CIE AS/A Level Maths-4.1 Forces and equilibrium - Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-4.1 Forces and equilibrium- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-4.1 Forces and equilibrium- Study Notes
Key Concepts:
- Identifying Forces in a Given Situation
- Vector Nature of Force, Components, and Resultants
- Equilibrium of Forces
- Contact Forces: Normal and Frictional Components
- Limiting Friction and Limiting Equilibrium
- Newton’s Third Law
Identifying Forces in a Given Situation
Identifying Forces in a Given Situation
General Approach:
- To analyze forces, we must identify all external forces acting on an object.
- Draw a free-body diagram (FBD), showing the object as a point or box, and represent all forces with arrows.
- The length of each arrow indicates the relative magnitude of the force, and the arrow points in the direction the force acts.
Common Types of Forces:
- Weight (W): The gravitational force acting downward, \(W = mg\).
- Normal Force (N): The reaction force perpendicular to a surface.
- Tension (T): Force transmitted through a rope or string.
- Friction (f): A force opposing motion, acting parallel to the surface.
- Air Resistance / Drag: A resistive force opposing motion through a fluid.
- Applied Force (F): Any push or pull exerted externally.
- Buoyant Force (Upthrust, U): An upward force on objects in fluids.
Steps to Draw a Force Diagram (Free-Body Diagram):
- Represent the object as a point or small box.
- Draw arrows from the object to represent each force.
- Label each arrow with the type of force (e.g., \(W, N, T, f\)).
- Ensure directions are physically correct (e.g., weight always downward, normal perpendicular to surface).
Example:
A block of mass \(m\) rests on a horizontal rough surface. Identify the forces acting on it.
▶️ Answer/Explanation
Step 1: Forces acting:
Weight: \(W = mg\), vertically downward.
Normal force: \(N\), vertically upward from the surface.
Frictional force: \(f\), opposing motion (if an applied force acts).
Step 2: Free-body diagram:
(Block as a box/point; arrow downward labeled \(W\); arrow upward labeled \(N\); arrow to the left or right for friction if motion is attempted.)
\(\boxed{\text{Forces are } W, N, f}\)
Example:
A block is suspended by a string. Identify the forces acting on it and draw a free-body diagram.
▶️ Answer/Explanation
Step 1: Forces acting:
Weight: \(W = mg\), downward.
Tension: \(T\), upward through the string.
Step 2: Free-body diagram:
(Point representing the block; arrow down labeled \(W\); arrow up labeled \(T\).)
\(\boxed{\text{Forces are } W \text{ and } T}\)
Example:
A ball is thrown vertically upward. Identify the forces acting on it during its flight.
▶️ Answer/Explanation
Step 1: Forces acting:
Weight: \(W = mg\), downward at all times.
Air resistance (if present): upward while moving upward, downward while falling down.
Step 2: Free-body diagram:
(Dot representing the ball; arrow down for \(W\); smaller arrow opposite motion for air resistance.)
\(\boxed{\text{Forces are } W \text{ and air resistance}}\)
Vector Nature of Force, Components, and Resultants
Vector Nature of Force, Components, and Resultants
Vector Nature of Force:
- Force is a vector quantity it has both magnitude and direction.
- When several forces act on a body, they combine to give a resultant force.
- If the resultant force is zero, the body is in equilibrium.
Components of a Force:
A force at an angle can be resolved into two perpendicular components, usually along the x- and y-axes.
For a force \(F\) making an angle \(\theta\) with the horizontal:
\(F_x = F \cos \theta \;\;\;\; (x\text{-component})\)
\(F_y = F \sin \theta \;\;\;\; (y\text{-component})\)
These components can be used in calculations of motion, equilibrium, or resolving into resultants.
Resultant of Two Perpendicular Forces:
If two forces act at right angles:
\(R = \sqrt{F_x^2 + F_y^2}\)
\(\tan \theta = \dfrac{F_y}{F_x}\)
where \(R\) is the resultant, and \(\theta\) is the angle it makes with the x-axis.
