Seeing if something is Dimensionally correct.

[Sirsh]

V_t = (2mg/pA)^1/2

where:
V_t = Terminal Velocity (ms^-1) as velocity = s/t = m/s(seconds, not distance).
m = mass in kg.
g = gravity in ms^-2.
p = fluid density in kgm^-3
A = Area in m²

Show that equation is dimesionally correct.

I seem to get my answer wrong on the end product; ill show my working up too where I've got my answer incorrect.

V_t = (2mg/pA)^1/2

m/s = (2(kg)(m/s²)/(kg/m³)(m²))^1/2

(m/s)² = (2(kg)(m/s²)/(kg/m³)(m²))

(m/s)² = (2(kg)(m⁴)/((kg)(s²)(m²))

(m²/s²) = ((kg)(m²))/(s²)

As you can see the equation is quite simplified at this point, with the only problem being that both left and right sides do not equal, which shows that the equation is not dimensionally correct, i am unsure though how to get rid of this unit of kg in my equation though. can anyone help me out on this one, thanks alot!

 

[Pengwuino]

Why didn't you cancel out the kilograms at any point? From the get go, it cancels out.

 

[Sirsh]

If i cancle the kg out, then i am stuck with a '2' in the final equation instead of a kg unit. because if i try cancle the 2 out by halving that side, then the other side (left hand side) gets 2 in the denominator :/

 

[amy andrews]

I'm not completely sure about this, but does the "2" even matter in dimensional analysis? I believe dimensional analysis is just to check the units. The coefficient isn't involved or checked.

 

[Sirsh]

Good point! ill check that up!