Ch 1 · 4.1 Forces and Equilibrium · A Level Mathematics 9709
Topic 1 of 5 in Mechanics — A Level Mathematics 9709 — 91 lessons.
Introduction
A picture hanging from two wires, a ladder against a wall, a parcel resting on a sloping conveyor: none of them move, so the forces on each must balance exactly. Mechanics models each object as a particle (all forces act at one point) and finds unknown forces by resolving. Throughout Paper 4 take g = 10 m s⁻². (Syllabus section 4.1, Paper 4.)
Key ideas — Forces you will meet
Always start with a clear force diagram: the particle, every force as an arrow, and the angles.
🧑🏫 Teacher's explanation
Imagine you are trying to push a heavy wooden crate across a carpeted floor. You pull on it with a certain force, but it doesn't move. Why? Because the floor is pushing back against the crate (Normal Reaction) and the carpet is rubbing against the bottom of the crate (Friction).
📖 Definitions
🔍 Worked example — Easy
A block of mass 5 kg rests on a smooth horizontal table. Calculate the magnitude of the normal reaction force acting on the block. (Use g = 9.8 m/s^2)
🔍 Worked example — Standard
A mass of 10 kg is pulled up a smooth plane inclined at an angle of 30^° to the horizontal by a force T acting parallel to the plane. Find the magnitude of T required to keep the mass in equilibrium.
🔍 Worked example — Exam-style
A crate of mass 20 kg rests on a rough horizontal floor. The coefficient of friction between the crate and the floor is μ = 0.3. Calculate the minimum horizontal force P required to cause the crate to move (i.e., to reach limiting equilibrium).
⚠️ Common errors
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🎯 Exam question
Key ideas — Components and resultants
A force F at angle θ to a direction has component Fcosθ along it and Fsinθ perpendicular to it. To find a resultant, add all the components in two perpendicular directions to get X and Y, then
🧑🏫 Teacher's explanation
Imagine you are playing tug-of-war with two friends. If you both pull in exactly the same direction with the same strength, the rope moves easily. But if you pull North and your friend pulls East, the rope moves diagonally. This 'diagonal' movement is the resultant force. In mechanics, we don't just care about how hard someone pulls, but in which direction that pull is actually helping or hindering motion.
📖 Definitions
🔍 Worked example — Easy
A force of 10 N acts at an angle of 30^° to the horizontal. Find the horizontal and vertical components.
🔍 Worked example — Standard
Two forces, 8 N acting due North and 6 N acting due East, act on a particle. Calculate the magnitude and direction of the resultant force.
🔍 Worked example — Exam-style
A particle is acted upon by three forces: 10 N due East, 15 N at 120^° from the East (counter-clockwise), and a third force F. If the particle is in equilibrium, find the magnitude and direction of F.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Equilibrium
A particle is in equilibrium when the resultant force is zero. Equivalently, the sum of the components in any direction is zero. Choose two convenient perpendicular directions (horizontal/vertical, or along/perpendicular to a slope), resolve, and solve the two equations. (A triangle of forces or Lami's theorem is acceptable but never required.)
🧑🏫 Teacher's explanation
Imagine you are playing tug-of-war. If both teams pull with exactly the same strength in opposite directions, the rope doesn't move. This state of 'no movement' or 'constant velocity' is what we call equilibrium. In physics, we don't just look at the forces; we look at the *resultant* force. If the resultant is zero, the object is in equilibrium.
📖 Definitions
🔍 Worked example — Easy
A 5 kg mass is suspended by a light, inextensible string. Calculate the tension T in the string. (Assume g = 9.8 m/s^2 and ignore air resistance).
🔍 Worked example — Standard
A block of mass 10 kg rests on a smooth inclined plane at an angle of 30^° to the horizontal. Find the component of the weight acting down the slope.
