A spur gear with 20° full depth teeth is transmitting 20 kW at 200 rad/s. The pitch circle diameter of the gear is 100 mm. The magnitude of the force applied on the gear in the radial direction is

The correct option is 0.73 kN.

Step-by-Step Explanation:

1. Identify the given parameters:
Power transmitted by the spur gear, $P=20\text{kW}=20\times {10}^3\text{W}$
Angular velocity of the gear, $\omega =200\text{rad/s}$
Pitch circle diameter, $d=100\text{mm}=0.1\text{m}$
Pressure angle for 20° full depth teeth, $\phi =20°$

2. Calculate the pitch circle radius:
The pitch circle radius $r$ is half of the pitch circle diameter $d$:
$$r=\frac{d}{2}=\frac{0.1\text{m}}{2}=0.05\text{m}$$

3. Calculate the transmitted torque ($T$):
Power transmitted is related to torque and angular velocity by the formula:
$$P=T·\omega$$
Rearranging the formula to find torque:
$$T=\frac{P}{\omega }=\frac{20\times {10}^3\text{W}}{200\text{rad/s}}=100\text{N·m}$$

4. Calculate the tangential force ($F_t$):
The torque is also the product of the tangential force and the pitch circle radius:
$$T=F_t·r$$
Rearranging to find the tangential force:
$$F_t=\frac{T}{r}=\frac{100\text{N·m}}{0.05\text{m}}=2000\text{N}=2\text{kN}$$

5. Calculate the radial force ($F_r$):
The radial component of the gear force is related to the tangential force by the tangent of the pressure angle:
$$F_r=F_t·\tan \left(\phi \right)$$
Substitute the values into the equation:
$$F_r=2\text{kN}\times \tan (20°)$$
Since $\tan (20°)\approx 0.36397$:
$$F_r\approx 2\text{kN}\times 0.36397\approx 0.7279\text{kN}$$
Rounding to two decimal places, we get:
$$F_r\approx 0.73\text{kN}$$