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9_ch 13 Mechanical Design budynas_SM_ch13

9_ch 13 Mechanical Design budynas_SM_ch13 -...

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Chapter 13 341 13-22 N 5 = 12 + 2(16) + 2(12) = 68 teeth Ans . Let gear 2 be first, n F = n 2 = 320 rev/min. Let gear 5 be last, n L = n 5 = 0 e = 12 16 16 12 12 68 = 3 17 , e = n L n A n F n A 320 n A = 17 3 (0 n A ) n A = − 3 14 (320) = − 68 . 57 rev/min The negative sign indicates opposite of n 2 n A = 68 . 57 rev/min cw Ans . Alternatively, n A = n 2 N 2 2( N 3 + N 4 ) = 320(12) 2(16 + 12) = 68 . 57 rev/min cw Ans . 13-23 Let n F = n 2 then n L = n 7 = 0 . e = − 24 18 18 30 36 54 = − 8 15 e = n L n 5 n F n 5 = − 8 15 0 5 n 2 5 = − 8 15 n 2 = 5 + 15 8 (5) = 14 . 375 turns in same direction 13-24 (a) Let n F = n 2 = 0, then n L = n 5 . e = 99 100 101 100 = 9999 10 000 , e = n L n A n F n A = n L n A 0 n A n L n A = − en A n L = n A ( e + 1)
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