Moments — Physics (IAL) Revision & Questions

Summary — What This Chapter Covers

Moment of a force = force × perpendicular distance from the pivot. This is what makes things rotate rather than just slide.
If the force isn't perpendicular to the pivot arm, you need the perpendicular component of the distance (using cosθ) — not the raw distance.
Centre of gravity is the single point where you can imagine all of an object's weight acting.
An object's stability depends on the position of its centre of gravity relative to its base — wide base + low centre of gravity = stable.
The Principle of Moments: for a system in equilibrium, total clockwise moments about a point = total anticlockwise moments about that same point.

1. Calculating the Moment of a Force

Think about trying to loosen a stuck bolt with a spanner. If you push right next to the bolt, it barely budges. If you push at the far end of the spanner handle, it turns much more easily — even though you're using exactly the same amount of force. That "extra turning power" you get by pushing further from the pivot is exactly what a moment measures.

A moment is defined as the turning effect of a force. It happens whenever a force causes (or tries to cause) an object to rotate about some fixed point, called the pivot. The pivot could be a hinge, a nail, a see-saw's fulcrum, or any point you choose to take moments about.

The Basic Formula (Force Perpendicular to the Pivot Arm)

Formula Moment (N m) = Force (N) × perpendicular distance from the pivot (m)

In plain English: multiply how hard you push by how far away from the pivot you're pushing. Bigger force OR bigger distance both give you a bigger moment.

←────────── x ──────────→ │ ┌──┐ ────┴───────────────────────┤ │ △ (pivot) └┬─┘ │ ↓ F MOMENT = F × x (force is perpendicular to the arm)

The SI unit is the newton metre (N m). Sometimes you'll see newton centimetre (N cm) used instead — that's fine, just make sure the force and distance units match whatever the question gives you, and stay consistent throughout your working.

When the Force Isn't Perpendicular

Here's where a lot of students trip up. The formula above only works cleanly when the force is applied at right angles to the line joining the pivot to the point where the force acts. In real life — like turning a spanner at an angle — the force is often applied at some other angle θ to that line.

In that case, the distance x in the diagram is not the perpendicular distance to the force — it's just the distance from the pivot to where the force is applied. To find the true turning effect, you need to find the component of that distance which is perpendicular to the force.

╱│ ╱ │ F ╱ θ │ ╱______│ pivot x·cos(θ) ← this is the perpendicular component MOMENT = F × x·cos(θ)

General Formula (Non-Perpendicular Force) Moment = F × x cos(θ)

where x is the distance from the pivot to the point of application, and θ is the angle between the force and the arm.

💡 Why cosθ and not sinθ? It depends entirely on how θ is defined in the diagram. If θ is measured between the force and the arm (as shown above), you use cos(θ) to get the component of the arm that's perpendicular to the force. Always sketch the right-angle triangle yourself rather than memorising "always cos" — that's how marks get lost in exams.

Real-World Example: Why Door Handles Are Where They Are

Ever wonder why door handles are placed on the far side of the door from the hinges, rather than right next to them? It's moments in action. The hinge is the pivot. By placing the handle as far from the hinge as possible, you maximise the perpendicular distance x in the formula. For the same force from your hand, this gives you a much bigger moment — meaning the door swings open with much less effort. Try pushing a door open right next to its hinge and you'll feel exactly why this design choice matters.

Worked Example

A uniform metre rule is pivoted at the 50 cm mark. A 0.5 kg weight is suspended at the 80 cm mark, causing the rule to rotate about the pivot. Assuming the weight of the rule is negligible, what is the turning moment about the pivot?

Step 1 — State the formula

Moment = Force × perpendicular distance from the pivot

Step 2 — Identify the force

The only force here is the weight of the mass acting downwards.
Weight = mg = 0.5 × 9.81 = 4.905 N ≈ 5 N

Step 3 — Identify the perpendicular distance

Distance from the rule's pivot to where the mass hangs: 80 cm − 50 cm = 30 cm. Since the weight acts straight down and the rule is horizontal, this distance is already perpendicular to the force — no cosθ needed here.

Step 4 — Substitute into the formula

Moment = 5 N × 30 cm = 150 N cm

📌 Examiner Tip If forces aren't already drawn on the diagram, sketch them in yourself. This makes it much easier to spot which distances are actually perpendicular to which forces. Exam questions often deliberately include extra forces that provide no turning effect at all — just to test whether you can tell the difference. Don't just multiply everything in sight!

Practice Question 1

A force of 12 N is applied perpendicular to a spanner at a distance of 0.15 m from the centre of a bolt. Calculate the moment of the force about the bolt.

Practice Question 2

A force of 20 N is applied to a lever at a point 0.4 m from the pivot, but at an angle of 60° to the lever arm (not perpendicular to it). Calculate the moment of the force about the pivot.

