Solving for tan γ: Understanding the Relationship between d/l and h/l"

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The discussion clarifies the reasoning behind expressing tan γ in terms of dimensionless ratios d/l and h/l instead of using the variables d and h directly. The author emphasizes that all terms in the problem are dimensionless, including θ (in radians) and constants α and β, which scale h and l. By normalizing d and h with a common length l, the problem maintains consistency in units, enhancing clarity and mathematical rigor.

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Expat09
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This is more a question about how the problem is written, rather than a question about how to do the problem

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Why do they ask for tan γ in terms of d/l and h/l, rather than d and h? Is there some relation that I'm just not seeing? I'm pretty sure I know how to do the problem in terms of d and h, but the d/l and h/l terms are throwing me off
 
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Welcome to PF, Expat09.

If you look at all the rest of the terms, all of them are dimensionless. By this I mean, they have no units.

[itex]\theta[/itex] is in radians, which are dimensionless
[itex]\alpha[/itex] and [itex]\beta[/itex] are just constants to scale h and l.

Therefore, to make h and d dimensionless, the author chose a common length, l, to make it so there are no units for d/l and h/l.

Does this help at all?
 

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