Statics | NEB Class 11 Mathematics Notes

Chapter Overview

This chapter is the mathematical bedrock of mechanical stability — analyzing how systems of forces interact to keep bodies in perfect rest and equilibrium. Originating from ancient Archimedean principles and formalized by Newtonian mechanics, statics governs how modern bridges stand, how ladders lean, and how suspended structures distribute weight safely. In NEB board exams, Statics is a highly systematic and scoring unit, featuring regularly across all sections — from 1-mark conceptual MCQs to 5-mark analytic derivations and geometric proofs. Master these core force vectors, and the mechanics section of coordinate and physical mathematics becomes intuitive.

What this chapter covers

Concept of Rigid Bodies, Particle, and Force VectorsParallelogram Law of ForcesAnalytical Determination of Resultant Magnitude and DirectionMaximum and Minimum ResultantsResolution of a Force into Rectangular ComponentsResultant of Any Number of Coplanar ForcesConverse of the Triangle and Polygon Law of ForcesLami's Theorem and Its ApplicationsAnalytical Conditions for Equilibrium of Coplanar ForcesTension in Strings and Thrust in Solid Rods

Learning Objectives

1

Calculate the magnitude and direction of the resultant of concurrent forces using the Parallelogram Law

2

Resolve force vectors into orthogonal components along any set of perpendicular coordinate axes

3

State and analytically derive Lami's Theorem for three coplanar forces in equilibrium

4

Apply the conditions of static equilibrium (∑X = 0, ∑Y = 0) to solve systems with multiple concurrent forces

5

Analyze real-world static systems involving suspended weights, strings, smooth walls, and spherical contacts

6

Identify and solve past NEB questions on concurrent forces and string tensions with high precision

Every Equilibrium System reduces to ∑X = 0 and ∑Y = 0

Key Insight

No matter how many concurrent, coplanar forces are acting on a particle at rest, the net effect must be zero. This means that if we project every single force onto a set of perpendicular axes (usually horizontal and vertical), the sum of the components along each axis must individually equal zero.

∑X = 0 (No net horizontal tendency)
∑Y = 0 (No net vertical tendency)

Whether you are using Lami's Theorem, the Triangle Law, or direct vector resolution, they are all mathematically equivalent pathways to verifying this exact state of zero net acceleration.

In NEB exams: When a question states that the resultant of two forces P and Q is perpendicular to P, immediately write down the condition P + Q cosα = 0. This single step instantly simplifies the system to help solve for unknown parameters.

Key Definitions

Statics

The branch of mechanics that deals with the state of rest of physical bodies under the action of balanced systems of forces.

Rigid Body

An idealized physical body in which the distance between any two given points remains completely constant over time, regardless of the external forces applied to it.

Force

An external agency or vector quantity that changes, or tends to change, the state of rest or of uniform motion of a body in a straight line.

Resultant Force

A single force whose effect on a body is equivalent to the combined effect of two or more individual forces acting on that same body.

$$R^2 = P^2 + Q^2 + 2PQ \cos \alpha$$

Equilibrium

The state of a body at rest, or in uniform motion, where the vector sum of all external forces and torques acting on it is exactly zero.

Tension (T)

The pulling force exerted by a stretched string, rope, or cable on the bodies attached to its ends. Its direction is always directed away from the body along the string.

Thrust (Th)

The compressive or pushing force exerted throughout the length of a rigid rod when compressed. Its direction is directed towards the supported bodies.

Lami's Theorem

If three coplanar forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.

$$\frac{P}{\sin \beta} = \frac{Q}{\sin \gamma} = \frac{R}{\sin \theta}$$

Tension in a String vs Thrust in a Rod

Understanding the physical mechanics and directions of forces within flexible strings versus rigid structures is crucial for formulating correct equilibrium equations.

Tension (String / Rope)

$$\vec{T} \text{ (Pulling Direction)}$$

Acts strictly as a pulling force along the axis of the flexible element

Its direction is always directed away from the body being supported

Strings can only sustain tension; they collapse under compression (thrust is zero)

Assumed uniform throughout the length for light, frictionless strings

Keywords: 'suspended by strings', 'tension in cables', 'tied to hooks'

VS

Thrust (Rigid Rod / Strut)

$$\vec{Th} \text{ (Pushing Direction)}$$

Acts primarily as a pushing force when under compression

Its direction is directed towards the body or support point

Rigid rods can sustain both tension (stretching) and thrust (compression)

Maintains structural spacing between connected bodies under load

Keywords: 'supported by a light rod', 'thrust in struts', 'compressive load'

Quick check: Strings pull, rods push. When setting up vectors, draw Tension arrows pointing away from the joint, and Thrust arrows pointing towards the joint.

Complete Formula Sheet with Derivations

Resultant of Two Concurrent Forces

R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}

P and Q are force magnitudes, α is the angle between their lines of action.

1Let $P$ act along $OA$ (horizontal direction) and $Q$ act along $OB$ at an angle $\alpha$ to $OA$. Complete the parallelogram $OACB$.

2Produce $OA$ to $D$ and draw $CD$ perpendicular to $OA$ produced.

3In right-angled triangle $ADC$: $AD = AC \cos \alpha = Q \cos \alpha$ and $CD = AC \sin \alpha = Q \sin \alpha$.

4In right-angled triangle $ODC$, apply Pythagoras' theorem: $OC^2 = OD^2 + CD^2 \Rightarrow R^2 = (OA + AD)^2 + CD^2$.

5Substitute values: $R^2 = (P + Q \cos \alpha)^2 + (Q \sin \alpha)^2 = P^2 + 2PQ \cos \alpha + Q^2 \cos^2 \alpha + Q^2 \sin^2 \alpha$.

6Simplify: $R^2 = P^2 + Q^2(\cos^2 \alpha + \sin^2 \alpha) + 2PQ \cos \alpha$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$:, we get: $$R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$$

Direction of the Resultant Force

\tan \theta = \frac{Q \sin \alpha}{P + Q \cos \alpha}

θ is the angle that the resultant R makes with the line of action of force P.

1In the same right-angled triangle $ODC$, let angle $\angle COA$ be $\theta$ (the angle $R$ makes with $P$).

2By basic trigonometry: $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{CD}{OD}$.

3Since $OD = OA + AD$:, we substitute the values: $CD = Q \sin \alpha$ and $OD = P + Q \cos \alpha$.

4This directly gives: $$\tan \theta = \frac{Q \sin \alpha}{P + Q \cos \alpha}$$

Resultant of Equal Forces

R = 2P \cos \frac{\alpha}{2}

When P = Q, the resultant bisects the angle between the two forces.

1Substitute $Q = P$ into the standard resultant formula: $R^2 = P^2 + P^2 + 2P(P) \cos \alpha = 2P^2 + 2P^2 \cos \alpha$.

2Factor out $2P^2$: $R^2 = 2P^2(1 + \cos \alpha)$.

3Apply trigonometric identity: $1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$.

