I'm doing the MIT Physics 1 : Classical Mechanics course, offered by OpenCourseware. I'm watching the first lecture and reviewing the slides, and don't seem to understand this question on Dimensional Analysis:

The speed of a sail-boat or other craft that does not plane is limited by the wave it makes – it can’t climb uphill over the front of the wave. What is the maximum speed you’d expect? Hint: relevant quantities might be the length l of the boat, the density ρ of the water, and the gravitational acceleration g.

Here is a link to the question - LINK- It is on slide 15.

Any guidance to help solve this problem will be much appreciated!

  • $\begingroup$ A displacement boat creates a standing wave between it's bow and stern. Try looking at what point the craft has to start climbing it's own wave. $\endgroup$ – user6972 Sep 1 '13 at 17:20

In an equality with probable physical meaning, the dimension of both sides must be coincide. The velocity $v$ has dimension $L/T$. Similarly, length $l$ has $L$, density $\rho$ has $M/L^3$, acceleration $g$ has $\cdot L/T^2$. The equality implies

For length dimension, $$1=x-3y+z$$ For mass dimension, $$0=y$$ For time dimension, $$-1=-2z$$ Thus a probable hypothesis is $$v=C\sqrt{gl}$$

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On Google or Wikipedia look into something called the Froude number.

The question on the slide is asking how you to pick $x,y$, and $z$ so that the dimensions on both sides of the equation $ v = l^x\rho ^yg^z$ match.

$[v] = L/T$ (i.e. speed has dimensions of length over time)

$[l] = L$

$[\rho] = M/L^3$

$[g] = L/T^2$

Let's just try multiplying the three quantities on the righthand side together and seeing what the units are. This is equivalent to setting x, y, and z to 1:

$[l\rho g] = L^2M/(L^3T^2) = M/(LT^2) $

That does not have the dimensions $L/T$ so that can't be the answer. Notice $\rho$ is the only quantity on the rhs with dimensions of mass. Speed doesn't have mass as a dimension, so $y$ must be $0$;the top speed of the boat is independent of the water density.

Let's try $x = 1, y =0, z = 1$

$[l^1\rho^0 g^1] = L^2/T^2 $

We're close. The answer must be $x = 1/2, y =0, z = 1/2$.

There a more systematic approach as well. One can writes down a system of algebraic equations based on the exponents of each dimension on the two sides of the equation:

Mass: $0 = y $

Length: $1 = x + z$

Time: $-1 = z^{-.5}$

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  • $\begingroup$ In accordance with our homework policy, I'm temporarily deleting this. $\endgroup$ – Qmechanic Sep 1 '13 at 16:43

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