Resultant of Two Forces at Any Angle:
If two forces \(F_1\) and \(F_2\) act at an angle \(\theta\), then:
\(R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos \theta}\)
\(\tan \phi = \dfrac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}\)
where \(\phi\) is the angle between \(R\) and \(F_1\).
Note: In Diagram \(\vec{A} = F_1\) and \(\vec{B} =F_2\) , \(\phi=\beta\)
Example:
A force of \(50 \ \text{N}\) acts at an angle of \(30^\circ\) above the horizontal. Find its horizontal and vertical components.
▶️ Answer/Explanation
Step 1: Use component formulas:
\(F_x = F \cos \theta = 50 \cos 30^\circ = 50 \times 0.866 = 43.3 \ \text{N}\)
\(F_y = F \sin \theta = 50 \sin 30^\circ = 50 \times 0.5 = 25.0 \ \text{N}\)
\(\boxed{F_x = 43.3 \ \text{N}, \; F_y = 25.0 \ \text{N}}\)
Example:
A body is acted on by a \(60 \ \text{N}\) force horizontally and a \(80 \ \text{N}\) force vertically. Find the magnitude and direction of the resultant.
▶️ Answer/Explanation
Step 1: Use Pythagoras for resultant:
\(R = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \ \text{N}\)
Step 2: Find angle with horizontal:
\(\tan \theta = \dfrac{80}{60} = \dfrac{4}{3}\)
\(\theta = \tan^{-1}\left(\dfrac{4}{3}\right) \approx 53.1^\circ\)
\(\boxed{R = 100 \ \text{N}, \; \theta = 53.1^\circ \text{ above horizontal}}\)
Example:
Two forces of \(10 \ \text{N}\) and \(15 \ \text{N}\) act at an angle of \(120^\circ\). Find the magnitude of the resultant.
▶️ Answer/Explanation
Step 1: Apply the cosine rule:
\(R = \sqrt{10^2 + 15^2 + 2(10)(15)\cos 120^\circ}\)
Since \(\cos 120^\circ = -0.5\):
\(R = \sqrt{100 + 225 + 2(10)(15)(-0.5)}\)
\(R = \sqrt{325 – 150} = \sqrt{175} \approx 13.2 \ \text{N}\)
\(\boxed{R \approx 13.2 \ \text{N}}\)
Equilibrium of Forces
Equilibrium of Forces
A particle is said to be in equilibrium if it remains at rest (or moves with constant velocity). This means that the net or resultant force acting on it is zero.
Principle of Equilibrium:
The vector sum of all forces acting on the particle is zero:
\(\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots = \vec{0}\)
Equivalently, the sum of components in any chosen directions is zero:
\(\Sigma F_x = 0 \quad \text{and} \quad \Sigma F_y = 0\)
Methods of Solving Equilibrium Problems
Resolving forces into components: Forces are split into horizontal and vertical components. Then apply \(\Sigma F_x = 0\) and \(\Sigma F_y = 0\).
Triangle of forces (not required but acceptable): If three forces act on a particle in equilibrium, they must form the sides of a closed triangle when drawn head-to-tail.
Lami’s Theorem (optional method): If three forces act on a particle in equilibrium, then:
\(\dfrac{F_1}{\sin \alpha} = \dfrac{F_2}{\sin \beta} = \dfrac{F_3}{\sin \gamma}\)
where \(\alpha, \beta, \gamma\) are the angles opposite to the forces.
Example:
A particle is in equilibrium under the action of two forces of magnitudes \(10 \, \text{N}\) and \(6 \, \text{N}\), inclined at right angles to each other. Find the third force acting on the particle.
▶️ Answer/Explanation
Step 1: For equilibrium, the resultant of the first two forces must be equal and opposite to the third force.
Step 2: Resultant of two perpendicular forces:
\(R = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.66 \, \text{N}\)
Step 3: Hence, the third force must have magnitude:
\(\boxed{11.66 \, \text{N}}\)
and must act in the opposite direction to the resultant of the two forces.