🔍 Worked example — Exam-style
A particle is held in equilibrium on a rough plane inclined at 25^° to the horizontal. The coefficient of friction between the particle and the plane is μ = 0.4. Determine if the particle is in limiting equilibrium (on the point of slipping) by comparing the required friction to the maximum possible friction.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Contact forces: normal and friction
The force a surface exerts can be split into a normal component R (perpendicular) and a frictional component F (along the surface). A smooth surface is a model in which F = 0; it is a simplification, since real surfaces always have some friction.
🧑🏫 Teacher's explanation
Imagine you are pushing a heavy wooden crate across a carpeted floor. If you push it perfectly horizontally, the floor doesn't push you 'up' into the floor; it pushes back against your hand to prevent you from sinking into the floor. This perpendicular push is the 'normal' force. At the same time, the carpet's fibers catch against the bottom of the crate, resisting your movement. This resistance is friction.
📖 Definitions
🔍 Worked example — Easy
A block of mass 5 kg rests on a smooth horizontal plane. Calculate the normal reaction force R. (Take g = 9.8 m/s^2)
🔍 Worked example — Standard
A crate of mass 10 kg is placed on a rough inclined plane making an angle of 30^° to the horizontal. The coefficient of friction between the crate and the plane is μ = 0.2. Calculate the normal reaction force R.
🔍 Worked example — Exam-style
A block of mass 4 kg is on a rough plane inclined at 25^° to the horizontal. The block is 'about to slip' downwards. Find the coefficient of friction μ between the block and the plane.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Friction and limiting equilibrium
Friction only grows as large as it needs to be, up to a maximum:
🧑🏫 Teacher's explanation
Imagine you are trying to push a heavy wooden crate across a carpeted floor. At first, you push with a little force, but the crate doesn't budge. Why? Because friction is 'smart'—it only pushes back with exactly as much force as you apply, just enough to keep the crate still. If you push harder, friction increases to match you.
📖 Definitions
🔍 Worked example — Easy
A block of mass 5 kg rests on a horizontal smooth surface. A horizontal force of 10 N is applied. If the coefficient of friction μ = 0.4, determine if the block moves.
🔍 Worked example — Standard
A 10 kg crate is on a rough plane inclined at 30^° to the horizontal. The crate is in limiting equilibrium. Find the coefficient of friction μ. (Take g = 9.8 m/s^2)
🔍 Worked example — Exam-style
A particle of mass m is held in equilibrium on a rough plane inclined at angle θ. The particle is on the point of slipping downwards. Show that μ = tan θ.
⚠️ Common errors
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🎯 Exam question
Key ideas — Newton's third law
If body A exerts a force on body B, then B exerts an equal and opposite force on A. The force of a particle on the ground equals in size, and is opposite to, the force of the ground on the particle.
🧑🏫 Teacher's explanation
Imagine you are standing on a skateboard and you push against a wall. You don't move forward because the wall pushes back on you with the exact same amount of force. This is Newton's Third Law in action. It is not just about 'pushing'; it is about the interaction between two objects.
📖 Definitions
🔍 Worked example — Easy
A book of mass 2 kg rests on a smooth horizontal table. State the magnitude and direction of the force exerted by the book on the table.
🔍 Worked example — Standard
A particle is pulled by a rope with a force of 10 N at an angle of 30^° to the horizontal. State the force exerted by the particle on the rope.
🔍 Worked example — Exam-style
A block is being pushed against a vertical wall by a horizontal force P. The wall exerts a normal reaction force R on the block. If P = 50 N, find the magnitude and direction of the force exerted by the block on the wall.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — On an inclined plane
For a particle of mass m on a plane inclined at α to the horizontal, resolve parallel and perpendicular to the plane. The weight has components mgsinα down the slope and mgcosα into the slope. With no other perpendicular forces, R = mgcosα.
🧑🏫 Teacher's explanation
Imagine a heavy wooden crate sitting on a ramp. Gravity pulls the crate straight down towards the center of the Earth, but the ramp prevents the crate from falling through it. Because the ramp is tilted, gravity doesn't just pull the crate 'down'; it pulls it 'down and into the ramp'.
📖 Definitions
🔍 Worked example — Easy
A block of mass 5 kg rests on a smooth plane inclined at 30^° to the horizontal. Calculate the magnitude of the normal reaction force.
