Master vector resolution, calculate moments and couples, apply equilibrium conditions, and locate the centre of mass of objects.
AQA A-Level Physics · Unit 4: MechanicsPhysical quantities are classified as either scalars or vectors.
Vectors are added using the triangle rule (tip-to-tail) or the parallelogram rule. The resultant is the single vector that has the same effect as all the original vectors combined.
For two vectors at right angles, the magnitude of the resultant R is found using Pythagoras: R = √(A² + B²), and the direction is θ = tan⁻¹(B/A).
The opposite of vector addition is resolution: splitting a single vector into two perpendicular components. This is almost always done horizontally (x) and vertically (y):
where θ is the angle between the vector and the horizontal. Resolution into components makes all subsequent calculations far simpler, especially in mechanics problems.
A moment is the turning effect of a force about a pivot point. It depends on both the size of the force and its perpendicular distance from the pivot.
| Symbol | Quantity | Unit |
|---|---|---|
| M | Moment (torque) | N m |
| F | Force | N |
| d | Perpendicular distance from pivot to line of action of force | m |
The critical word is perpendicular: d must be the shortest distance from the pivot to the line of action of the force — not just the straight-line distance to where the force is applied. If the force is applied at an angle, you must find the perpendicular component or the perpendicular distance.
Moments are described as clockwise or anticlockwise. By convention, anticlockwise is usually taken as positive.
A couple is a pair of equal and opposite forces that are not acting through the same point. A couple produces a pure turning effect (torque) with no resultant force.
where F is the magnitude of one force and d is the perpendicular distance between the two forces (not the distance from one force to a pivot).
Examples of couples: turning a steering wheel, using a spanner with two hands, the forces on a compass needle in a magnetic field.
For an object to be in static equilibrium, two conditions must both be satisfied:
The Principle of Moments states: for a body in equilibrium, the sum of clockwise moments about any pivot equals the sum of anticlockwise moments about the same pivot.
The centre of mass (CoM) of an object is the single point at which the entire weight of the object appears to act. For a uniform object, the centre of mass is at its geometric centre (midpoint for a rod, centre of a circle, etc.).
For a composite object made of several uniform parts, the position of the overall CoM is found by taking moments of weight:
where x₁, x₂, … are the distances of each component's own CoM from a chosen reference point.
Experimental method to find the CoM of an irregular lamina: suspend the object freely from different points — the CoM lies along the vertical through the suspension point. The intersection of two such vertical lines gives the CoM location.
A body is stable if a vertical line through its CoM passes within its base of support. Wider bases and lower centres of mass give greater stability.
Q1. Which of the following is a vector quantity?
Q2. A force of 30 N acts at 60° to the horizontal. What is the vertical component of this force?
Q3. State both conditions for a rigid body to be in static equilibrium.
Q4. A spanner applies a force of 25 N at a perpendicular distance of 0.18 m from a nut. What is the moment applied?
Q5. A uniform plank of length 3.0 m and mass 4.0 kg sits on a pivot at its centre. A 6.0 kg mass is placed 1.0 m from the left end. How far from the right end must a 3.0 kg mass be placed to achieve balance?
Challenge Q1. A ladder of mass 20 kg and length 5.0 m leans against a smooth vertical wall at 70° to the horizontal. A person of mass 70 kg stands 2.0 m up the ladder. Calculate the reaction forces at the wall and ground, assuming the ground is rough.
Challenge Q2. A composite object consists of a solid uniform square plate (side 0.4 m, mass 3.0 kg) with a solid uniform circle (radius 0.1 m, mass 1.5 kg) attached at its top-right corner. Taking the bottom-left corner of the square as origin, find the x and y coordinates of the overall centre of mass.
Challenge Q3. Explain why a racing car has a very wide wheelbase and very low centre of mass. Use the concept of centre of mass and toppling to justify your answer.