The derivative of $f(x) = 5^{x^2}$ with respect to $x$ is:

B) $2x \cdot 5^{x^2} \cdot \ln 5$

The sum of the first n terms of an arithmetic sequence (S_n) is formulated as:

S_n = n / 2 * [ 2a + (n - 1) * d ] = n / 2 * [ a + l ]

The sum to infinity (S_infinity) of a geometric series exists and converges if and only if the common ratio satisfies:

|r| < 1 (i.e. -1 < r < 1), and S_infinity = a / (1 - r)

كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

$Mathematics cover$

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Home/ Courses/ Mathematics/ Secondary 3 Mathematics 2026/2027 — further lessons/ Parallel Coplanar Forces: Resultant and Equilibrium

DABOUR HUB / LESSON 2026-2027

Parallel Coplanar Forces: Resultant and Equilibrium

Q: كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

Course: Mathematics Unit: Secondary 3 Mathematics 2026/2027 — further lessons

🧭 Lesson points — pick a point and study it with its question bank

1. 2-1 (a) Parallel forces acting on a rigid bodyExplanation

2. 2-1 (b) The resultant of two like parallel forcesExplanation

3. 2-1 (c) The resultant of two unlike parallel forcesExplanation

4. 2-1 (d) Finding one force from the resultantExplanation

5. 2-1 (e) The moment theorem (Varignon)Explanation

6. 2-1 (f) The resultant of a system of parallel forcesExplanation

7. 2-1 (g) Theoretical resultsExplanation

8. 2-2 (a) Conditions of equilibriumExplanation

9. 2-2 (b) Beams on supportsExplanation

10. 2-2 (c) Tilting and limiting positionsExplanation

11. 2-2 (d) A rod on a rough surface and a smooth wedgeExplanation

12. 2-2 (e) A ladder against a smooth wallExplanation

This lesson covers lessons 2-1 and 2-2 of the statics unit "Parallel Coplanar Forces". After it you should be able to:

find the resultant of two like or two unlike parallel forces, and of a whole system of parallel forces;
use the moment theorem to locate the line of action of the resultant;
apply the conditions of equilibrium to rods, boards and ladders.

🌳 Concept tree

Parallel Coplanar Forces: Resultant and Equilibrium
├── 2-1 (a) Parallel forces acting on a rigid body
│   ├── Why a new idea is needed
│   └── A balancing experiment
├── 2-1 (b) The resultant of two like parallel forces
│   ├── Rule
│   └── Where the rule comes from
├── 2-1 (c) The resultant of two unlike parallel forces
│   └── Rule
├── 2-1 (d) Finding one force from the resultant
│   └── Method
├── 2-1 (e) The moment theorem (Varignon)
│   ├── Theorem
│   ├── Proof for two like forces
│   └── Consequence
├── 2-1 (f) The resultant of a system of parallel forces
│   ├── Procedure
│   └── Balance (centre of parallel weights)
├── 2-1 (g) Theoretical results
│   ├── Moving one of two like forces
│   └── Forces given as vectors
├── 2-2 (a) Conditions of equilibrium
│   ├── The two conditions
│   └── In components
├── 2-2 (b) Beams on supports
│   └── Procedure
├── 2-2 (c) Tilting and limiting positions
│   └── Idea
├── 2-2 (d) A rod on a rough surface and a smooth wedge
│   └── Method
├── 2-2 (e) A ladder against a smooth wall
│   └── Setting
└── 2-2 (f) Summary
    ├── Comparison
    └── Checklist

01 2-1 (a) Parallel forces acting on a rigid body

Why a new idea is needed

For forces acting on one particle the lines of action all pass through a single point, and their resultant passes through that same point.
For forces acting on a rigid body the lines of action need not meet. Parallel forces never meet, so we must also find where the resultant acts.
Like forces are parallel and in the same direction. Unlike forces are parallel and in opposite directions.

