The problem asks us to determine the train value of the gear system. Given that gear A rotates at 1750 rpm clockwise, we need to compute the speed and direction of rotation of gear E. The number of teeth for each gear is given as follows: $N_A = 20$, $N_B = 70$, $N_C = 18$, and $N_E = 54$. Gears A and B are on different shafts and mesh, Gears C and E are on different shafts and mesh. Gears B and C are on the same shaft.

1. Problem Description

The problem asks us to determine the train value of the gear system. Given that gear A rotates at 1750 rpm clockwise, we need to compute the speed and direction of rotation of gear E. The number of teeth for each gear is given as follows: $N_A = 20$, $N_B = 70$, $N_C = 18$, and $N_E = 54$. Gears A and B are on different shafts and mesh, Gears C and E are on different shafts and mesh. Gears B and C are on the same shaft.

2. Solution Steps

The gear ratio between two meshing gears is the inverse ratio of their number of teeth. The speed of gear B ($N_B$) relative to gear A ($N_A$) is:

$N_A \cdot RPM_A = N_B \cdot RPM_B$

$RPM_B = RPM_A \cdot \frac{N_A}{N_B}$

$RPM_B = 1750 \cdot \frac{20}{70} = 1750 \cdot \frac{2}{7} = 500$ rpm

Since gear B meshes with gear A, gear B rotates in the opposite direction, i.e., counter-clockwise.

Since gear B and gear C are on the same shaft, they rotate at the same speed and in the same direction. Therefore, $RPM_C = RPM_B = 500$ rpm. Gear C rotates counter-clockwise.

The speed of gear E ($N_E$) relative to gear C ($N_C$) is:

$N_C \cdot RPM_C = N_E \cdot RPM_E$

$RPM_E = RPM_C \cdot \frac{N_C}{N_E}$

$RPM_E = 500 \cdot \frac{18}{54} = 500 \cdot \frac{1}{3} = \frac{500}{3} \approx 166.67$ rpm

Since gear E meshes with gear C, gear E rotates in the opposite direction, i.e., clockwise.

3. Final Answer

The speed of gear E is $\frac{500}{3}$ rpm (approximately 166.67 rpm). The direction of rotation of gear E is clockwise.