Answered: If an object is light, it may be…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

If an object is light, it may be supported on the surface of a fluid. The weight of the object W, supportable by the fluid, depends on the perimeter S of the object, fluid density r, surface tension s, and gravitational acceleration constant g. Determine the dimensionless groups to describe this problem.

Expert Solution

Step 1: Step 1

Step 1: First, identify the parameters and their dimensions. We have five parameters, weight W, Perimeter S, Fluid density r, surface density r, surface tension s, and gravitational acceleration constant g. The dimensions of these parameters are as follows:

$open square brackets W close square brackets equals M L T to the power of negative 2 end exponent open square brackets S close square brackets equals L open square brackets r close square brackets equals M L to the power of negative 2 end exponent open square brackets s close square brackets equals M T to the power of negative 2 end exponent open square brackets g close square brackets equals L T to the power of negative 2 end exponent$

Step 2: We apply the Buckingham Pi theorem.We have n=5 parameters and r=3 primary dimensions (M,L,T) so we should have n-r=2 dimensionless groups.

Step 3: We choose g,r and S as our repeating variables because they contain all the primary dimensions.

We then form two dimensionless group n₁ and n_2. .

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