4Substitute: $R^2 = 2P^2 \left( 2 \cos^2 \frac{\alpha}{2} \right) = 4P^2 \cos^2 \frac{\alpha}{2}$.

5Take the positive square root: $$R = 2P \cos \frac{\alpha}{2}$$

Resolution of a Force in Perpendicular Directions

X = R \cos \theta, \quad Y = R \sin \theta

Used to break any vector into horizontal (X) and vertical (Y) components.

1Let a force $R$ act at an angle $\theta$ with the horizontal x-axis.

2Construct a right-angled triangle with $R$ as the hypotenuse, and components $X$ along the x-axis and $Y$ along the y-axis.

3By definition: $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{X}{R} \Rightarrow X = R \cos \theta$.

4Similarly: $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{Y}{R} \Rightarrow Y = R \sin \theta$.

Lami's Theorem

\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}

P, Q, R are three coplanar forces in equilibrium; α, β, γ are the angles opposite to P, Q, R respectively.

1Let three concurrent, coplanar forces $P$, $Q$, $R$ act on a particle in equilibrium.

2By the Triangle Law of Forces, these forces can be represented in magnitude and direction by the sides of a triangle $ABC$ taken in order.

3Let sides $BC$, $CA$, $AB$ represent $P$, $Q$, $R$ respectively.

4Apply the Sine Rule to triangle $ABC$: $\frac{BC}{\sin A} = \frac{CA}{\sin B} = \frac{AB}{\sin C}$.

5Since the interior angles of the triangle are supplementary to the angles between the forces: $A = 180^\circ - \alpha$, $B = 180^\circ - \beta$, $C = 180^\circ - \gamma$.

6Using $\sin(180^\circ - \theta) = \sin \theta$:, substitution yields: $$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$$

Static Force Analysis Decision Tree

Based on the details provided in a statics problem, follow this logic pathway to select the ideal mathematical tool immediately.

Are there exactly TWO concurrent forces acting at a point?

Yes

Use the Parallelogram Law of Forces: R = √(P² + Q² + 2PQ cosα)

No

Continue ↓

Are there exactly THREE forces maintaining a particle in equilibrium?

Yes

Use Lami's Theorem: P/sinα = Q/sinβ = R/sinγ (fastest method)

No

Continue ↓

Are there MORE THAN THREE coplanar forces acting at a point?

Yes

Resolve all forces: X = ∑P cosθ, Y = ∑P sinθ. Resultant R = √(X² + Y²)

No

Check if forces are parallel. If so, apply the Principle of Moments about a convenient point.

Lami's Theorem and Angle Relationships

Lami's Theorem simplifies equilibrium problems involving exactly three coplanar, concurrent forces by relating their magnitudes to the sines of their opposing angles.

$$\frac{\text{Force}_1}{\sin(\text{Angle between Force}_2 \text{ and Force}_3)} = \text{Constant}$$

Why it works (Derivation):

1In many standard NEB questions, a weight $W$ is suspended by strings or resting on surfaces, with three distinct tension, reaction, or gravitational forces intersecting.

2To find the angles between the forces, locate a reference horizontal or vertical line at the point of concurrency.

3The angle between tension $T_1$ and the vertical gravity vector is often expressed as $180^\circ - \theta$ or $90^\circ + \theta$ depending on how the system geometry is defined.

4Once all three angular separations are identified, write down the three-part ratio and solve for the unknown forces.

Example: A weight of 10N is suspended by two strings making angles of 30° and 60° with the vertical. The tension forces are $T_1$, $T_2$, and the gravity force is $10$N downwards. The angle opposite to $10$N is $30^\circ + 60^\circ = 90^\circ$ (since $180^\circ - 30^\circ - 60^\circ = 90^\circ$). The angle opposite to $T_1$ is $180^\circ - 60^\circ = 120^\circ$. The angle opposite to $T_2$ is $180^\circ - 30^\circ = 150^\circ$. Applying Lami's: $\frac{T_1}{\sin 150^\circ} = \frac{T_2}{\sin 120^\circ} = \frac{10}{\sin 90^\circ}$.

When to use: Use this method for any system with three forces in equilibrium to avoid setting up complicated systems of simultaneous equations.

Memory Tricks — Never Forget These

Trick: Three reliable visual and mathematical tricks to handle statics concepts in exam situations.

Steps to Remember:

Resultant Cosine Sign — 'Plus for Sum': Do not confuse the resultant formula with the Cosine Rule from trigonometry. In mechanics, forces combine to create a larger resultant, so the sign in front of the cosine term is always **positive** ($+2PQ \cos \alpha$).
Lami's Angle Opposites: Think of Lami's theorem as a circular balance. The force vector points in one direction, while the angle in its denominator is the gap directly behind it, bounded by the other two forces.
Direction Ratio — 'Sin is on Top': In $\tan \theta = \frac{Q \sin \alpha}{P + Q \cos \alpha}$, the force $Q$ is the 'other' force (not the reference line $P$). Remember: the 'Other' force goes with $\sin$ on top, and $\cos$ on the bottom next to the reference force.

Mnemonic: Plus for Resultant | Opposing angles for Lami | Other force takes Sin on top for direction

Common Mistakes

Avoid these errors to protect your marks in the NEB exam.

Incorrect Reference Angle: Calculating the direction angle $\theta$ from the wrong vector. Always track whether $\theta$ is measured with respect to $P$ or $Q$.
Applying Lami's to Non-Equilibrium Systems: Attempting to use Lami's Theorem on three concurrent forces that are accelerating or have a non-zero resultant.
Ignoring Force Directions: Forgetting to assign negative signs to resolved components pointing left (negative x-axis) or downwards (negative y-axis).
Tension and Thrust Vector Arrow Reversal: Drawing the tension vector pointing towards the load instead of away from it. This changes the signs in your equations.
Mixing Sine and Cosine Components: Swapping components by default (e.g., assuming the horizontal is always $\cos$ and vertical is always $\sin$). If the reference angle is with the vertical, the vertical component is $R \cos \theta$.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Examiners regularly test these specific cases. Keep them in mind during your preparation.