A balancing experiment

A uniform ruler balances on a pivot at its centre.
If two equal stones sit at the ends, the pivot stays at the centre. If one stone is doubled, the pivot must move towards the heavier end.
The pivot is directly under the point where the resultant of the two weights acts.

02 2-1 (b) The resultant of two like parallel forces

Rule

Let $F_{1}$ and $F_{2}$ be like forces at points $A$ and $B$ of a rigid body.
Their resultant $R$ is parallel to them, in the same direction, with magnitude $R = F_{1} + F_{2}$
Its line of action cuts the line $A B$ at an inner point $C$ which divides $A B$ internally in the inverse ratio of the forces: $F_{1} \cdot A C = F_{2} \cdot B C or \frac{A C}{B C} = \frac{F_{2}}{F_{1}}$
So $C$ is nearer to the larger force.

Where the rule comes from

Add two equal and opposite forces $F$ along $A B$ at $A$ and at $B$ ; they do not change the effect.
Combine $F_{1}$ with the force at $A$ , and $F_{2}$ with the force at $B$ , by the parallelogram law. The two resultants meet at a point $O$ .
Resolve them again at $O$ : the added horizontal forces cancel and $F_{1}$ , $F_{2}$ act together along the line through $O$ .
Similar triangles then give $\frac{A C}{O C} = \frac{F}{F_{1}}$ and $\frac{B C}{O C} = \frac{F}{F_{2}}$ , so $F_{1} \cdot A C = F_{2} \cdot B C$ .

✍️ Worked examples

Example 1: Two like forces of 5 N and 7 N act at $A$ and $B$ with $A B = 36$ cm. Find their resultant.

Solution:

$R = 5 + 7 = 12$ N, in the direction of the two forces.
$\frac{A C}{C B} = \frac{7}{5}$ with $A C + C B = 36$ , so $A C = \frac{7}{12} \times 36 = 21$ cm.
The resultant acts 21 cm from $A$ (the end of the 5 N force) and 15 cm from $B$ .

03 2-1 (c) The resultant of two unlike parallel forces

Rule

Let $F_{1} > F_{2}$ act in opposite directions at $A$ and $B$ .
The resultant is parallel to them, in the direction of the larger force, with magnitude $R = F_{1} - F_{2}$
Its line of action cuts $A B$ at a point $C$ outside the segment, on the side of the larger force, and $F_{1} \cdot A C = F_{2} \cdot B C$
Here $C$ is again nearer to the larger force $F_{1}$ .
If $F_{1} = F_{2}$ there is no resultant force: the pair is a couple (see the next lesson).

✍️ Worked examples

Example 3: Unlike forces of 100 N (at $A$ ) and 40 N (at $B$ ) act at points 240 cm apart. Find the resultant.

Solution:

$R = 100 - 40 = 60$ N in the direction of the 100 N force.
$C$ lies on $B A$ produced beyond $A$ , with $B C = A C + 240$ .
$100 \cdot A C = 40 (A C + 240)$ , so $60 A C = 9600$ and $A C = 160$ cm.
The resultant acts 160 cm from $A$ outside the segment $A B$ .

04 2-1 (d) Finding one force from the resultant

Method

Take the unit vector along the resultant and use $R = F_{1} + F_{2}$ with signs.
Take moments about the point where the resultant acts; the sum of moments of the forces about that point is zero.

✍️ Worked examples

Example 4:

One of two parallel forces is 20 N.
The resultant is 35 N, and its line of action is 15 cm from the 20 N force.
Find the other force and its position in each case: (a) the 20 N force and the resultant have the same direction, (b) opposite directions.

Solution:

(a) $35 = 20 + F_{2}$ , so $F_{2} = 15$ N, in the same direction.
Moments about $C$ (point of the resultant): $20 \times 15 = 15 \times B C$ , so $B C = 20$ cm. The 15 N force is on the other side of $C$ , so $A B = 15 + 20 = 35$ cm.
(b) The resultant (35 N) is opposite to the 20 N force, so $35 = F_{2} - 20$ and $F_{2} = 55$ N in the direction of the resultant.
Moments about $C$ : $20 \times 15 = 55 \times B C$ , so $B C = \frac{60}{11}$ cm.
Both $C$ and the 55 N force are on the same side of the 20 N force, so $A B = 15 - \frac{60}{11} = \frac{105}{11}$ cm.

