8_ch 13 Mechanical Design budynas_SM_ch13

8_ch 13 Mechanical Design budynas_SM_ch13 - 16 35 180 =...

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340 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-20 Let gear 2 be frst, then n F = n 2 = 0. Let gear 6 be last, then n L = n 6 =− 12 rev/min. e = 20 30 µ 16 34 = 16 51 , e = n L n A n F n A (0 n A ) 16 51 =− 12 n A n A = 12 35 / 51 =− 17 . 49 rev/min (negative indicates cw) Ans . Alternatively, since N r , let v = Nn (crazy units). v = N 6 n 6 N 6 = 20 + 30 16 = 34 teeth v A N 4 = v N 4 N 5 v A = N 4 N 6 n 6 N 4 N 5 n A = v A N 2 + N 4 = N 4 N 6 n 6 ( N 2 + N 4 )( N 4 N 5 ) = 30(34)(12) (20 + 30)(30 16) = 17 . 49 rev/min cw Ans . 13-21 Let gear 2 be frst, then n F = n 2 = 180 rev/min. Let gear 6 be last, then n L = n 6 = 0 . e = 20 30 µ 16 34 = 16 51 , e = n L n A n F n A (180 n A ) 16 51 = (0 n A ) n A = µ
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Unformatted text preview: 16 35 ¶ 180 = − 82 . 29 rev/min The negative sign indicates opposite n 2 ∴ n A = 82 . 29 rev/min cw Ans . Alternatively, since N ∝ r , let v = Nn (crazy units). v A N 5 = v N 4 − N 5 = N 2 n 2 N 4 − N 5 v A = N 5 N 2 n 2 N 4 − N 5 n A = v A N 2 + N 4 = N 5 N 2 n 2 ( N 2 + N 4 )( N 4 − N 5 ) = 16(20)(180) (20 + 30)(30 − 16) = 82 . 29 rev/min cw Ans . 4 5 v 5 v 5 N 2 n 2 N 2 v A 2 4 5 v v 5 v A 2...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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