Resultant is Perpendicular to P: The question states $R \perp P$. Students often try to solve this with complex geometry. Instead, set $\tan 90^\circ = \infty$, which means the denominator of the direction formula must be zero: $P + Q \cos \alpha = 0$.
Forces Along Regular Polygon Sides: Questions involving forces acting along the sides of a regular hexagon. Remember that the angle of each successive side increases by $60^\circ$ relative to the positive x-axis ($0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$).
Sphere Resting on a Smooth Wall: When a sphere of weight $W$ is supported by a string on a wall, three concurrent forces act at its center: Weight (downward), Tension (along string), and the Normal Reaction of the wall (pointing horizontally away from the wall).
Maximum and Minimum Resultant Conditions: If given $R_{\max} = A$ and $R_{\min} = B$, solve the system of linear equations $P + Q = A$ and $P - Q = B$ to find the individual forces.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Two forces P and Q have maximum resultant R_max and minimum R_min	Set up equations P + Q = R_max and P − Q = R_min, then solve
Resultant of two forces is perpendicular to one of them (e.g., R ⊥ P)	Use the perpendicular condition: P + Q cosα = 0
Three forces in equilibrium with given angular relationships	Apply Lami's Theorem: P/sinα = Q/sinβ = R/sinγ
Forces acting along the sides of a triangle, square, or hexagon	Use the resolution method: sum horizontal (X) and vertical (Y) components
Sphere suspended by a string against a vertical wall	Identify concurrent forces at the center of the sphere, then apply Lami's Theorem
Weight suspended by two strings of given lengths and hook spacing	Use the triangle of lengths to find angles, then apply Lami's Theorem
Two equal forces P and P act at angle α	Use the simplified resultant formula: R = 2P cos(α/2)

Solved Examples (Easy → Hard)

1

Two forces of magnitude 6 N and 8 N act at a point at an angle of 90°. Find the magnitude of their resultant.

Easy

Solution

Identify parameters: $P = 6$ N, $Q = 8$ N, $\alpha = 90^\circ$.

Apply standard resultant formula:
$$R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$$
$$R = \sqrt{6^2 + 8^2 + 2(6)(8) \cos 90^\circ}$$

Since $\cos 90^\circ = 0$, the equation simplifies to:
$$R = \sqrt{36 + 64 + 0} = \sqrt{100} = 10 \text{ N}$$

∴ Resultant = 10 N

2

The maximum resultant of two concurrent forces is 16 N and their minimum resultant is 4 N. Find the magnitude of the two individual forces.

Easy

Solution

Formulate equations:
Let the two forces be $P$ and $Q$ (where $P \ge Q$).
• $R_{\max} = P + Q = 16$ ... (i)
• $R_{\min} = P - Q = 4$ ... (ii)

Solve simultaneously:
Add equation (i) and (ii):
$2P = 20 \Rightarrow P = 10 \text{ N}$.

Substitute $P = 10$ back into (i):
$10 + Q = 16 \Rightarrow Q = 6 \text{ N}$.

∴ Forces are 10 N and 6 N

3

The resultant of two forces $P$ and $Q$ acting at an angle $\alpha$ is $R$. If $Q$ is doubled, the new resultant is perpendicular to $P$. Prove that the magnitude of the new resultant is $\sqrt{4Q^2 - P^2}$.

Medium

Solution

Step 1 — Apply perpendicular direction condition:
The new forces are $P$ and $2Q$. Since their resultant is perpendicular to $P$, the denominator of the direction formula must be zero:
$$P + (2Q) \cos \alpha = 0 \implies \cos \alpha = -\frac{P}{2Q}$$

Step 2 — Calculate magnitude of the new resultant ($R'$):
$$R'^2 = P^2 + (2Q)^2 + 2P(2Q) \cos \alpha$$
$$R'^2 = P^2 + 4Q^2 + 4PQ \cos \alpha$$

Substitute $\cos \alpha = -\frac{P}{2Q}$ into the equation:
$$R'^2 = P^2 + 4Q^2 + 4PQ \left( -\frac{P}{2Q} \right)$$
$$R'^2 = P^2 + 4Q^2 - 2P^2 = 4Q^2 - P^2$$
$$R' = \sqrt{4Q^2 - P^2}$$

∴ New Resultant = √(4Q² − P²) [Proved]

4

Two forces $P$ and $Q$ have a resultant $R$. If $Q$ is reversed in direction, the resultant becomes $R'$. Prove that $R^2 + R'^2 = 2(P^2 + Q^2)$.

Medium

Solution

Step 1 — Write expression for R:
When forces are $P$ and $Q$ at angle $\alpha$:
$$R^2 = P^2 + Q^2 + 2PQ \cos \alpha \quad \text{--- (i)}$$

Step 2 — Write expression for R':
Reversing $Q$ means the angle between $P$ and $-Q$ becomes $180^\circ - \alpha$.
$$R'^2 = P^2 + Q^2 + 2PQ \cos(180^\circ - \alpha)$$
Since $\cos(180^\circ - \alpha) = -\cos \alpha$:
$$R'^2 = P^2 + Q^2 - 2PQ \cos \alpha \quad \text{--- (ii)}$$

Step 3 — Add equations (i) and (ii):
$$R^2 + R'^2 = (P^2 + Q^2 + 2PQ \cos \alpha) + (P^2 + Q^2 - 2PQ \cos \alpha)$$
$$R^2 + R'^2 = 2P^2 + 2Q^2 = 2(P^2 + Q^2)$$.

∴ R² + R'² = 2(P² + Q²) [Proved]

5

A smooth sphere of weight W and radius r rests against a smooth vertical wall. It is supported by a string of length l attached to its surface and to a point on the wall. Find the tension in the string.

Hard

Solution

Step 1 — Analyze concurrent forces at the center O:
Let $O$ be the center of the sphere. Three forces act concurrently at $O$:
1. Weight $W$ acting vertically downwards.
2. Reaction of the wall $R$ acting horizontally away from the wall.
3. Tension $T$ acting along the string at an angle $\theta$ with the vertical wall.

Step 2 — Set up right triangle OAC:
Let $C$ be the point of attachment on the wall, and $A$ the point of contact of the sphere with the wall. $OA = r$, $OC = r + l$.
In right-angled triangle $OAC$:
$$\sin \theta = \frac{OA}{OC} = \frac{r}{r+l}$$
$$\cos \theta = \frac{AC}{OC} = \frac{\sqrt{(r+l)^2 - r^2}}{r+l} = \frac{\sqrt{l^2 + 2rl}}{r+l}$$

Step 3 — Apply equilibrium along the vertical axis:
Resolving forces vertically:
$$T \cos \theta = W \implies T = \frac{W}{\cos \theta}$$
Substitute the value of $\cos \theta$:
$$T = \frac{W}{\frac{\sqrt{l^2+2rl}}{r+l}} = \frac{W(r+l)}{\sqrt{l^2 + 2rl}}$$.

∴ Tension = W(r+l) / √(l² + 2rl)

MCQ Practice

Mixed Level — MCQ

1

If two equal forces P and P act at a point, and their resultant is also equal to P, what is the angle between them?

Correct!

Incorrect. Correct: 120°

Solution

Using $R^2 = P^2 + Q^2 + 2PQ \cos \alpha$ with $Q = P$ and $R = P$:
$$P^2 = P^2 + P^2 + 2P^2 \cos \alpha \Rightarrow P^2 = 2P^2(1 + \cos \alpha)$$
$$\frac{1}{2} = 1 + \cos \alpha \Rightarrow \cos \alpha = -\frac{1}{2} \Rightarrow \alpha = 120^\circ$$.

∴ 120°

2

The maximum and minimum resultants of two forces are 20 N and 4 N respectively. The magnitude of the larger force is:

Correct!

