p0294 - 2.94. CHAPTER 2, PROBLEM 94 241 2.94 Chapter 2,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
2.94. CHAPTER 2, PROBLEM 94 241 2.94 Chapter 2, Problem 94 Figure 2.24: Chihuahua with powerful rear-leg muscles. 2.94(a): The dimensional quantities and their dimensions are [ E ]= F · L = ML/T 2 · L = ML 2 T 2 [ H L, [ M M, [ g L T 2 There are 4 dimensional quantities and 3 independent dimensions ( M,L,T ) , so that the number of dimensionless groupings is 1. Using the indicial method, the appropriate dimensional equation is [ H ]=[ M ] a 1 [ g ] a 2 [ E ] a 3 Substituting the dimensions for each quantity yields L = M a 1 L a 2 T 2 a 2 M a 3 L 2 a 3 T 2 a 3 = M a 1 + a 3 L a 2 +2 a 3 T 2 a 2 2 a 3 Thus, equating exponents, we arrive at the following three equations: 0= a 1 + a 3 1= a 2 +2 a 3 2 a 2 2 a 3 We can solve immediately for a 2 and a 3 from the first and third equations, viz., a 1 = a 3 and a 2 = a 3
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
242 CHAPTER 2. DIMENSIONAL ANALYSIS Substituting into the second equation yields 1= a 3 +2 a 3 = a 3 = a 3 =1 Consequently, a 1 = 1a n d a 2 = 1 Substituting back into the dimensional equation, we have [ H ]=[ M ] 1 [ g ] 1 [ E ] Therefore, the dimensionless grouping is MgH E Using E. S. Taylor’s method, we have the following. ML T H 01 0 M 10 0 g 01 2 E 12 2 MLT H 010 M/E 0 22 g 2 E 2 LT H 10 g
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

p0294 - 2.94. CHAPTER 2, PROBLEM 94 241 2.94 Chapter 2,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online