05 2-1 (e) The moment theorem (Varignon)

Theorem

The algebraic sum of the moments of a system of parallel coplanar forces about any point in their plane equals the moment of their resultant about the same point.
Write $\sum M_{O} = M_{O} (R)$ .

Proof for two like forces

Draw the common perpendicular through $O$ meeting the lines of action at $A$ , $B$ and $C$ (for $R$ ).
Moment sum about $O$ : $- F_{1} \cdot O A - F_{2} \cdot O B$ (with $O A = O C - A C$ and $O B = O C + C B$ in the appropriate arrangement).
Using $F_{1} \cdot A C = F_{2} \cdot C B$ the sum becomes $- (F_{1} + F_{2}) O C = - R \cdot O C$ , which is the moment of $R$ .
For unlike forces the same method gives $(F_{1} - F_{2}) O C = R \cdot O C$ .
A system of many forces is handled by combining forces in pairs.

Consequence

If the resultant is zero, the sum of moments about any point is zero or constant (the system is either in equilibrium or equivalent to a couple).
The theorem is the tool for locating the line of action of a resultant: take moments about a convenient point.

✍️ Worked examples

Example 5:

Parallel forces act perpendicular to a line at the points $A$ , $C$ , $E$ , $B$ , which are 0, 4, 9 and 16 cm from $A$ .
The force at $A$ is 5 N, at $C$ it is 6 N, at $E$ it is 4 N in the opposite sense to the force at $C$ , and at $B$ it is 2 N in the same sense as the force at $C$ .
Find the algebraic sum of the moments about $A$ (take the sense of the force at $C$ as positive).

Solution:

The 5 N force at $A$ has zero moment about $A$ .
Moments about $A$ : $6 \times 4 - 4 \times 9 + 2 \times 16 = 24 - 36 + 32 = 20$ N·cm.

06 2-1 (f) The resultant of a system of parallel forces

Procedure

Choose a positive direction (a unit vector $\hat{c}$ along the forces) and add the forces with signs: $R = \sum F$ .
If $R \neq 0$ , take moments about a convenient point $O$ (often the first force): $R \cdot x = \sum F \cdot d$ where $x$ is the distance from $O$ to the line of action of $R$ , and $d$ the signed distance of each force.
State: the magnitude, direction and position of the resultant (where it cuts the line through the points).

✍️ Worked examples

Example 6: $A, B, C, D, E$ lie on a line with $A B : B C : C D : D E = 2 : 3 : 4 : 7$ . Like forces of 30, 50, 20, 70, 40 N act at $A, B, C, D, E$ . Find the resultant.

Solution:

$R = 30 + 50 + 20 + 70 + 40 = 210$ N, in the common direction.
Let $A B = 2 x$ , so the distances from $A$ are $0, 2 x, 5 x, 9 x, 16 x$ .
Moments about $A$ : $R \cdot A O = 50 (2 x) + 20 (5 x) + 70 (9 x) + 40 (16 x) = 100 x + 100 x + 630 x + 640 x = 1470 x$ .
$A O = \frac{1470 x}{210} = 7 x$ , so $\frac{A O}{A E} = \frac{7 x}{16 x} = \frac{7}{16}$ .
The resultant divides $A E$ internally in the ratio $7 : 16$ from $A$ .

Balance (centre of parallel weights)

The weights of particles on a light rod have a resultant equal to the total weight, acting at the balance point.
Example: a light rod 1 m long carries 2 N at one end and 3 N at the other; the balance point is $\frac{2}{5} \times 100 = 40$ cm from the 3 N end.