Incorrect. Correct: 12 N

Solution

Let forces be $P$ and $Q$. $P+Q=20$ and $P-Q=4$. Adding these gives $2P = 24 \Rightarrow P = 12$ N.

∴ 12 N

3

If the resultant of two forces P and Q acting at an angle α is perpendicular to P, then the value of cosα is:

Correct!

Incorrect. Correct: −P/Q

Solution

Since the resultant is perpendicular to $P$, $\theta = 90^\circ$.
$\tan 90^\circ = \frac{Q \sin \alpha}{P + Q \cos \alpha} = \infty \Rightarrow P + Q \cos \alpha = 0 \Rightarrow \cos \alpha = -\frac{P}{Q}$.

∴ −P/Q

4

Two perpendicular forces of magnitude P and P√3 act at a point. The angle made by their resultant with force P is:

Correct!

Incorrect. Correct: 60°

Solution

With $\alpha = 90^\circ$, $Q = P\sqrt{3}$.
$$\tan \theta = \frac{Q \sin 90^\circ}{P + Q \cos 90^\circ} = \frac{P\sqrt{3}(1)}{P + 0} = \sqrt{3} \Rightarrow \theta = 60^\circ$$

∴ 60°

5

Lami's Theorem is applicable for which of the following systems?

Correct!

Incorrect. Correct: Exactly three coplanar forces in equilibrium acting at a point

Solution

Lami's Theorem states that if exactly three coplanar concurrent forces are in equilibrium, each force is proportional to the sine of the angle between the other two.

∴ Exactly three coplanar forces in equilibrium acting at a point

6

If three concurrent forces are in equilibrium, they can be represented in magnitude and direction by the sides of a:

Correct!

Incorrect. Correct: Triangle taken in order

Solution

According to the Triangle Law of Forces, if three forces acting at a point are in equilibrium, they can be represented by the sides of a triangle taken in order.

∴ Triangle taken in order

7

A weight W is suspended by two strings of equal length making an angle θ with the vertical. The tension in each string is:

Correct!

Incorrect. Correct: W / (2 cosθ)

Solution

Resolving vertically: $2T \cos \theta = W \Rightarrow T = \frac{W}{2 \cos \theta}$.

∴ W / (2 cosθ)

8

The horizontal component of a force F acting at an angle θ with the horizontal is:

Correct!

Incorrect. Correct: F cosθ

Solution

By orthogonal resolution, the component of force adjacent to angle $\theta$ is $F \cos \theta$.

∴ F cosθ

9

If two forces P and Q act at an angle of 180°, their resultant is:

Correct!

Incorrect. Correct: |P − Q|

Solution

At $\alpha = 180^\circ$, $\cos 180^\circ = -1$.
$R = \sqrt{P^2+Q^2-2PQ} = \sqrt{(P-Q)^2} = |P-Q|$.

∴ |P − Q|

10

Which of the following is a condition for the equilibrium of a system of coplanar concurrent forces?

Correct!

Incorrect. Correct: Both ∑X = 0 and ∑Y = 0

Solution

For complete static equilibrium in a plane, the net resolved parts in both horizontal ($\sum X$) and vertical ($\sum Y$) directions must independently equal zero.

∴ Both ∑X = 0 and ∑Y = 0

Short Questions (2 Marks Each)

1

Define a rigid body and explain why it is considered an idealized concept.

[2 marks]

Solution

A rigid body is one in which the distance between any two of its points remains completely unchanged when external forces are applied.
In reality, all materials undergo some degree of deformation under load. Therefore, a perfectly rigid body does not exist, making it an idealized mathematical concept.

∴ A body with zero deformation under load; it is idealized because all real bodies deform slightly.

2

State the Parallelogram Law of Forces.

[2 marks]

Solution

If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point.

∴ Resultant of two forces is represented by the diagonal of a parallelogram formed by them.

3

State Lami's Theorem.

[2 marks]

Solution

Lami's Theorem states that if three coplanar forces acting at a point are in equilibrium, then each force is directly proportional to the sine of the angle between the other two forces:
$$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$$

∴ Each force is proportional to the sine of the angle between the other two forces for three forces in equilibrium.

4

Two forces of 5 N and 12 N act at right angles to each other at a point. Find the magnitude of their resultant.

[2 marks]

Solution

Here, $P = 5$ N, $Q = 12$ N, $\alpha = 90^\circ$.
Since the forces are perpendicular, the resultant is:
$$R = \sqrt{P^2 + Q^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ N}$$

∴ Resultant = 13 N

5

Resolve a force of 100 N into two perpendicular components, where one component makes an angle of 60° with the force.

[2 marks]

Solution

Let $R = 100$ N and $\theta = 60^\circ$.
• The component adjacent to the angle is:
$$X = R \cos \theta = 100 \cos 60^\circ = 100 \times 0.5 = 50 \text{ N}$$
• The perpendicular component is:
$$Y = R \sin \theta = 100 \sin 60^\circ = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \text{ N} \approx 86.6 \text{ N}$$

∴ Components are 50 N and 50√3 N

6

If the maximum resultant of two forces is 20 N and the minimum resultant is 4 N, find the magnitude of the forces.

[2 marks]

Solution

Let the forces be $P$ and $Q$ ($P \ge Q$).
1. $P + Q = 20$
2. $P - Q = 4$
Adding both equations: $2P = 24 \Rightarrow P = 12$ N.
Subtracting both equations: $2Q = 16 \Rightarrow Q = 8$ N.

∴ Forces are 12 N and 8 N

7

State the converse of the Triangle Law of Forces.

[2 marks]

Solution

The converse of the Triangle Law of Forces states that if three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the sides of a triangle taken in order.

∴ Three forces in equilibrium can be represented by the sides of a closed triangle taken in order.

8

Prove that the resultant of two equal forces P acts along the bisector of the angle between them.

[2 marks]

Solution

Let the two forces be $P$ and $P$ acting at an angle $\alpha$.
The direction angle $\theta$ is given by:
$$\tan \theta = \frac{P \sin \alpha}{P + P \cos \alpha} = \frac{P \sin \alpha}{P(1 + \cos \alpha)} = \frac{\sin \alpha}{1 + \cos \alpha}$$
Substitute trigonometric identities:
$$\tan \theta = \frac{2 \sin(\alpha/2) \cos(\alpha/2)}{2 \cos^2(\alpha/2)} = \tan \frac{\alpha}{2} \implies \theta = \frac{\alpha}{2}$$
Since $\theta = \alpha/2$, the resultant bisects the angle.

∴ θ = α/2, showing the resultant bisects the angle.

9

Define tension in a string and thrust in a rod, explaining their directions of action.

[2 marks]

Solution

Tension: The pulling force exerted by a stretched string on a body. Its direction is always directed along the string and away from the body.
Thrust: The pushing force exerted by a compressed rod. Its direction is directed along the rod and towards the body under compression.

∴ Tension pulls (away from the body); Thrust pushes (towards the body).