🔍 Worked example — Standard
A mass m is on a rough plane inclined at α. The coefficient of friction is μ. Find the maximum force F acting down the plane such that the mass remains in limiting equilibrium.
🔍 Worked example — Exam-style
A particle of mass 2 kg is on a rough plane inclined at 25^°. The coefficient of friction μ is 0.3. Calculate the magnitude of the frictional force when the particle is in limiting equilibrium.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
(a) Calculate the magnitude of the normal reaction force. (b) Determine whether the block is in limiting equilibrium or if it will slip down the plane. [5 marks]
Worked examples — Example 1: two strings
A particle of weight 12 N hangs from two strings making angles of 30^° and 60^° with the horizontal. Find the tensions.
🔍 Worked example — Easy
A particle of weight 10 N is suspended by two strings. String A makes an angle of 45^° with the horizontal, and String B makes an angle of 45^° with the horizontal. Find the tension in each string.
🔍 Worked example — Standard
A particle of weight 12 N hangs from two strings making angles of 30^° and 60^° with the horizontal. Find the tensions T_1 and T_2 (where T_1 is at 30^° and T_2 is at 60^°).
⚠️ Common errors
Worked examples — Example 2: a resultant
Forces of 6 N along the positive x-axis and 8 N at 120^° to it act on a particle. X = 6 + 8cos 120^° = 2, Y = 8sin 120^° = 6.93. Resultant = √(4 + 48) = 7.21 N at tan^-1(6.93/2) = 73.9^° to the x-axis.
🔍 Worked example — Easy
A force of 10 N acts due East and a force of 10 N acts due North. Find the magnitude and direction of the resultant force.
🔍 Worked example — Standard
Two forces act on a particle: 6 N along the positive x-axis and 8 N at 120^° to the x-axis. Find the resultant force.
🔍 Worked example — Exam-style
A particle is acted upon by three forces: 5 N at 0^°, 12 N at 90^°, and a third force F at 210^°. Find the magnitude of F required for the particle to be in equilibrium.
⚠️ Common errors
Worked examples — Example 3: about to slip on a slope
A 2 kg block rests on a rough plane inclined at 30^° and is about to slip down. Find μ.
🔍 Worked example — Easy
A 5 kg block rests on a rough plane inclined at 20^° to the horizontal. The block is about to slip. Calculate the coefficient of friction μ. (Take g = 9.8 m/s^2)
🔍 Worked example — Standard
A block of mass m is on a rough plane inclined at θ. If the coefficient of friction is μ, find the expression for the angle θ at which the block is about to slip.
⚠️ Common errors
Worked examples — Example 4: pulling at an angle
A 5 kg box on a rough horizontal floor (μ = 0.4) is pulled by a force P at 20^° above the horizontal. Find P when the box is about to move.
🔍 Worked example — Easy
A 10 kg block rests on a smooth horizontal plane. A force P is applied at an angle of 30^° to the horizontal. Find the vertical component of P if the block is in equilibrium.
🔍 Worked example — Standard
A 5 kg box on a rough horizontal floor (μ = 0.4) is pulled by a force P at 20^° above the horizontal. Find P when the box is about to move.
🔍 Worked example — Exam-style
A block of mass m is pulled by a force P at an angle θ above the horizontal on a rough surface with coefficient μ. Derive the expression for the minimum P required to cause motion.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) X = 5 - 4cos 45^° = 2.17, Y = 7 - 2.83 = 4.17; resultant 4.70 N at 62.5^° north of east (bearing 027.5^°). (2) P = 30sin 25^° = 12.7 N; R = 30cos 25^° = 27.2 N. (3) F = 15 N; μ R = 20 N > 15 N, so not limiting. (4) μ = tanα = 0.75. (5) R = 60cos 30^° = 51.96, μ R = 15.59; mgsin 30^° = 30. Least P = 30 - 15.59 = 14.4 N; greatest P = 30 + 15.59 = 45.6 N.