Question

I'm having problems with the very first Challenge Question (Module 1, Problem 1, http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/units-and-dimensional-analysis/MIT8_01SC_problems01.pdf). I'm confused about how I'm supposed to connect it all. Help, please? :)

Answer 1

Hello.
The parameters present here are : ρ, A1, A2, h and g.
so the time that the liquid takes to flow out should be proportional to these parameters.
\[t\alphaρ^{\alpha}.g^{\beta}.h^{\gamma}.A1^{\delta}.A2^{\epsilon}\]
Then, we turn this into dimensions equation.
We have 3 dimensions (Mass, Length and Time) and 5 unknowns :\[\alpha, \beta, \gamma, \delta, \epsilon\]
we will have :
\[\beta = -1/2\]
\[\alpha=0\]
\[\beta+\gamma+2(\epsilon+\zeta)=0\]
So, it's obvious that we still need two other equations.
We will make two approximations.
the first, is that the flow remains constant.. so we can write:
\[\phi=v.A2\] and\[\phi=V/t\]
where v is the speed and V is the total volume flowing out = h.A1
Thus, h.A1/t=v.A2 => t=(h/v).(A1/A2)
Since we have no hint about the relationship between h and v we will just say that :
\[t \alpha A1/A2\]Thus, we have the two missing equations:
\[2\epsilon=-2\zeta=1\] and so,
\[\beta=-\alpha=1/2\]

Finaly, replace these results in the first equation you will find:
\[t=C.(h/g)^{1/2}.A1/A2\]

Answer 2

I hope that this Helps.

Answer 3

how did we get
0 =V
1 = -2X ?

Answer 4

I can understand we assume the density of the fluid has no impact on the flux and therefore α=0, but why do you say β=−1/2 for the gravitational acceleration?

Thanks in advance!