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

Designate n_{F} = n_{2} = −100 rev/min, and n_{L} = n_{5} = 0. Unlocking gear 5 and holding the arm stationary, in our imagination, we find

e = − \left( \frac{20}{30}\right) \left( \frac{30}{80}\right)= −0.25

Substituting this value in Eq. (13–32) gives

e =\frac{n_{L} − n_{A}}{n_{F} − n_{A}}                (13–32)

−0.25 =\frac {0 − n_{A}}{(−100) − n_{A}}

or

n_{A} = −20 rev/min

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

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)

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