Mass falls in unit time

1. Dec 11, 2008

physiker99

Assume there is an area S. Then we have dust particles falling over it having velocity v and density rho.
Is is correct to state mass falling over the S in unit time equals (rho)*(S)*v?

I guess it should be correct, because mass shouldn't have velocity component if none other cancels it.

2. Dec 12, 2008

tiny-tim

Welcome to PF!

Hi physiker99! Welcome to PF!

Yes.
If you mean that, dimensionally, the 1/L3 in the M/L3 in rho has to be converted to 1/T, then that's correct.

3. Dec 12, 2008

physiker99

I did a typo there. I don't understand how velocity is used to calculate mass.

4. Dec 12, 2008

D.R.

It's not purely to calculate the mass. It also determines the rate per unit time through an area S. So you can rewrite S and v as:
$$Sv = S\frac{dr}{dt} = \frac{dV}{dt}$$
, with V a volume

The physical interpretation is that a volume V moves through an area S in dt time. See it as moving a cube through a square hole in dt time. The density is defined as mass per volume. So you can rewrite:
$$\rho \frac{dV}{dt} = \frac{d\rho V}{dt} = \frac{dM}{dt}$$
, with M the mass

Last edited: Dec 12, 2008
5. Dec 12, 2008

tiny-tim

Welcome to PF!

I'll just add this to what D.R. says:

dimensionally, rho is M/L3, S is L2, and V is L/T

so multiply them all together, and you get M/L3 x L2 x L/T, = M/T, which is mass per time.

oh, and welcome to PF, D.R.!

6. Dec 12, 2008

physiker99

It's understood. Thanks a lot.