Asuquo's Law · Resultant Force

Law of Composition of Resultant Force, R

A unified formula discovered by Nigerian scientist and mathematician Uwem E. Asuquo during laboratory experiments in physics.

AuthorUwem E. Asuquo

NationalityNigerian

DisciplinesPhysics · Mathematics

Exam SourceWASSCE 2007–2022

Statement

Asuquo's Law of Composition of Resultant

Formal Statement

Provided two or more coplanar non-parallel forces act at a point, the resultant R of the system of forces is given by the vector sum R = ΣFsinθ / ΣFcosθ, where angles θ are measured in clockwise sense from the north or y-axis.

Unlike the conventional approach that tracks force components through four separate sign rules, Asuquo's law unifies all cases into one consistent formula. The sign of each component (positive or negative) emerges naturally from evaluating the sine and cosine of the bearing angle θ.

Applicability

Who Uses Asuquo's Law?

The law is applicable across any field requiring force resolution or vector addition:

Physics Students

Mathematicians

Physics Teachers

Engineers

Physicists

Related Professionals

Derivation

Mathematical Proof

Asuquo's law operates trigonometrically by considering the quadrant system of angles in the reverse direction from the North or y-axis. Let θ₁, θ₂, θ₃, θ₄, … indicate the directions of forces F₁, F₂, F₃, F₄, … measured clockwise from north. Let α₁, α₂, α₃, α₄, … represent the corresponding acute angles in the right-angled triangles.

Component Table

Quadrant	Horizontal (Fₓ = F sinθ)	Vertical (F_y = F cosθ)
1st Quadrant	+F₁ sinα₁ = F₁ sinθ₁	+F₁ cosα₁ = F₁ cosθ₁
2nd Quadrant	+F₂ sin(180°−α₂) = F₂ sinθ₂	−F₂ cos(180°−α₂) = F₂ cosθ₂
3rd Quadrant	−F₃ sinα₃ = F₃ sin(180°+α₃) = F₃ sinθ₃	−F₃ cosα₃ = F₃ cos(180°+α₃) = F₃ cosθ₃
4th Quadrant	−F₄ sinα₄ = F₄ sin(360°−α₄) = F₄ sinθ₄	+F₄ cosα₄ = F₄ cos(360°−α₄) = F₄ cosθ₄

From the table, the resultant of forces F₁, F₂, F₃, F₄ is:

Vector Sum (Proved)

R = (F₁sinθ₁ + F₂sinθ₂ + F₃sinθ₃ + F₄sinθ₄) / (F₁cosθ₁ + F₂cosθ₂ + F₃cosθ₃ + F₄cosθ₄) = ΣFsinθ / ΣFcosθ ✓

Reference

The Quadrant System

One complete revolution of the arm ON gives a full circle divided into 4 equal parts, called Asuquo's quadrants. These differ from conventional quadrants in that angles are measured clockwise from the North (y-axis) rather than counter-clockwise from the East (x-axis).

Conventional (CAST)

S
2nd

A
1st

T
3rd

C
4th

Angles from +x, counter-clockwise

Asuquo's Quadrants

4th
−x,+y

1st
+x,+y

3rd
−x,−y

2nd
+x,−y

Angles from North, clockwise

Direction Rules

Sign of ΣFsinθ	Sign of ΣFcosθ	Resultant Quadrant	Direction θ
+	+	1st	θ = α
+	−	2nd	θ = 180° − α
−	−	3rd	θ = 180° + α
−	+	4th	θ = 360° − α

Part 1 · Multiple Choice

Objective Test Questions

Click an answer option to check it. Toggle the solution to see step-by-step working using Asuquo's Law.

WASSCE June 2007

Two forces act upon a body: F₁ = (5N, 060°) and F₂ = (10N, 180°). Find the magnitude of the resultant force.

Solution — Asuquo's Law

R = (ΣFsinθ / ΣFcosθ)

ΣFsinθ = 5 sin 60° + 10 sin 180° = 4.33 + 0.00 = 4.33
ΣFcosθ = 5 cos 60° + 10 cos 180° = 2.50 + (−10.00) = −7.50

|R| = √(4.33² + (−7.50)²)
    = √(18.75 + 56.25)
    = √75 ≈ 8.66 N

Answer: D

02

Practice Problem

Three forces F₁ = (8N, 300°), F₂ = (6N, 090°) and F₃ = (4N, 180°) act on a particle. Find the vertical component of the resultant force.
Solution — Vertical Component = ΣFcosθ
```
ΣFcosθ = 8 cos 300° + 6 cos 90° + 4 cos 180°
       = 4 + 0 + (−4)
       = 0

Vertical component of R = 0  →  Answer: A
```

WASSCE June 2008

Two forces F₁ = (10N, 020°) and F₂ = (7N, 200°) act on a particle. Find the resultant force.