07 2-1 (g) Theoretical results

Moving one of two like forces

✍️ Worked examples

Example 9:

Like forces $F_{1}$ and $F_{2}$ act at $A$ and $B$ with resultant $R$ .
If $F_{2}$ moves parallel to itself a distance $x$ along $A B$ , prove that the resultant moves a distance $\frac{F_{2}}{F_{1} + F_{2}} x$ in the same direction.

Solution:

Moments about $A$ : $R \cdot A C = F_{2} \cdot A B$ .
After the move the point of action of $F_{2}$ is $B^{'}$ with $A B^{'} = A B + x$ , and the resultant acts at $C^{'}$ : $R \cdot A C^{'} = F_{2} \cdot A B^{'}$ .
Subtract: $R (A C^{'} - A C) = F_{2} x$ , so $C C^{'} = \frac{F_{2}}{R} x = \frac{F_{2}}{F_{1} + F_{2}} x$ .

Forces given as vectors

✍️ Worked examples

Example 10: ${\vec{F}}_{1} = 2 \hat{i} - 3 \hat{j}$ acts at $A (1, 3)$ and ${\vec{F}}_{2} = 4 \hat{i} - 6 \hat{j}$ acts at $B (4, 9)$ . Find the resultant and the point where its line of action meets $A B$ .

Solution:

$\vec{R} = {\vec{F}}_{1} + {\vec{F}}_{2} = 6 \hat{i} - 9 \hat{j}$ .
${\vec{F}}_{2} = 2 {\vec{F}}_{1}$ , so the forces are parallel and in the same direction.
$\frac{A C}{C B} = \frac{F_{2}}{F_{1}} = \frac{2}{1}$ , so $C$ divides $A B$ in the ratio $2 : 1$ from $A$ .
$C = (\frac{2 (4) + 1 (1)}{3}, \frac{2 (9) + 1 (3)}{3}) = (3, 7)$ .

08 2-2 (a) Conditions of equilibrium

The two conditions

A rigid body acted on by coplanar forces is in equilibrium if and only if:
the resultant vector is zero: $\vec{R} = \vec{0}$ ;
the sum of moments about any point is zero: $\sum M = 0$ .

In components

Take $\hat{i}$ and $\hat{j}$ in the plane and $\hat{k}$ perpendicular to it.
$\vec{R} = X \hat{i} + Y \hat{j}$ and the moment is along $\hat{k}$ , so the conditions become $X = 0, Y = 0, \sum M = 0$
For parallel forces (say vertical), the first reduces to one equation: the forces up equal the forces down.
The moment equation can be taken about any point; the best choice is a point where an unknown force acts, to eliminate it.

09 2-2 (b) Beams on supports

Procedure

Draw all forces: weights (a uniform rod's weight acts at its midpoint), loads, reactions of supports, tensions.
Write "vertical forces balance".
Take moments about one support to find the other reaction directly.

✍️ Worked examples

Example 11:

A uniform wooden board, 6 m long, has mass 30 kg per metre.
It rests horizontally on supports at its ends $A$ and $B$ and a box of 240 kg·wt is on the board 4 m from $B$ .
Find the reaction at each support.

Solution:

Mass of board $= 30 \times 6 = 180$ kg, so its weight is 180 kg·wt at the middle (3 m from each end).
Vertical balance: $R_{A} + R_{B} = 180 + 240 = 420$ .
Moments about $B$ : $R_{A} \times 6 = 180 \times 3 + 240 \times 4 = 540 + 960 = 1500$ .
$R_{A} = 250$ kg·wt and $R_{B} = 420 - 250 = 170$ kg·wt.
As the box moves closer to $A$ , $R_{A}$ increases and $R_{B}$ decreases.

10 2-2 (c) Tilting and limiting positions

Idea

A beam on two supports is about to tilt about one support when the reaction at the other support becomes zero.
Take moments about the support that remains in contact.

✍️ Worked examples

Example 14:

A uniform beam $A B$ of length 6 m and weight 40 kg·wt rests on supports at $C$ (1 m from $A$ ) and $D$ (1 m from $B$ ).
A man of weight 100 kg·wt walks from the middle towards $B$ .
How far from $A$ is he when the beam is about to tilt?

