Formula of forces

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  • #1
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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)
D58crH5.jpg

BDZcGcq.jpg

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 ...
 

Answers and Replies

  • #2
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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?
 
  • #3
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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
 
  • #4
haruspex
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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.
 
  • #5
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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?
 
  • #6
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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 ??
 
  • #7
haruspex
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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.
 
  • #8
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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??
 
  • #9
haruspex
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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?
 

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