Solution

ΣFsinθ = 10 sin 20° + 7 sin 200° = 3.42 + (−2.39) = 1.03
ΣFcosθ = 10 cos 20° + 7 cos 200° = 9.40 + (−6.58) = 2.82

|R| = √(1.03² + 2.82²) = √9.013 ≈ 3 N

Both ΣFsinθ and ΣFcosθ are +ve → 1st quadrant
tanα = 1.03/2.82 ≈ 0.365  →  α ≈ 20°

R = (3 N, 020°)  →  Answer: A

WASSCE June 2015 (Fig. 1.3: 8N at 45°, 5N at 300°)

Forces of magnitude 8N and 5N act on a body as shown. Calculate, to two decimal places, the resultant force acting at O.

Solution

F₁ = (8N, 045°), F₂ = (5N, 300°)

ΣFsinθ = 8 sin 45° + 5 sin 300° = 5.66 + (−4.33) = 1.33
ΣFcosθ = 8 cos 45° + 5 cos 300° = 5.66 + 2.50  = 8.16

|R| = √(1.33² + 8.16²) = √(1.769 + 66.586) = √68.355 ≈ 8.26 N

Answer: D

10

WASSCE June 2015

Forces F₁(8N, 30°) and F₂(10N, 150°) act on a particle. Find the horizontal component of the resultant force.
Solution — Horizontal = ΣFsinθ
```
ΣFsinθ = 8 sin 30° + 10 sin 150°
       = 4.00 + 5.00
       = 9.00 N

Answer: C
```
11

WASSCE June 2022

A particle is acted on by F = (10N, 060°), P = (15N, 120°) and Q = (12N, 200°). Find the force (in xi + yj form) that will keep the particle in equilibrium.
Solution
```
ΣFsinθ = 10 sin 60° + 15 sin 120° + 12 sin 200°
       = 8.66 + 12.99 + (−4.10) = 17.55

ΣFcosθ = 10 cos 60° + 15 cos 120° + 12 cos 200°
       = 5.00 + (−7.50) + (−11.28) = −13.78

R = (17.55i − 13.78j)

Equilibrant E = −R = −17.55i + 13.78j

Answer: C
```

Part 2 · Essay

Essay Problems

Full worked solutions for selected essay questions are available in the original document. Problems sourced from WASSCE examinations 2008–2015.

Question 15

Essential Further Maths — Example 37.8

Three forces of magnitudes 10N, 20N, and 12N act at a point O in the directions F₁ = (10N, 045°), F₂ = (20N, 210°), F₃ = (12N, 300°). Find the:

Components of each of the forces
Magnitude and direction of the resultant force, R

Solution

F₁ = (10N, 45°):  Fx = 10 sin 45° ≈ 7.07 N,  Fy = 10 cos 45° ≈ 7.07 N
F₂ = (20N, 210°): Fx = 20 sin 210° = −10 N,  Fy = 20 cos 210° ≈ −17.32 N
F₃ = (12N, 300°): Fx = 12 sin 300° ≈ −10.39 N, Fy = 12 cos 300° = 6 N

By Asuquo's Law:
ΣFsinθ = 7.07 + (−10.00) + (−10.39) = −13.32
ΣFcosθ = 7.07 + (−17.32) + 6.00   = −4.25

|R| = √(13.32² + 4.25²) = √195.49 ≈ 13.98 N

Both components negative → 3rd quadrant → θ = 180° + α
tanα = 13.32/4.25 ≈ 3.134  →  α ≈ 72.3°
θ = 180° + 72.3° = 252.3°

R = (13.98 N, 252.3°)

Question 16

WASSCE June 2008

Forces 5N, 5√3 N, 10N, 5√3 N and 5N act on body P of mass 5 kg (each 30° apart, starting from 030°). Find the:

Magnitude of the resultant force
Acceleration of the body

Solution

Applying Asuquo's Law for all five forces (at 030°, 060°, 090°, 120°, 150°):

ΣFsinθ = 2.5 + 7.5 + 10 + 7.5 + (−7.5) = 30
ΣFcosθ = 2.5√3 + 2.5√3 + 0 + (−2.5√3) + (−2.5√3) = 0

|R| = √(30² + 0²) = 30 N

By Newton's 2nd Law: a = R/m = 30/5 = 6 ms⁻²

Question 21

WASSCE June 2013

A particle is under the action of forces P = (4N, 030°) and R = (10N, 300°). Find the force that will keep the particle in equilibrium.

Solution

ΣFsinθ = 4 sin 30° + 10 sin 300° = 2.00 + (−8.66) = −6.66 (→ 4th quadrant)
ΣFcosθ = 4 cos 30° + 10 cos 300° = 3.46 + 5.00  =  8.46

Equilibrant E = −R:
E = (6.66i − 8.46j)

|E| = √(6.66² + 8.46²) = √116 ≈ 10.77 N

Acts in Asuquo's 2nd quadrant, opposite to R.

Questions 17–28

The remaining essay questions (17–28) cover topics including: expressing forces as column vectors, finding equilibrants, momentum change, string tensions, particle acceleration under multiple forces, and the unit vector direction. All follow the same Asuquo's Law framework of computing ΣFsinθ and ΣFcosθ, then finding magnitude and quadrant direction.

Refer to the original document (pages 4–6) for complete diagrams and full worked solutions to questions 17–28.