the version in his Principia:

$\mathbf{F}=\dot{\mathbf{p}}.$
• May 5th 2011, 07:33 AM
hazeleyes
Quote:

Originally Posted by TheEmptySet
You need to show that the units match on both sides of the equation.

Force has units

$F=\frac{(\text{kg})(\text{m})}{\text{s}^2}$

$m=\text{kg}$

$v=\frac{\text{m}}{\text{s}}$

Now we need to take the derivative on the right hand side

$\frac{d}{dt}(mv)=\left( v\frac{dm}{dt}+m\frac{dv}{dt}\right)$

So now analyze the units

$\frac{\text{m}}{\text{s}}\cdot \frac{\text{kg}}{\text{s}}+\text{kg}\cdot \frac{\text{m}}{\text{s}^2}$

So the units on both sides are the same

but isn't Volume flow rate(dv/dt)= $m^3s^{-1}$
• May 5th 2011, 07:39 AM
Ackbeet
In Newton's Second Law, v is velocity, not volume. So it would have units of m/s. Usually, volume is an upper-case V (math is case-sensitive!).
• May 5th 2011, 07:43 AM
hazeleyes
but why dm/dt is kg/s?
• May 5th 2011, 07:49 AM
TheEmptySet
Quote:

Originally Posted by hazeleyes
but why dm/dt is kg/s?

It is the change is mass with respect to time so the units are mass over time or kg/s
• May 5th 2011, 07:52 AM
hazeleyes
Thanks so much!!
• May 5th 2011, 07:59 AM
Ackbeet
Quote:

Originally Posted by TheEmptySet
It is the change is mass with respect to time so the units are mass over time or kg/s

More generally, if I have a function $y=f(x),$ and the units of $y$ are $[y],$ and the units of $x$ are $[x],$ then the units of

$\frac{dy}{dx}$ are $\frac{[y]}{[x]}.$

Similarly, the units of

$\int y(x)\,dx$ are $[y][x].$