Solution:

At the limit the reaction at $C$ is zero, so take moments about $D$ (5 m from $A$ ).
The beam's weight is 2 m to the left of $D$ : anticlockwise moment $40 \times 2 = 80$ .
The man at distance $x$ from $A$ is $(x - 5)$ to the right of $D$ : clockwise moment $100 (x - 5)$ .
$100 (x - 5) = 80$ , so $x = 5.8$ m from $A$ (0.2 m from the end $B$ ).

11 2-2 (d) A rod on a rough surface and a smooth wedge

Method

Forces on the rod: weight $W$ , reaction of the wedge $S$ (perpendicular to the rod, since the wedge is smooth), the ground reaction $R$ and friction $F$ at the lower end.
Use $X = 0$ , $Y = 0$ and moments about the foot of the rod.
At the point of slipping, $F = μ R$ .

✍️ Worked examples

Example 16:

A uniform bar $A B$ of weight 5 kg·wt and length 30 cm rests with $A$ on rough ground and touches a smooth wedge at $C$ , 12.5 cm above the ground, at $3 0^{\circ}$ to the horizontal.
It is about to slip.
Find the reaction of the wedge and the coefficient of friction.

Solution:

$A C = \frac{12.5}{\sin 3 0^{\circ}} = 25$ cm. The weight acts 15 cm from $A$ .
Moments about $A$ : $5 \times 15 \cos 3 0^{\circ} = S \times 25$ , so $S = 3 \cos 3 0^{\circ} = \frac{3 \sqrt{3}}{2} \approx 2.60$ kg·wt.
The wedge reaction is perpendicular to the bar, so it makes $3 0^{\circ}$ with the vertical. Horizontal: $F = S \sin 3 0^{\circ} = \frac{3 \sqrt{3}}{4}$ .
Vertical: $R + S \cos 3 0^{\circ} = 5$ , so $R = 5 - \frac{3 \sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = 5 - \frac{9}{4} = \frac{11}{4}$ .
$μ = \frac{F}{R} = \frac{3 \sqrt{3} / 4}{11 / 4} = \frac{3 \sqrt{3}}{11} \approx 0.47$ .

12 2-2 (e) A ladder against a smooth wall

Setting

A uniform ladder of weight $W$ and length $ℓ$ rests with its foot $A$ on rough horizontal ground and its top $B$ against a smooth vertical wall.
Forces: weight $W$ at the middle; vertical ground reaction $R$ and friction $F$ at $A$ ; the wall reaction $S$ (horizontal, because the wall is smooth) at $B$ .
Conditions: $R = W$ and $F = S$ ; moments about $A$ give $S$ .
Limiting equilibrium: $F = μ R$ .

✍️ Worked examples

Example 17:

A uniform ladder of weight $W$ makes an angle $θ$ with the ground and is about to slip.
The coefficient of friction at the ground is $μ$ .
Find $\tan θ$ .

Solution:

$R = W$ and $S = F = μ W$ .
Moments about the foot: $S ℓ \sin θ = W \frac{ℓ}{2} \cos θ$ .
$μ W \sin θ = \frac{W}{2} \cos θ$ , so $\tan θ = \frac{1}{2 μ}$ .
With $μ = 0.5$ the ladder is on the point of slipping at $θ = 4 5^{\circ}$ ; it is safe for larger angles.

13 2-2 (f) Summary

Comparison

Point of comparison	Like forces	Unlike forces
Direction of $R$	same as the forces	that of the larger force
Magnitude of $R$	$F_{1} + F_{2}$	$F_{1} - F_{2}$
Position of $R$	between the forces	outside, beyond the larger force
Relation	$F_{1} \cdot A C = F_{2} \cdot B C$	$F_{1} \cdot A C = F_{2} \cdot B C$
Equal forces	resultant at the midpoint	no resultant (a couple)

Checklist

Choose a positive direction before adding forces.
Take moments about a point where an unknown force acts.
Check the signs of clockwise and anticlockwise moments.
Check results: the sum of moments about a second point must also be zero.

