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p0145

# p0145 -

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Unformatted text preview: 52 CHAPTER 1. INTRODUCTION 1.45 Chapter 1, Problem 45 Problem: A ball moving with velocity V = V i approaches a surface inclined to the x axis at an angle φ as shown. To analyze this problem, it is convenient to work in the ξζ coordinate system, where the ξ and ζ axes are parallel to and normal to the surface, respectively. Working with unit vectors, write V in terms of unit vectors i and k parallel to the ξ and ζ axes, respectively. . .. ζ ......... . . ... .. .... .... .... .... . . .. .. .... .... . . .. ...... ..... ............. . ............ .............. . .. . . ..... ......... .. .................... . .. . .................. .. . ............................... . .. . ..................................... .. ....................................... .. . ........................................... . . .. .. .. . ............................ . ..................................................... . .. ....................................................... . ................ ............... ... .. . . . .............................. ................................................................ . ................................................................ . . .. ................................................................. . ..................................................................... . .. .. . ..... . .......... . .. ................................................. . .. . .................................................................................... . . .... . ......................................................................................... . . .. ......................................................................................... . .. . ...... . . ................................................................................................. . . . .. ........................................... .... .. . ........................................................................................................ ... .. . ............................................................................................................. ... . .. . . ... ..................................................... . . .................................. . ......................................................................... . ................................ .. ....................................................................................................................... . ........................................................................................................................ . .......................................................................................................................... . . .. . . .. . .... . .. . ........ ............. ........................................................................................ ...................................................................................................................................... ...................................................................................................................................... .. . . .. .. . ....................................................................................................................................................... ... ....... .......... .. . . .................................................................................. ..................................................... .................................................................................................................................................... . ... .. ............................................................................................................................................................ .................................................................................................................................................................. . ..... . ... .................................................................................... . .......... . ................................................................ .. . . . . . . . z ..... . . . . ξ • V g = −g k x φ Solution: We must express unit vector i in terms of i and k as follows. i = iξ i + iζ k Now, we can use the dot product to determine the components iξ and iζ as follows. iξ = i · i = cos φ Therefore, unit vector i is Thus, the ball’s velocity vector is i = i cos φ − k sin φ V = V cos φ i − V sin φ k and iζ = k · i = − sin φ ...
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