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## Q. 8.7

For the gear train shown in Figure 8–42, if the input shaft rotates at 1750 rpm clockwise, compute the speed of the output shaft and its direction of rotation. ## Verified Solution

We can find the output speed if we can determine the train value:
$TV = n_A/n_D$ = input speed/output speed

Then
$n_D = n_A/TV$
But

$TV=(VR_1)(VR_2)=\frac{N_B}{N_A} \frac{N_D}{N_C}=\frac{70}{20} \frac{54}{18}=\frac{3.5}{1} \frac{3.0}{1}=10.5$

Now
$n_D = n_A/TV = (1750$ rpm$)/10.5 = 166.7$ rpm
Gear $A$ rotates clockwise; gear $B$ rotates counterclockwise.
Gear $C$ rotates counterclockwise; gear $D$ rotates clockwise.
Thus, the train in Figure 8–42 is a positive train.