Question 13.6: In Fig. 13–30 the sun gear is the input, and it is driven cl...

Let n_F = n_2 = -100 rev/min, and n_L = n_5 = 0. For e, unlock gear 5 and fix the arm. Then, planet gear 4 and ring gear 5 rotate in the same direction, opposite of sun gear 2. Thus, e is negative and

e=-\left(\frac{N_2}{N_4} \right)\left(\frac{N_4}{N_5} \right)=-\left(\frac{20}{30} \right)\left(\frac{30}{80} \right)=-0.25

Substituting this value in Eq. (13–32)

-0.25=\frac{n_L-n_A}{n_F-n_A}                      (13.32)

gives

-0.25=\frac{0-n_A}{(-100)-n_A}

or

n_A = -20 rev/min = 20 rev/min clockwise

To obtain the speed of gear 4, we follow the procedure outlined by Eqs. (b), (c), and (d).

n_{23} = n_2 – n_3                           (b)

n_{53} = n_5 – n_3                           (c)

\frac{n_{53}}{n_{23}} =\frac{n_5-n_3}{n_2-n_3}                           (d)

Thus

n_{43} = n_4 – n_3                      n_{23} = n_2 – n_3

and so

\frac{n_{43}}{n_{23}} =\frac{n_4-n_3}{n_2-n_3}                     (1)

But

\frac{n_{43}}{n_{23}} =-\frac{20}{30}=-\frac{2}{3}                       (2)

Substituting the known values in Eq. (1) gives

-\frac{2}{3}=\frac{n_4-(-20)}{(-100)-(-20)}

Solving gives

n_4=+33\frac{1}{3} rev/min = 33\frac{1}{3} rev/min counter-clockwise