Contact Forces: Normal and Frictional Components
Contact Forces: Normal and Frictional Components
When two solid surfaces are in contact, the total contact force can be resolved into two perpendicular components:
Normal Force (\(R\)) Acts perpendicular to the surface.
- It prevents the objects from penetrating each other.
- For a horizontal surface, \(R = mg\) if there are no other vertical forces.
Frictional Force (\(F\)) Acts parallel to the surface, opposing relative motion or the tendency to slide.
- At limiting equilibrium, \(F = \mu R\), where \(\mu\) is the coefficient of friction.
- Before slipping, \(F \leq \mu R\).
Total Contact Force:
\(\vec{C} = \vec{R} + \vec{F}\)
Thus, the contact force is the vector sum of the normal and frictional components. In problems, it is often easier to work with these components rather than the resultant directly.
Example:
A block of mass \(5 \, \text{kg}\) rests on a rough horizontal surface. The coefficient of friction between the block and the surface is \(\mu = 0.3\). A horizontal force of \(10 \, \text{N}\) is applied to the block. Determine whether the block moves, and find the contact force acting between the block and the surface.
▶️ Answer/Explanation
Step 1: Weight of block: \(W = mg = 5 \times 9.8 = 49 \, \text{N}\).
Step 2: Normal force: \(R = W = 49 \, \text{N}\) (since the surface is horizontal).
Step 3: Maximum possible friction: \(F_{\text{max}} = \mu R = 0.3 \times 49 = 14.7 \, \text{N}\).
Step 4: Since the applied force is \(10 \, \text{N} < F_{\text{max}} = 14.7 \, \text{N}\), the block does not move. Friction adjusts to balance the applied force, so \(F = 10 \, \text{N}\).
Step 5: The contact force is:
\(\vec{C} = \sqrt{R^2 + F^2} = \sqrt{49^2 + 10^2} = \sqrt{2401 + 100} = \sqrt{2501} \approx 50.0 \, \text{N}\)
Final Answer: The block does not move, and the contact force has magnitude \(\boxed{50.0 \, \text{N}}\).
Model of a Smooth Contact
In mechanics, surfaces are often assumed to be smooth unless stated otherwise. This is a modelling assumption that simplifies problems.
- Definition: A smooth surface is one in which no frictional force acts between the two surfaces in contact.
- Implication: The only force that the surface exerts is the normal reaction force (\(R\)), which acts perpendicular to the surface at the point of contact.
- Simplification: This assumption reduces problems to considering only the normal force, making equilibrium and motion analysis easier.
Limitations of the Smooth Contact Model
- In reality, most surfaces have some degree of friction, so assuming a smooth contact may not accurately represent physical situations.
- The model cannot explain situations where an object actually comes to rest or resists motion due to friction.
- Predictions using this model may be unrealistic if friction plays a significant role (e.g., cars braking, blocks on inclined planes).
- It is best used for theoretical analysis or where friction is negligible compared to other forces.
Example:
A particle of mass \(2 \, \text{kg}\) rests on a smooth inclined plane that makes an angle of \(30^\circ\) with the horizontal. Find the normal reaction force exerted by the plane on the particle.
▶️ Answer/Explanation
Step 1: Weight of the particle: \(W = mg = 2 \times 9.8 = 19.6 \, \text{N}\).
Step 2: On a smooth plane, only the normal force acts from the surface (no friction).
Step 3: The normal reaction balances the perpendicular component of the weight: \(R = W \cos 30^\circ = 19.6 \times \dfrac{\sqrt{3}}{2} = 16.97 \, \text{N}\).
Final Answer: The normal reaction force is \(\boxed{16.97 \, \text{N}}\).
Limiting Friction and Limiting Equilibrium
Limiting Friction and Limiting Equilibrium
- Frictional Force (\(F\)): A resistive force that opposes relative motion (or the tendency of motion) between two surfaces in contact.