10

Under what conditions do two forces P and Q yield their maximum and minimum resultants?

[2 marks]

Solution

The resultant is given by $R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$.
• Maximum: Occurs when $\cos \alpha = 1 \Rightarrow \alpha = 0^\circ$ (forces act in the same direction). $R_{\max} = P + Q$.
• Minimum: Occurs when $\cos \alpha = -1 \Rightarrow \alpha = 180^\circ$ (forces act in opposite directions). $R_{\min} = |P - Q|$.

∴ Max at α = 0° (R = P+Q); Min at α = 180° (R = |P−Q|).

Long Questions (4–5 Marks Each)

1

Derive analytically the expression for the magnitude and direction of the resultant of two forces P and Q acting at an angle α.

[5 marks]

Solution

Step 1 — Geometric Representation:
Let $P$ and $Q$ be represented in magnitude and direction by the sides $OA$ and $OB$ of a parallelogram $OACB$ with angle $\angle AOB = \alpha$. The diagonal $OC$ represents the resultant $R$.

Step 2 — Projection setup:
Produce $OA$ to a point $D$ and draw $CD$ perpendicular to $OA$ produced. Since $AC$ is parallel to $OB$, $AC = Q$ and $\angle CAD = \alpha$.
In right-angled triangle $ADC$:
$$AD = AC \cos \alpha = Q \cos \alpha$$
$$CD = AC \sin \alpha = Q \sin \alpha$$

Step 3 — Derivation of Magnitude (R):
In right-angled triangle $ODC$, applying Pythagoras' theorem:
$$OC^2 = OD^2 + CD^2 \implies R^2 = (OA + AD)^2 + CD^2$$
$$R^2 = (P + Q \cos \alpha)^2 + (Q \sin \alpha)^2$$
$$R^2 = P^2 + 2PQ \cos \alpha + Q^2 \cos^2 \alpha + Q^2 \sin^2 \alpha$$
$$R^2 = P^2 + Q^2(\cos^2 \alpha + \sin^2 \alpha) + 2PQ \cos \alpha$$
Since $\cos^2 \alpha + \sin^2 \alpha = 1$:
$$R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$$

Step 4 — Derivation of Direction (θ):
Let the resultant $R$ make an angle $\theta$ with force $P$ ($OA$). In right-angled triangle $ODC$:
$$\tan \theta = \frac{CD}{OD} = \frac{CD}{OA + AD} = \frac{Q \sin \alpha}{P + Q \cos \alpha}$$

∴ R = √(P² + Q² + 2PQ cosα) and tanθ = (Q sinα) / (P + Q cosα)

2

Prove Lami's Theorem analytically.

[5 marks]

Solution

Step 1 — Setup and Diagram:
Let three coplanar concurrent forces $P$, $Q$, and $R$ act on a particle at point $O$ in equilibrium. Let the angles opposite to $P$, $Q$, and $R$ be $\alpha$, $\beta$, and $\gamma$ respectively.

Step 2 — Vector Representation:
By the Triangle Law of Forces, since the particle is in equilibrium, these three forces can be represented in magnitude and direction by the sides of a triangle $ABC$ taken in order.
Let side $BC$ represent $P$, $CA$ represent $Q$, and $AB$ represent $R$.

Step 3 — Relate interior and exterior angles:
The direction of forces means the interior angles of triangle $ABC$ are supplementary to the angles between the forces:
• Angle $A$ of the triangle is opposite to side $BC$ (representing $P$). The angle between $Q$ and $R$ is $\alpha$. Hence, $A = 180^\circ - \alpha$.
• Similarly, $B = 180^\circ - \beta$.
• Similarly, $C = 180^\circ - \gamma$.

Step 4 — Apply the Sine Rule to the triangle:
By the Sine Rule in triangle $ABC$:
$$\frac{BC}{\sin A} = \frac{CA}{\sin B} = \frac{AB}{\sin C}$$
Substitute the force representations and interior angle values:
$$\frac{P}{\sin(180^\circ - \alpha)} = \frac{Q}{\sin(180^\circ - \beta)} = \frac{R}{\sin(180^\circ - \gamma)}$$
Since $\sin(180^\circ - x) = \sin x$:
$$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$$.

∴ P/sinα = Q/sinβ = R/sinγ [Proved]

3

The sum of two forces is 18 N and their resultant is 12 N. If the resultant is perpendicular to the smaller force, find the magnitude of the two forces.

[5 marks]

Solution

Step 1 — Define variables and write given values:
Let the smaller force be $P$ and the larger force be $Q$.
• $P + Q = 18 \implies Q = 18 - P$ ... (i)
• Resultant $R = 12$ N.
• Resultant is perpendicular to $P \implies \theta = 90^\circ$.

Step 2 — Apply perpendicular direction condition:
$$P + Q \cos \alpha = 0 \implies Q \cos \alpha = -P$$ ... (ii)

Step 3 — Substitute into resultant formula:
$$R^2 = P^2 + Q^2 + 2PQ \cos \alpha$$
$$12^2 = P^2 + Q^2 + 2P(Q \cos \alpha)$$
Substitute $Q \cos \alpha = -P$ from (ii):
$$144 = P^2 + Q^2 + 2P(-P)$$
$$144 = P^2 + Q^2 - 2P^2 \implies 144 = Q^2 - P^2$$
$$144 = (Q - P)(Q + P)$$ ... (iii)

Step 4 — Solve for forces P and Q:
Substitute $Q + P = 18$ into (iii):
$$144 = (Q - P) \times 18 \implies Q - P = \frac{144}{18} = 8$$ ... (iv)
We now have two linear equations:
1. $Q + P = 18$
2. $Q - P = 8$
Adding these: $2Q = 26 \pm Q = 13$ N.
Subtracting these: $2P = 10 \Rightarrow P = 5$ N.

∴ Forces are 5 N and 13 N

4

Forces of magnitude 1, 2, 3, 4, 5, and 6 N act along the sides AB, BC, CD, DE, EF, and FA of a regular hexagon taken in order. Find the magnitude and direction of the resultant.

[5 marks]

Solution

Step 1 — Set up coordinate axes and angles:
Let the side $AB$ lie along the positive x-axis. The angles that the sides make with the x-axis are:
$AB \to 0^\circ$, $BC \to 60^\circ$, $CD \to 120^\circ$, $DE \to 180^\circ$, $EF \to 240^\circ$, $FA \to 300^\circ$.

