Question:
Two meshed involute gears transmit 1.0 kW power. The pressure angle is \(20^\circ\) and the pitch circle diameter of the large gear is 20 cm. If only a pair of teeth meshes at a time, the normal force acting between the meshed teeth in N will be \(\underline{\hspace{1cm}}\) (round off to one decimal place).
Two meshed involute gears transmit 1.0 kW power. The pressure angle is \(20^\circ\) and the pitch circle diameter of the large gear is 20 cm. If only a pair of teeth meshes at a time, the normal force acting between the meshed teeth in N will be \(\underline{\hspace{1cm}}\) (round off to one decimal place).
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Normal force \(F = \frac{P}{v}\), where velocity \(v = \omega \times radius\), and \(\omega\) is the angular velocity.
Updated On: Dec 22, 2025
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Correct Answer: 165
Solution and Explanation
Power transmitted: \(P = 1.0\ \text{kW} = 1000\ \text{W}\)
Pitch circle diameter of the large gear: \(D = 20\ \text{cm} = 0.2\ \text{m}\)
Gear speed:
\[ N = \frac{P}{\text{torque}} = \frac{1000}{F \times R} \] Normal force:
\[ F = \frac{P}{v} \] Where \(v\) is the velocity of the mesh, calculated using the gear's radius. Hence, the normal force is approximately: \(\boxed{165.0}\ \text{N}\).
Pitch circle diameter of the large gear: \(D = 20\ \text{cm} = 0.2\ \text{m}\)
Gear speed:
\[ N = \frac{P}{\text{torque}} = \frac{1000}{F \times R} \] Normal force:
\[ F = \frac{P}{v} \] Where \(v\) is the velocity of the mesh, calculated using the gear's radius. Hence, the normal force is approximately: \(\boxed{165.0}\ \text{N}\).
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