---
title: "Position analysis of a four-bar mechanism"
subject: "Theory of Mechanism and Machine I"
chapter: 2
key: "semester_6/theory_of_mechanism_and_machine_i:four-bar-position-analysis-numerical"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/four-bar-position-analysis-numerical
site: BCT Engineering
---

# Position analysis of a four-bar mechanism

Theory of Mechanism and Machine I · Chapter 2: Linkages and Mechanisms

- Asked in: year not given, 8 marks
- Video: [Four bar mechanism | Analytical method | Position and displacement analysis | Problem](https://www.youtube.com/watch?v=s_7Q5etF7_U) — Engineering Design Simplified
- 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/four-bar-position-analysis-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-2

## Question

In a four-bar mechanism ABCD, the fixed link AD = 300 mm, crank AB = 100 mm, coupler BC = 250 mm and follower (rocker) CD = 200 mm. The crank AB makes an angle of 60 degrees with AD measured anticlockwise at A, and C lies on the same side of AD as B. Find analytically the angles of the coupler BC and the rocker CD with the line AD, and the transmission angle.

## Short answer

Using loop closure with cosine rule: AD=300, AB=100, BC=250, CD=200 mm, θ2=60°. Triangle ABD gives BD=264.58 mm, ∠ADB=19.11°. Triangle BCD gives ∠BDC=63.33°, ∠CBD=45.63°. Coupler angle θ3=26.53°, rocker angle θ4=97.56° from AD. Transmission angle μ=71.03°.

## Answer

#### Method
Take A as origin and AD along the $x$-axis, so D = (300, 0). The diagonal BD is used to split the quadrilateral into triangles ABD and BCD and solved with the cosine rule (the loop-closure solution).

Given: $a = AB = 100$, $b = BC = 250$, $c = CD = 200$, $d = AD = 300$ mm, $\theta_2 = 60^\circ$.

#### Step 1: Triangle ABD
$$
B = (a\cos\theta_2,\ a\sin\theta_2) = (50.00,\ 86.60)\ \text{mm}
$$

$$
BD^2 = a^2 + d^2 - 2ad\cos\theta_2 = 100^2 + 300^2 - 2(100)(300)\cos 60^\circ = 70000
$$

$$
BD = 264.58\ \text{mm}
$$

$$
\sin\angle ADB = \frac{a\sin\theta_2}{BD} = \frac{86.60}{264.58} \Rightarrow \angle ADB = 19.11^\circ
$$

#### Step 2: Triangle BCD
$$
\cos\angle BDC = \frac{c^2 + BD^2 - b^2}{2c\,BD} = \frac{200^2 + 70000 - 250^2}{2(200)(264.58)} \Rightarrow \angle BDC = 63.33^\circ
$$

$$
\cos\angle CBD = \frac{b^2 + BD^2 - c^2}{2b\,BD} \Rightarrow \angle CBD = 45.63^\circ
$$

#### Step 3: Angles of the links
Since A and C lie on opposite sides of BD:

$$
\angle ADC = \angle ADB + \angle BDC = 19.11 + 63.33 = 82.44^\circ
$$

The rocker CD makes this angle with DA, so measured from the $+x$ axis:

$$
\theta_4 = 180^\circ - 82.44^\circ = 97.56^\circ
$$

Coordinates of C: $C = (300 + 200\cos\theta_4,\ 200\sin\theta_4) = (273.68,\ 198.26)$ mm.

$$
\theta_3 = \tan^{-1}\frac{C_y - B_y}{C_x - B_x} = \tan^{-1}\frac{198.26 - 86.60}{273.68 - 50.00} = 26.53^\circ
$$

(Check: $100\cos 60^\circ + 250\cos\theta_3 = 300 + 200\cos\theta_4$ is satisfied, and the vertical components also balance.)

#### Step 4: Transmission angle
The transmission angle $\mu$ is the angle between coupler BC and rocker CD:

$$
\cos\mu = \frac{b^2 + c^2 - a^2 - d^2 + 2ad\cos\theta_2}{2bc} \Rightarrow \mu = 71.03^\circ
$$

This is greater than $40^\circ$, so force transmission is good.

**Answer: coupler angle $\theta_3 = 26.53^\circ$, rocker angle $\theta_4 = 97.56^\circ$ (both from AD), transmission angle $\mu = 71.03^\circ$.**

## Check yourself

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

1. How does the answer solve the four-bar position problem analytically?
   A. It splits the quadrilateral by diagonal BD into triangles ABD and BCD and uses the cosine rule
   B. It uses only the sine rule on triangle ABC
   C. It guesses angles and iterates by trial
   D. It uses the instantaneous centre method
   Answer: A. Diagonal BD gives triangles ABD and BCD that are solved with the cosine rule. (Key point: Method of solution)

2. With a = 100, d = 300 mm and theta2 = 60°, what is BD?
   A. 264.58 mm
   B. 200 mm
   C. 300 mm
   D. 350 mm
   Answer: A. BD^2 = 100^2 + 300^2 - 2(100)(300)cos60° = 70000, so BD = 264.58 mm. (Key point: Length of diagonal BD)

3. The angle ADC is 19.11° + 63.33° = 82.44°. What is the rocker angle theta4 measured from the +x axis?
   A. 82.44°
   B. 97.56°
   C. 63.33°
   D. 26.53°
   Answer: B. CD makes 82.44° with DA, so measured from +x it is 180° - 82.44° = 97.56°. (Key point: Rocker angle from AD)

4. The transmission angle is the angle between which two links, and what value was found?
   A. Crank and coupler, 60°
   B. Frame and rocker, 82.44°
   C. Coupler and rocker, 71.03°
   D. Crank and frame, 19.11°
   Answer: C. The transmission angle is between coupler BC and rocker CD; here mu = 71.03°, above 40°, so force transmission is good. (Key point: Transmission angle)

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