Step 2 — Resolve forces horizontally (X):
$$X = 1 \cos 0^\circ + 2 \cos 60^\circ + 3 \cos 120^\circ + 4 \cos 180^\circ + 5 \cos 240^\circ + 6 \cos 300^\circ$$
$$X = 1(1) + 2(0.5) + 3(-0.5) + 4(-1) + 5(-0.5) + 6(0.5)$$
$$X = 1 + 1 - 1.5 - 4 - 2.5 + 3 = -3 \text{ N}$$

Step 3 — Resolve forces vertically (Y):
$$Y = 1 \sin 0^\circ + 2 \sin 60^\circ + 3 \sin 120^\circ + 4 \sin 180^\circ + 5 \sin 240^\circ + 6 \sin 300^\circ$$
$$Y = 1(0) + 2\left(\frac{\sqrt{3}}{2}\right) + 3\left(\frac{\sqrt{3}}{2}\right) + 4(0) + 5\left(-\frac{\sqrt{3}}{2}\right) + 6\left(-\frac{\sqrt{3}}{2}\right)$$
$$Y = 0 + \sqrt{3} + 1.5\sqrt{3} + 0 - 2.5\sqrt{3} - 3\sqrt{3} = -3\sqrt{3} \text{ N}$$

Step 4 — Calculate resultant magnitude (R) and direction:
$$R = \sqrt{X^2 + Y^2} = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \text{ N}$$
Direction $\theta$:
$$\tan \theta = \frac{Y}{X} = \frac{-3\sqrt{3}}{-3} = \sqrt{3}$$
Since both $X$ and $Y$ are negative, $\theta$ lies in the third quadrant:
$$\theta = 180^\circ + 60^\circ = 240^\circ \text{ with } AB$$.

∴ Resultant = 6 N acting at 240° to AB

5

A body of weight W is supported by two strings of lengths 3 cm and 4 cm tied to two points in a horizontal line at a distance of 5 cm. Find the tension in each string.

[5 marks]

Solution

Step 1 — Verify triangle geometry:
The lengths of the strings are $AC = 3$ cm and $BC = 4$ cm. The horizontal spacing is $AB = 5$ cm.
Since $3^2 + 4^2 = 5^2$, triangle $ABC$ is a right-angled triangle with $\angle C = 90^\circ$.

Step 2 — Identify angles with the vertical:
Let $T_1$ be the tension in string $AC$ and $T_2$ be the tension in $BC$.
Let string $AC$ make angle $\theta$ with the vertical. Then string $BC$ makes angle $90^\circ - \theta$ with the vertical.
From triangle geometry:
$$\cos \theta = \frac{AC}{AB} = \frac{3}{5} \quad \text{and} \quad \sin \theta = \frac{BC}{AB} = \frac{4}{5}$$

Step 3 — Apply equilibrium equations:
• Resolving forces horizontally:
$$T_1 \sin \theta = T_2 \cos \theta \implies T_1 \left( \frac{4}{5} \right) = T_2 \left( \frac{3}{5} \right) \implies 4T_1 = 3T_2 \implies T_2 = \frac{4}{3} T_1$$
• Resolving forces vertically:
$$T_1 \cos \theta + T_2 \sin \theta = W$$
$$T_1 \left( \frac{3}{5} \right) + \left( \frac{4}{3} T_1 \right) \left( \frac{4}{5} \right) = W$$
$$T_1 \left( \frac{3}{5} + \frac{16}{15} \right) = W \implies T_1 \left( \frac{9 + 16}{15} \right) = W \implies T_1 \left( \frac{25}{15} \right) = W$$
$$T_1 = \frac{3}{5} W = 0.6W$$

Step 4 — Calculate T2:
$$T_2 = \frac{4}{3} \left( \frac{3}{5} W \right) = \frac{4}{5} W = 0.8W$$.

∴ Tension 1 = 3W/5, Tension 2 = 4W/5

6

Three forces P, Q, and R acting on a particle are in equilibrium. The angle between P and Q is 120°, and that between Q and R is 150°. Find the ratio of the forces P : Q : R.

[4 marks]

Solution

Step 1 — Determine the third angle:
The sum of the angles around a point is $360^\circ$.
Angle between $P$ and $Q = 120^\circ$.
Angle between $Q$ and $R = 150^\circ$.
Angle between $R$ and $P = 360^\circ - (120^\circ + 150^\circ) = 360^\circ - 270^\circ = 90^\circ$.

Step 2 — Apply Lami's Theorem:
$$\frac{P}{\sin(\text{angle between } Q \text{ and } R)} = \frac{Q}{\sin(\text{angle between } R \text{ and } P)} = \frac{R}{\sin(\text{angle between } P \text{ and } Q)}$$
$$\frac{P}{\sin 150^\circ} = \frac{Q}{\sin 90^\circ} = \frac{R}{\sin 120^\circ}$$

Step 3 — Evaluate sines and write the ratio:
• $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = 0.5$
• $\sin 90^\circ = 1.0$
• $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$
$$\frac{P}{1/2} = \frac{Q}{1} = \frac{R}{\sqrt{3}/2}$$
Multiply all terms by 2:
$$P : Q : R = 1 : 2 : \sqrt{3}$$.

∴ P : Q : R = 1 : 2 : √3

7

Forces P, 2P, 3P, and 4P act along the sides AB, BC, CD, and DA of a square ABCD taken in order. Find the magnitude and direction of the resultant.

[4 marks]

Solution

Step 1 — Set up coordinate directions:
Let $AB$ be along the positive x-axis. Since $ABCD$ is a square, the directions of the forces are:
• $P$ along $AB$ (along $+x$ direction) $\to \theta_1 = 0^\circ$
• $2P$ along $BC$ (along $+y$ direction) $\to \theta_2 = 90^\circ$
• $3P$ along $CD$ (along $-x$ direction) $\to \theta_3 = 180^\circ$
• $4P$ along $DA$ (along $-y$ direction) $\to \theta_4 = 270^\circ$

Step 2 — Resolve forces horizontally (X) and vertically (Y):
$$X = P \cos 0^\circ + 2P \cos 90^\circ + 3P \cos 180^\circ + 4P \cos 270^\circ = P + 0 - 3P + 0 = -2P$$
$$Y = P \sin 0^\circ + 2P \sin 90^\circ + 3P \sin 180^\circ + 4P \sin 270^\circ = 0 + 2P + 0 - 4P = -2P$$

Step 3 — Find magnitude (R) and direction (θ):
$$R = \sqrt{X^2 + Y^2} = \sqrt{(-2P)^2 + (-2P)^2} = \sqrt{4P^2 + 4P^2} = \sqrt{8P^2} = 2\sqrt{2}P$$
Direction $\theta$:
$$\tan \theta = \frac{Y}{X} = \frac{-2P}{-2P} = 1$$
Since both $X$ and $Y$ are negative, $\theta$ lies in the third quadrant:
$$\theta = 180^\circ + 45^\circ = 225^\circ \text{ with respect to } AB$$.

∴ Resultant magnitude = 2√2 P; Direction = 225° to AB

8

If the resultant R of two forces P and Q acting at an angle α makes an angle θ with P, and if P is replaced by P + R while Q remains unchanged, prove that the new resultant makes an angle θ/2 with the new force.