- Limiting Friction: The maximum possible value of the frictional force before motion begins. Beyond this value, the body cannot remain at rest.
- Limiting Equilibrium: A situation where a body is on the point of moving. At this stage, the frictional force has reached its maximum (limiting) value.
Coefficient of Friction (\(\mu\))
- Definition: A dimensionless constant that measures how “rough” a surface is.
- Formula: \(\mu = \dfrac{F}{R}\) when the body is at limiting equilibrium.
- Here, \(R\) is the normal reaction force and \(F\) is the frictional force.
Mathematical Relationships
- If the body is in motion or about to move (limiting case):
\(F = \mu R\)
- If the body is still at rest but friction is preventing motion (not yet at the limit):
\(F \leq \mu R\)
Example:
A block of weight \(50 \, \text{N}\) rests on a rough horizontal surface. The coefficient of friction is \(\mu = 0.4\). Find the maximum frictional force.
▶️ Answer/Explanation
Step 1: Normal reaction \(R = 50 \, \text{N}\) (since surface is horizontal).
Step 2: Maximum (limiting) frictional force: \(F = \mu R = 0.4 \times 50 = 20 \, \text{N}\).
Final Answer: The maximum frictional force is \(\boxed{20 \, \text{N}}\).
Example:
A horizontal force of \(12 \, \text{N}\) acts on the same block in Example 1. Will the block move?
▶️ Answer/Explanation
Step 1: Maximum frictional force available: \(F_{\text{max}} = 20 \, \text{N}\).
Step 2: Applied force is \(12 \, \text{N}\), which is less than \(F_{\text{max}}\).
Step 3: Hence, friction simply balances the applied force: \(F = 12 \, \text{N} \leq 20 \, \text{N}\).
Final Answer: The block will remain at rest (static equilibrium).
Example:
A block of weight \(100 \, \text{N}\) rests on a rough horizontal surface. A horizontal force of \(60 \, \text{N}\) is applied. The block is about to slip. Given that the normal reaction is \(100 \, \text{N}\), find the coefficient of friction between the block and the surface.
▶️ Answer/Explanation
Step 1: Since the block is about to slip, it is in limiting equilibrium.
Step 2: In limiting equilibrium, \(F = \mu R\).
Step 3: The frictional force equals the applied force at this instant: \(F = 60 \, \text{N}\).
Step 4: Normal reaction: \(R = 100 \, \text{N}\).
Step 5: Coefficient of friction: \(\mu = \dfrac{F}{R} = \dfrac{60}{100} = 0.6\).
Final Answer: The coefficient of friction is \(\boxed{0.6}\).
Newton’s Third Law
Newton’s Third Law of Motion
Newton’s Third Law states:
If body A exerts a force on body B, then body B exerts an equal and opposite force on body A.
- The forces are always equal in magnitude and opposite in direction.
- They act on different bodies, not on the same object.
- They form what is called an action-reaction pair.
- The forces are of the same type (e.g., both contact forces, or both gravitational).
Key Example of Contact Force:
If a particle of weight \(W\) rests on the ground:
- The particle exerts a downward force \(W\) on the ground (action).
- The ground exerts an equal upward normal reaction \(R = W\) on the particle (reaction).
Example:
A box of weight \(200 \, \text{N}\) rests on the floor. State the action-reaction pair of forces according to Newton’s third law.
▶️ Answer/Explanation
Step 1: The box pushes down on the floor with a force equal to its weight.
Step 2: By Newton’s third law, the floor pushes up on the box with an equal and opposite force.
Step 3: Identify the action–reaction pair:
- Action: The box exerts a downward force of \(200 \, \text{N}\) on the floor.
- Reaction: The floor exerts an upward normal force of \(200 \, \text{N}\) on the box.
Final Answer: The action–reaction pair is \(\boxed{\text{Box on floor (200 N down), Floor on box (200 N up)}}\).