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

thanks

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

thanks - May 4th 2011, 09:14 PMTheEmptySet
You need to show that the units match on both sides of the equation.

Force has units

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

$\displaystyle m=\text{kg}$

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

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

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

So now analyze the units

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

So the units on both sides are the same - May 4th 2011, 11:06 PMbugatti79
- May 4th 2011, 11:14 PMbugatti79
- May 5th 2011, 01:56 AMAckbeet
- May 5th 2011, 07:33 AMhazeleyes
- May 5th 2011, 07:39 AMAckbeet
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 AMhazeleyes
but why dm/dt is kg/s?

- May 5th 2011, 07:49 AMTheEmptySet
- May 5th 2011, 07:52 AMhazeleyes
Thanks so much!!

- May 5th 2011, 07:59 AMAckbeet
More generally, if I have a function $\displaystyle y=f(x),$ and the units of $\displaystyle y$ are $\displaystyle [y],$ and the units of $\displaystyle x$ are $\displaystyle [x],$ then the units of

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

Similarly, the units of

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