Solving Formula of Forces Homework

[foo9008]

Homework Statement

from the notes , the author stated that force has the formula of ρ(L^2)(v^2) and also = ρ(L^3)
i think there's something wrong with the ρ(L^3)

Homework Equations

The Attempt at a Solution

IMO , ρ(L^2)(v^2) can also be written as ρ(L^4)(T^2) , so Force is proportional to L^4 , am i right . ? IMO, the prototype force should be [ (100^4) ] x 0.12 N

correct me if i am wrong ...

 

[tommyxu3]

I'm not sure, but in your picture, one is $\rho$ and the other is $\rho_r.$ Maybe there is detailed definitions in the problem?

 

[foo9008]

I'm not sure, but in your picture, one is $\rho$ and the other is $\rho_r.$ Maybe there is detailed definitions in the problem?

rho r is actually r , it's the pi buckingham theorem

 

[haruspex]

ρ(L^2)(v^2) can also be written as ρ(L^4)(T^2) , so Force is proportional to L^4

I guess you mean ρ(L^4)(T^-2), but either way it does not follow. The whole point of figuring out these dimensionless expressions is that you cannot change the scale of L and assume T fixed, etc. There are physical constants in the system, like viscosity, which impose a relationship between the fundamental dimensions. Changing L keeping T etc. fixed will therefore change the viscosity.
Similarly, density fixes a relationship between M and L.

 

[foo9008]

I guess you mean ρ(L^4)(T^-2), but either way it does not follow. The whole point of figuring out these dimensionless expressions is that you cannot change the scale of L and assume T fixed, etc. There are physical constants in the system, like viscosity, which impose a relationship between the fundamental dimensions. Changing L keeping T etc. fixed will therefore change the viscosity.
Similarly, density fixes a relationship between M and L.

Then, what is the correct formula of force in this question?

 

[foo9008]

I guess you mean ρ(L^4)(T^-2), but either way it does not follow. The whole point of figuring out these dimensionless expressions is that you cannot change the scale of L and assume T fixed, etc. There are physical constants in the system, like viscosity, which impose a relationship between the fundamental dimensions. Changing L keeping T etc. fixed will therefore change the viscosity.
Similarly, density fixes a relationship between M and L.

so , the author is correct ? it is (rho)(L^2)(L ) , where (v^2) = L ??

 

[haruspex]

so , the author is correct ? it is (rho)(L^2)(L ) , where (v^2) = L ??

I'm hampered by not knowing what Fr stands for in the first line.
I presume F=ρL²V² comes from some earlier work.
If we accept both of those equations, the rest follows.

 

[foo9008]

I'm hampered by not knowing what Fr stands for in the first line.
I presume F=ρL²V² comes from some earlier work.
If we accept both of those equations, the rest follows.

So, the authors working is correct??

 

[haruspex]

So, the authors working is correct??

I'm not sure. I don't know where the first line of equations comes from, or what Fr represents. I'm surprised to see any reference to g here. How is gravity relevant? If gravity were to increase but the densities, masses, and lengths stay the same, those equations seem to say the velocity would increase. Why?