# Math Help - dimensional homogeneity

1. ## dimensional homogeneity

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

thanks

2. 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

3. 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?

4. 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

5. 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:

$\mathbf{F}=\dot{\mathbf{p}}.$

6. 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)= $m^3s^{-1}$

7. 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!).

8. but why dm/dt is kg/s?

9. 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

10. Thanks so much!!

11. 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 $[y][x].$