1

2

3

4

5 2. Después, mediante el teorema de los ejes paralelos, obtenemos el momento de inercia de cada componente de la manivela… I 1 = I G + md 2 donde: I G =1/12 ml 2 I 1 = 1/12 (0.123307 Kg)(0.05m) 2 + (0.123307 Kg)(0.12m) 2 I 1 = 1.8012 x 10 -3 kg. m 2 I 2 = I G + md 2 donde: I G =1/12 m(a 2 +b 2 ) I 2 = 1/12(0.08488 kg)(9x10 -4 m 2 + 0.0324 m 2 )+(0.08488 kg)(0.06m) 2 I 2 =5.411 x 10 -3 kg. m 2 I 3 = I G + md 2 donde: I G =0 + m(0) 2 I 3 =0 I=(1.8012 x 10 -3 kg. m 2 ) + (5.411 x 10 -3 kg. m 2 ) I=7.21 x 10 -3 kg. m 2

6

7 + +  = tan -1 (3/5)  = 30.96° + + DE=16.32 KN (C) DC=8.39 KN (T)

8  = tan -1 (5/3)  = 59.03° + + EA= 8.85 KN (C) EC= 6.22 KN (C) + AB= 3.11 KN (T) + AF= 6.2 KN (T)

9 BC= 2.2 KN (T) + + BF= 6.2 KN (C) + CF= 8.77 KN (T)

10

11 a = 0.0157 m/s 2 + +

12 N A = 72.1 N N B =71.9 N

13

14

15 + + a = 4.94 ft/s 2