📖

Comprehensive Curriculum Encyclopedia & Core Terminology — Parallel Coplanar Forces: Resultant and Equilibrium

📌 Comprehensive Scientific Terminology & Definitions:

Parallel Coplanar Forces (القوى المتوازية المستوية): Forces whose lines of action lie in the same geometric plane and are mutually parallel (same direction or opposite directions).

Resultant Magnitude & Direction (محصلة القوى المتوازية): For like parallel forces (same direction): R = F₁ + F₂ in the same direction; For unlike parallel forces (opposite directions, F₁ > F₂): R = F₁ - F₂ acting in the direction of the larger force F₁.

Point of Action of Resultant (نقطة تأثير المحصلة): The resultant acts at a point C dividing the segment AB internally for like forces (F₁ · AC = F₂ · BC) and externally for unlike forces on the side of the larger force.

Varignon's Theorem of Moments (نظرية فارينون للعزوم): The algebraic sum of the moments of a system of coplanar forces about any point in their plane equals the moment of their resultant about that same point: Σ M_O = M_R.

Equilibrium of Parallel Forces (اتزان القوى المتوازية): A system of parallel coplanar forces is in complete static equilibrium if and only if two simultaneous conditions hold: 1. Vector sum of forces equals zero (Σ F = 0, meaning upward forces = downward forces); 2. Sum of moments about any arbitrary point equals zero (Σ M = 0).

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

('Like Parallel Forces: R = F₁ + F₂; Point C lies on AB between A and B such that: F₁ · AC = F₂ · BC.',)

('Unlike Parallel Forces (F₁ > F₂): R = F₁ - F₂; Point C lies on ray BA outside AB (beyond A) such that: F₁ · AC = F₂ · BC.',)

('General Equilibrium Conditions: Σ F_y = 0 and Σ M_any point = 0.',)

🔍 In-Depth Explanations, Deductions & Core Concepts:

Why the resultant of unlike parallel forces acts outside the segment AB: Because the moment of the resultant must balance the opposing moments of F₁ and F₂. Since F₁ > F₂, to satisfy F₁ · AC = F₂ · BC, AC must be smaller than BC, requiring C to be outside AB closer to F₁.
Rod on the verge of tilting / rotating about a support (على وشك الدوران): When a heavy uniform rod resting on two knife-edge supports A and B is loaded so heavily that it is on the verge of rotating about support B, the reaction at support A drops to zero (R_A = 0).
Finding unknown loads using smart moment centers: Always take moments about an unknown support reaction (e.g. Σ M_B = 0) to eliminate that unknown force from the torque equation.

📊 Summary of Like vs Unlike Parallel Coplanar Forces:

1Resultant Magnitude

Like Parallel Forces (Same Direction)

R = F₁ + F₂

Unlike Parallel Forces (Opposite Directions, F₁ > F₂)

R = F₁ - F₂

2Resultant Direction

Like Parallel Forces (Same Direction)

Same direction as both forces

Unlike Parallel Forces (Opposite Directions, F₁ > F₂)

Direction of the larger force F₁

3Point of Action C

Like Parallel Forces (Same Direction)

Internal division on segment AB

Unlike Parallel Forces (Opposite Directions, F₁ > F₂)

External division outside segment AB (closer to F₁)

4Moment Ratio Formula

Like Parallel Forces (Same Direction)

F₁ · AC = F₂ · BC (C between A and B)

Unlike Parallel Forces (Opposite Directions, F₁ > F₂)

F₁ · AC = F₂ · BC (AC < BC, C outside on side of F₁)

💡 High-Yield Exam Traps & Key Takeaways:

Exam Secret: 'On the verge of lifting off support A' means immediately set Reaction R_A = 0 in your equilibrium system.
Weight of Uniform Rod: The weight of a uniform rod always acts at its geometric midpoint.