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 ( Kg)(0.05m) 2 + ( Kg)(0.12m) 2 I 1 = x kg. m 2 I 2 = I G + md 2 donde: I G =1/12 m(a 2 +b 2 ) I 2 = 1/12( kg)(9x10 -4 m m 2 )+( kg)(0.06m) 2 I 2 =5.411 x kg. m 2 I 3 = I G + md 2 donde: I G =0 + m(0) 2 I 3 =0 I=( x kg. m 2 ) + (5.411 x kg. m 2 ) I=7.21 x kg. m 2
+ + = tan -1 (3/5) = 30.96° + + DE=16.32 KN (C) DC=8.39 KN (T)
= 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)
BC= 2.2 KN (T) + + BF= 6.2 KN (C) + CF= 8.77 KN (T)
a = m/s 2 + +
N A = 72.1 N N B =71.9 N
+ + a = 4.94 ft/s 2