We are given a gear train with four gears A, B, C, and D. Gears A and B are meshed, gears C and D are meshed, and gears B and C are on the same shaft. The number of teeth on each gear is: $N_A = 20$, $N_B = 70$, $N_C = 18$, $N_D = 54$. The input shaft (shaft 1) rotates at 1750 rpm clockwise. We want to compute the speed of the output shaft (shaft 3) and its direction of rotation.

1. Problem Description

We are given a gear train with four gears A, B, C, and D. Gears A and B are meshed, gears C and D are meshed, and gears B and C are on the same shaft. The number of teeth on each gear is: $N_A = 20$, $N_B = 70$, $N_C = 18$, $N_D = 54$. The input shaft (shaft 1) rotates at 1750 rpm clockwise. We want to compute the speed of the output shaft (shaft 3) and its direction of rotation.

2. Solution Steps

First, we find the speed of gear B, which is on shaft

2. The gear ratio between A and B is:

$N_A / N_B = n_B / n_A$

where $n_A$ and $n_B$ are the speeds of gear A and gear B respectively. Thus,

$n_B = n_A * (N_A / N_B) = 1750 * (20 / 70) = 1750 * (2/7) = 500$ rpm

Since gear A rotates clockwise, gear B rotates counter-clockwise.

Gears B and C are on the same shaft, so they have the same speed. Thus $n_C = n_B = 500$ rpm.

Since gear B rotates counter-clockwise, gear C also rotates counter-clockwise.

Now, we find the speed of gear D, which is on shaft

3. The gear ratio between C and D is:

$N_C / N_D = n_D / n_C$

where $n_C$ and $n_D$ are the speeds of gear C and gear D respectively. Thus,

$n_D = n_C * (N_C / N_D) = 500 * (18 / 54) = 500 * (1/3) = 166.67$ rpm

Since gear C rotates counter-clockwise, gear D rotates clockwise.

3. Final Answer

The speed of the output shaft (shaft 3) is 166.67 rpm.

The direction of rotation of the output shaft is clockwise.