[5 marks]

Solution

Step 1 — Represent vectors geometrically:
Let $\vec{OA} = \vec{P}$ and $\vec{OB} = \vec{Q}$. Complete the parallelogram $OACB$. Diagonal $\vec{OC} = \vec{R}$.
Angle $\angle COA = \theta$.

Step 2 — Add R to P:
Produce $OA$ to a point $A'$ such that $AA' = OC = R$. Thus, $\vec{OA'} = \vec{P} + \vec{R}$.
The new forces are represented by $\vec{OA'}$ and $\vec{OB}$ (which is equal to $\vec{AC}$).
The new resultant $\vec{R'}$ is given by the diagonal of the parallelogram formed by $\vec{OA'}$ and $\vec{AC}$.
This resultant is represented by $\vec{OC'}$, where $\vec{CC'} = \vec{AA'} = \vec{R}$ and is parallel to $OA$.

Step 3 — Analyze triangle ACC':
In triangle $ACC'$:
• $AC = Q$ (parallel to $OB$)
• $CC' = R$
• $\angle CAC' = \theta$ because $CC'$ is parallel to $OA$.
Since $AC = CC' = R$, triangle $ACC'$ is isosceles with $\angle CAC' = \angle CC'A = \theta$.
The exterior angle of the triangle at $A$ is $\theta/2$, which corresponds to the direction of the new resultant relative to the horizontal axis.

∴ Geometric vector proof shows the direction angle is halved to θ/2 [Proved]

9

Show that the resultant of two equal forces P acting at an angle α is 2P cos(α/2) and find its direction.

[4 marks]

Solution

Step 1 — Apply equal forces to resultant magnitude:
$$R = \sqrt{P^2 + P^2 + 2P^2 \cos \alpha} = \sqrt{2P^2(1 + \cos \alpha)}$$
Using $1 + \cos \alpha = 2 \cos^2(\alpha/2)$:
$$R = \sqrt{2P^2 \left( 2 \cos^2 \frac{\alpha}{2} \right)} = \sqrt{4P^2 \cos^2 \frac{\alpha}{2}} = 2P \cos \frac{\alpha}{2}$$

Step 2 — Find direction:
$$\tan \theta = \frac{P \sin \alpha}{P + P \cos \alpha} = \frac{\sin \alpha}{1 + \cos \alpha}$$
Using double angle formulas:
$$\tan \theta = \frac{2 \sin(\alpha/2) \cos(\alpha/2)}{2 \cos^2(\alpha/2)} = \tan \frac{\alpha}{2} \implies \theta = \frac{\alpha}{2}$$
This confirms the resultant acts along the angle bisector.

∴ R = 2P cos(α/2) acting along the angle bisector

10

Three coplanar forces of magnitudes 5N, 10N, and 15N act on a particle. Is it possible for these three forces to maintain the particle in static equilibrium? Explain.

[4 marks]

Solution

Step 1 — Understand the condition for equilibrium:
For three forces to maintain a particle in equilibrium, they must be capable of forming a closed triangle when represented as vectors taken in order.
This requires the magnitudes to satisfy the triangle inequality theorem: **the sum of any two sides must be strictly greater than or equal to the third side**.

Step 2 — Test inequality with given values:
Let $a = 5$, $b = 10$, $c = 15$.
• $a + b = 5 + 10 = 15$. Here, the sum of the two smaller forces is exactly equal to the third force ($15 = 15$).

Step 3 — Analyze the edge case:
Since $5 + 10 = 15$, a triangle cannot be formed. However, equilibrium is still theoretically possible only if the forces are collinear:
The 5N and 10N forces must act in the exact same direction, while the 15N force acts in the directly opposite direction ($180^\circ$).
If the forces are non-collinear, equilibrium is impossible because no triangle can be closed.

∴ Yes, but only if they are collinear (acting in opposite directions); otherwise, they cannot maintain equilibrium.

Past NEB Board Questions

1

Two forces of magnitude 10 N and 15 N act at a point. If their resultant is 20 N, find the angle between the forces.

NEB 2079 [4 marks]

Solution

Step 1 — Identify parameters:
$P = 10$ N, $Q = 15$ N, $R = 20$ N.

Step 2 — Apply the resultant formula:
$$R^2 = P^2 + Q^2 + 2PQ \cos \alpha$$
$$20^2 = 10^2 + 15^2 + 2(10)(15) \cos \alpha$$
$$400 = 100 + 225 + 300 \cos \alpha$$
$$400 = 325 + 300 \cos \alpha$$

Step 3 — Solve for α:
$$75 = 300 \cos \alpha \implies \cos \alpha = \frac{75}{300} = 0.25$$
$$\alpha = \cos^{-1}(0.25) \approx 75.52^\circ$$.

∴ Angle between forces = 75.52°

2

A mass of 10 kg is suspended by two strings of lengths 6 m and 8 m from two points 10 m apart in a horizontal line. Find the tensions in the strings.

NEB 2078 [4 marks]

Solution

Step 1 — Verify triangle geometry:
The lengths of the strings are $AC = 6$ m and $BC = 8$ m. The spacing between the supports is $AB = 10$ m.
Since $6^2 + 8^2 = 10^2$, the system forms a right-angled triangle at the load point $C$ ($\angle C = 90^\circ$).
Let weight $W = 10 \text{ kg-wt} \approx 10g$ N.

Step 2 — Identify angles:
Let $T_1$ be the tension in string $AC$ and $T_2$ be the tension in $BC$.
Let string $AC$ make angle $\theta$ with the vertical. Then string $BC$ makes angle $90^\circ - \theta$ with the vertical.
From geometry:
$$\cos \theta = \frac{AC}{AB} = \frac{6}{10} = 0.6 \quad \text{and} \quad \sin \theta = \frac{BC}{AB} = \frac{8}{10} = 0.8$$

Step 3 — Apply equilibrium:
• Resolving horizontally:
$$T_1 \sin \theta = T_2 \cos \theta \implies 0.8T_1 = 0.6T_2 \implies T_2 = \frac{4}{3} T_1$$
• Resolving vertically:
$$T_1 \cos \theta + T_2 \sin \theta = 10g$$
$$T_1(0.6) + \left(\frac{4}{3}T_1\right)(0.8) = 10g \implies T_1 \left( 0.6 + \frac{3.2}{3} \right) = 10g$$
$$T_1 \left( \frac{1.8 + 3.2}{3} \right) = 10g \implies T_1 \left( \frac{5}{3} \right) = 10g \implies T_1 = 6g \text{ N} = 6 \text{ kg-wt}$$

Step 4 — Calculate T2:
$$T_2 = \frac{4}{3} (6g) = 8g \text{ N} = 8 \text{ kg-wt}$$.

∴ Tensions are 6 kg-wt and 8 kg-wt (or 58.8 N and 78.4 N)

3

State Lami's theorem. A string of length 13 m is tied to two horizontal hooks 5 m apart. A weight of 26 N is suspended from a point on the string 5 m from one hook. Find the tension in the two parts of the string.

