---
title: "Tooth forces on a spur gear pinion"
subject: "Theory of Mechanism and Machine I"
chapter: 8
key: "semester_6/theory_of_mechanism_and_machine_i:spur-gear-tooth-forces-numerical"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/spur-gear-tooth-forces-numerical
site: BCT Engineering
---

# Tooth forces on a spur gear pinion

Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms

- Asked in: year not given, 6 marks
- Video: [Gear Forces and Power Transmission of SPUR GEARS in Just Over 12 Minutes!](https://www.youtube.com/watch?v=UieUEBnO5ps) — Less Boring Lectures
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/spur-gear-tooth-forces-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8

## Question

A 20 degree involute spur pinion of 20 teeth and module 5 mm transmits 10 kW at 720 rpm. Draw the free-body diagram of forces acting on the pinion tooth and calculate the torque, the pitch line velocity, the tangential force, the radial (separating) force and the resultant normal force on the tooth.

## Short answer

Free-body: normal force W_n resolved into tangential W_t (transmitting torque) and radial W_r = W_t tanφ. Torque = 132.63 N m, pitch line velocity = 3.770 m/s, W_t = 2653 N, W_r = 965 N, W_n = 2823 N.

## Answer

#### Force on a spur gear tooth
The tooth force acts along the line of action (common normal), at the pressure angle $\phi$ to the common tangent of the pitch circles. The normal force $W_n$ is resolved into:
- tangential component $W_t$ (transmits the torque),
- radial component $W_r = W_t\tan\phi$ (separates the gears, loads the bearings).

```text
        W_r
         ^   W_n
         |  /
         | /  phi
   ------+------> W_t   (tangent at pitch point)
```

$$
W_t = \frac{T}{d/2}, \qquad W_r = W_t\tan\phi, \qquad W_n = \frac{W_t}{\cos\phi}
$$

#### Calculation
Angular speed: $\omega = \dfrac{2\pi \times 720}{60} = 75.398$ rad/s.

Torque:

$$
T = \frac{P}{\omega} = \frac{10000}{75.398} = 132.63\ \text{N m}
$$

Pitch circle diameter: $d = mz = 5 \times 20 = 100$ mm.

Pitch line velocity:

$$
v = \omega\frac{d}{2} = 75.398 \times 0.050 = 3.770\ \text{m/s}
$$

Tangential force:

$$
W_t = \frac{T}{d/2} = \frac{132.63}{0.050} = 2653\ \text{N} \quad \left(= \frac{P}{v} = \frac{10000}{3.770}\right)
$$

Radial force:

$$
W_r = W_t\tan20^\circ = 2653 \times 0.3640 = 965\ \text{N}
$$

Normal force:

$$
W_n = \frac{W_t}{\cos20^\circ} = \frac{2653}{0.9397} = 2823\ \text{N}
$$

The driven gear experiences equal and opposite forces, and each shaft bearing must support the vector sum of $W_t$ and $W_r$ (that is, $W_n$) in addition to the overhung weight.

**Answer: $T = 132.63$ N m, $v = 3.770$ m/s, $W_t = 2653$ N, $W_r = 965$ N, $W_n = 2823$ N.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the radial (separating) force related to the tangential force?
   A. W_r = W_t / tan(phi)
   B. W_r = W_t tan(phi)
   C. W_r = W_t cos(phi)
   D. W_r = W_t sin(phi)
   Answer: B. The normal force at pressure angle phi to the tangent gives W_r = W_t tan(phi). (Key point: Force components)

2. Torque on the pinion for 10 kW at 720 rpm is:
   A. 13.26 N m
   B. 75.4 N m
   C. 1326 N m
   D. 132.63 N m
   Answer: D. T = P/omega = 10000/75.398 = 132.63 N m. (Key point: Torque)

3. With d = m z = 100 mm, the tangential force is:
   A. 1326 N
   B. 2653 N
   C. 5306 N
   D. 965 N
   Answer: B. W_t = T/(d/2) = 132.63/0.05 = 2653 N, also P/v with v = 3.770 m/s. (Key point: Tangential force)

4. The resultant normal force on the tooth is:
   A. W_t cos(20) = 2493 N
   B. W_t tan(20) = 965 N
   C. W_t + W_r = 3618 N
   D. W_t / cos(20) = 2823 N
   Answer: D. W_n = W_t/cos(phi) = 2653/0.9397 = 2823 N. (Key point: Normal force)

## Related in Theory of Mechanism and Machine I

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