58in13.1

58in13.1 - 308 C H A P T E R 13 V E C T O R G E O M E T...

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308 CHAPTER 13 VECTOR GEOMETRY (ET CHAPTER 12) and substituting that into the frst equation gives 3 = 4 ( 2 5 s ) + 2 s = 8 18 s So 18 s = 5, so s = 5 / 18, and thus r = 11 / 18. In other words, u =h 2 , 3 i= 11 18 h 1 , 4 i+ 5 18 h 5 , 2 i as seen in this picture: y x v u w 58. Calculate the magnitude oF the Force on cables 1 and 2 in ±igure 27. 65 ° 25 ° Cable 1 Cable 2 50 lbs FIGURE 27 SOLUTION The three Forces acting on the point P are: The Force F oF magnitude 50 lb that acts vertically downward. The Forces F 1 and F 2 that act through cables 1 and 2 respectively. y x 25 ° 115 ° F 1 F F 2 P Since the point P is not in motion we have F 1 + F 2 + F = 0( 1 ) We compute the Forces. Letting k F 1 k= f 1 and k F 2 f 2 we have: F 1 = f 1 h cos 115 , sin 115 f 1 h− 0 . 423 , 0 . 906 i F 2 = f 2 h cos 25 , sin 25 f 2 h 0 . 906 , 0 . 423 i F 0 , 50 i Substituting the Forces in (1) gives f 1 h− 0 . 423 , 0 . 906 f 2 h 0 . 906 , 0 . 423 i+h 0 , 50 i=h 0 , 0 i h− 0 . 423 f 1 + 0 . 906 f 2 , 0 . 906 f 1 + 0 . 423 f 2 50 0 , 0 i We equate corresponding components and get 0 . 423 f 1 + 0 . 906 f 2 = 0 0 . 906 f 1 + 0 . 423 f 2 50 = 0 By the frst equation, f 2 = 0 . 467 f 1 . Substituting in the second equation and solving For
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58in13.1 - 308 C H A P T E R 13 V E C T O R G E O M E T...

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