NEB 2077 [5 marks]

Solution

Step 1 — Verify triangle geometry:
The horizontal hook distance is $AB = 5$ m.
The suspension point $C$ is $5$ m from hook $A$ ($AC = 5$ m).
The remaining part of the string is $BC = 13 - 5 = 8$ m.
Since $AC + AB = 5 + 5 = 10 > 8$, a closed triangle is formed. We can find the angles of triangle $ABC$ using the Cosine Rule:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 5^2 - 8^2}{2(5)(5)} = \frac{25 + 25 - 64}{50} = -\frac{14}{50} = -0.28$$
$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{8^2 + 5^2 - 5^2}{2(8)(5)} = \frac{64}{80} = 0.8$$
This gives $\sin A = \sqrt{1 - (-0.28)^2} \approx 0.96$ and $\sin B = 0.6$.

Step 2 — Apply Lami's Theorem:
The three forces acting at $C$ are $T_1$ (along $CA$), $T_2$ (along $CB$), and the weight $26$ N acting vertically downwards.
Using the horizontal reference line at $C$:
$$\frac{T_1}{\sin(90^\circ + B)} = \frac{T_2}{\sin(90^\circ + A)} = \frac{26}{\sin(180^\circ - A - B)}$$
Using identities: $\sin(90^\circ + x) = \cos x$ and $\sin(180^\circ - A - B) = \sin(A+B)$:
$$\frac{T_1}{\cos B} = \frac{T_2}{\cos A} \quad \text{[Analysis of angles gives vertical and horizontal equilibrium]}$$
Alternatively, resolving forces horizontally:
$$T_1 \sin A = T_2 \sin B \implies T_1(0.96) = T_2(0.6) \implies T_1 = 0.625 T_2$$
Resolving vertically:
$$T_1 \cos A + T_2 \cos B = 26 \implies T_1(-0.28) + T_2(0.8) = 26$$
Substitute $T_1 = 0.625 T_2$:
$$(0.625 T_2)(-0.28) + 0.8 T_2 = 26 \implies -0.175 T_2 + 0.8 T_2 = 26$$
$$0.625 T_2 = 26 \implies T_2 = 41.6 \text{ N}$$
Then, $T_1 = 0.625(41.6) = 26 \text{ N}$.

∴ Tensions are 26 N and 41.6 N

4

Show that the resultant of two equal forces P acting at an angle α is 2P cos(α/2).

NEB 2076 [4 marks]

Solution

Step 1 — Use the standard resultant formula:
$$R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$$

Step 2 — Substitute Q = P:
$$R = \sqrt{P^2 + P^2 + 2P^2 \cos \alpha} = \sqrt{2P^2(1 + \cos \alpha)}$$

Step 3 — Apply trigonometric identity:
Since $1 + \cos \alpha = 2 \cos^2(\alpha/2)$:
$$R = \sqrt{2P^2 \cdot 2 \cos^2 \frac{\alpha}{2}} = \sqrt{4P^2 \cos^2 \frac{\alpha}{2}}$$
Taking the square root gives:
$$R = 2P \cos \frac{\alpha}{2} \quad \text{[Proved]}$$.

∴ R = 2P cos(α/2) [Proved]

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes without looking at your notes. This matches the standard NEB exam format.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

The maximum resultant of two forces of magnitude 3 N and 4 N is:

Correct!

Incorrect. Correct: 7 N

Solution

The maximum resultant occurs when $\alpha = 0^\circ$ ($R_{\max} = P + Q = 3 + 4 = 7$ N).

∴ 7 N

2

The vertical component of a force F acting at an angle of 30° to the horizontal is:

Correct!

Incorrect. Correct: 0.5 F

Solution

The vertical component is $Y = F \sin 30^\circ = F \left(\frac{1}{2}\right) = 0.5 F$.

∴ 0.5 F

3

If the angle of the resultant with force P is 90°, the condition is:

Correct!

Incorrect. Correct: P + Q cosα = 0

Solution

Since the resultant is perpendicular to $P$, the denominator in the direction formula must be zero ($P + Q \cos \alpha = 0$).

∴ P + Q cosα = 0

2 Section B — Short Questions (2 Marks Each)

1

Find the components of a 50 N force acting at an angle of 45° to the horizontal.

[2 marks]

Solution

• Horizontal: $X = 50 \cos 45^\circ = 50 \left( \frac{1}{\sqrt{2}} \right) = 25\sqrt{2} \text{ N} \approx 35.36 \text{ N}$
• Vertical: $Y = 50 \sin 45^\circ = 50 \left( \frac{1}{\sqrt{2}} \right) = 25\sqrt{2} \text{ N} \approx 35.36 \text{ N}$

∴ Both components are 25√2 N (approx. 35.36 N)

2

Under what condition is the resultant of two forces equal to the difference of their magnitudes?

[2 marks]

Solution

This occurs when the angle between the two forces is $180^\circ$ (they act along the same line but in opposite directions). In this case, $\cos 180^\circ = -1$, simplifying the formula to $R = |P - Q|$.

∴ When the angle between the forces is 180°.

3 Section C — Long Questions (4–5 Marks Each)

1

State and prove the Parallelogram Law of Forces.

[5 marks]

Solution

See Section 6 and Long Question 1 for the detailed statement, diagram, and analytical proof of the magnitude and direction formulas.

∴ R = √(P² + Q² + 2PQ cosα) [Proved]

2

Find the resultant of forces of magnitude 1, 2, 3, 4, 5, and 6 N acting at the vertices of a regular hexagon taken in order.

[5 marks]

Solution

See Long Question 4 for the step-by-step resolution of horizontal and vertical components, showing that $R = 6$ N acting at an angle of $240^\circ$ with the first side $AB$.

∴ Resultant = 6 N at 240° to AB

One-Page Revision Cheat Sheet

1

Resultant Magnitude: $R = \sqrt{P^2 + Q^2 + 2PQ \cos \alpha}$ — do not use a negative sign here.

2

Direction of Resultant: $\tan \theta = \frac{Q \sin \alpha}{P + Q \cos \alpha}$ where $\theta$ is the angle made with force $P$.

3

Maximum Resultant: $R_{\max} = P + Q$ at $\alpha = 0^\circ$.

4

Minimum Resultant: $R_{\min} = |P - Q|$ at $\alpha = 180^\circ$.

5

Perpendicular Condition: If $R \perp P$, then the denominator is zero: $P + Q \cos \alpha = 0$.

6

Orthogonal Components: $X = F \cos \theta$ and $Y = F \sin \theta$ (when $\theta$ is measured from the x-axis).

7

Lami's Theorem: $\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$ for three coplanar forces in equilibrium.

8

String Tension: A pulling force acting along the string, directed away from the connected body.

9

Rod Thrust: A compressive force acting along the rod, directed towards the supported body.

10

Polygon Forces: Resolve all vectors along orthogonal horizontal and vertical axes to find the resultant: $R = \sqrt{(\sum X)^2 + (\sum Y)^2}$.