A 20° involute pinion with a pitch circle diameter of 192 mm is meshing gear. Calculate: The tangential force between the gears. The radial force between the gears. The normal force between the gears.

First, convert the pitch circle diameter to radius:

$$r = \frac{D}{2} = \frac{0.192 \text{ m}}{2} = 0.096 \text{ m}$$

Next, convert the rotational speed from rpm to angular velocity in rad/s:

$$\omega = \frac{2 \pi N}{60} = \frac{2 \pi (200 \text{ rpm})}{60 \text{ s/min}} = \frac{400 \pi}{60} \text{ rad/s} = \frac{20 \pi}{3} \text{ rad/s} \approx 20.944 \text{ rad/s}$$

Now, calculate the tangential velocity:

$$v = \omega r = \left(\frac{20 \pi}{3} \text{ rad/s}\right) (0.096 \text{ m}) = 0.64 \pi \text{ m/s} \approx 2.011 \text{ m/s}$$

Power transmitted ($P$) is the product of tangential force ($F_t$) and tangential velocity ($v$):

$$P = F_t \cdot v$$

Rearrange to solve for $F_t$:

$$F_t = \frac{P}{v} = \frac{18000 \text{ W}}{0.64 \pi \text{ m/s}}$$ $$F_t \approx \frac{18000}{2.0106} \text{ N} \approx 8952.55 \text{ N}$$

The tangential force between the gears is:

$$\boxed{\text{8950 N}}$$

The radial force is related to the tangential force and the pressure angle by the formula:

$$F_r = F_t \tan(\phi)$$ $$F_r = (8952.55 \text{ N}) \tan(20^\circ)$$ $$F_r \approx 8952.55 \text{ N} \times 0.36397$$ $$F_r \approx 3258.60 \text{ N}$$

The radial force between the gears is:

$$\boxed{\text{3260 N}}$$

The normal force is the resultant force acting along the line of action and is related to the tangential force and pressure angle by:

$$F_n = \frac{F_t}{\cos(\phi)}$$ $$F_n = \frac{8952.55 \text{ N}}{\cos(20^\circ)}$$ $$F_n \approx \frac{8952.55 \text{ N}}{0.93969}$$ $$F_n \approx 9527.90 \text{ N}$$

The normal force between the gears is:

$$\boxed{\text{9530 N}}$$