2. Centre of Gravity

Every object is made up of countless tiny particles, each with its own tiny bit of weight pulling it downward. Keeping track of every single one of these would be a nightmare for calculations — so physicists use a shortcut: the centre of gravity (sometimes called the centre of mass).

Definition Centre of gravity = the single point through which all of an object's weight can be considered to act

Instead of tracking weight spread across the whole object, you can treat the entire weight as if it acts at just this one point — which massively simplifies moment calculations.

Where Is It?

For a uniform, regular solid (same material throughout, symmetrical shape), the centre of gravity sits right at its geometric centre.

For a person standing upright, it's roughly in the middle of the body, just behind the navel.
For a sphere, it's exactly at the centre.
For any symmetrical object with uniform density, it sits at the point of symmetry — where all the lines of symmetry cross.

Triangle Ellipse Trapezium Parallelogram /\ ______ _______ ______ / \ / \ / \ / \ / .x \ ( .x ) / .x \ / .x \ /______\ \______/ /___________\ /_________\ (x marks the centre of gravity — found where symmetry lines intersect)

Stability — Why Some Objects Tip Over and Others Don't

The position of the centre of gravity directly determines how stable an object is. The rule is simple but powerful:

Stability Rule An object is stable as long as its centre of gravity lies directly above its base

Tip an object and its centre of gravity moves. The moment it drifts past the edge of the base, gravity creates a moment that topples the object over rather than tipping it back.

STABLE (upright) UNSTABLE (tilted too far) ┌────┐ ╱▔▔▔▔╲ │ x │ ╱ x ╲ │ ↓ │ ╱ ↓ ╲ └────┘ ╱________________╲ ← base edge, x is ────────── ╱ now past this edge base → object will topple!

This is exactly why furniture designers, engineers, and even Formula 1 car designers care so much about base width and how low they can keep the centre of gravity. Compare a tall, narrow bookshelf to a wide, squat one — push both with the same force, and the narrow one topples far more easily.

NARROW BASE WIDE BASE HIGH CENTRE OF GRAVITY LOW CENTRE OF GRAVITY ╱▔▔╲ ← push ╱▔▔╲ ← push ╱ x ╲ ╱_____╲ ╱______╲ │ x │ (tips easily) └───────┘ (hard to tip)

A wider base → lower centre of gravity relative to the tipping edge → more stable.
A narrower base → higher, more precarious centre of gravity → more likely to topple.

🚗 Real-world connection Racing cars are built wide and low for exactly this reason — it keeps the centre of gravity well within the "footprint" of the wheels even through sharp, high-speed corners, drastically reducing the risk of rolling over.

Practice Question 3

Explain, using the idea of centre of gravity, why a double-decker bus is more likely to tip over on a sharp bend than a single-decker bus of the same width.

3. The Principle of Moments

Think of a see-saw perfectly balanced with a small child sitting far from the pivot and a heavier adult sitting close to it. Neither side is rotating — the see-saw is in equilibrium. The Principle of Moments explains exactly why this balance happens.

The Principle of Moments For a system in equilibrium, the sum of clockwise moments about a point = the sum of anticlockwise moments about that same point

Every "twist" trying to rotate the object one way must be perfectly cancelled out by "twists" trying to rotate it the other way, otherwise the object would start spinning.

←──────── d₃ ────────→ ↑ F₃ ←─── d₁ ───→←── d₂ ──→ │ ┌──┐ ┌──┴──┐ │ │ ┌──┐ │pole │ └┬─┘ │ │ └──┬──┘ │ └┬─┘ △ (pivot) │ ↓F₁ ↓F₂ │ (F₃ acts up, at the end of a pole) F₂ supplies a CLOCKWISE moment F₁ and F₃ supply ANTICLOCKWISE moments F₂ × d₂ = (F₁ × d₁) + (F₃ × d₃)

Notice something important: this equation only balances the magnitudes of the moments on each side — you separately group all clockwise moments on one side of the equals sign, and all anticlockwise moments on the other. You never mix them together with plus and minus signs in one long sum; instead you keep two clearly labelled totals and set them equal.

🎯 How to approach ANY moments question

Pick a pivot point (the question usually tells you, or gives an obvious one like a hinge).
Identify every force acting and work out its perpendicular distance from that pivot.
Decide which forces create clockwise moments and which create anticlockwise moments.
Set: total clockwise moments = total anticlockwise moments.
Solve for the unknown.

Worked Example — Multiple Choice

A uniform beam of weight 40 N is 5 m long and is supported by a pivot situated 2 m from one end. When a load of weight W is hung from that end, the beam is in equilibrium. What is the value of W?

Options: A. 10 N B. 50 N C. 25 N D. 30 N

Step 1 — State the principle

Clockwise moments = Anticlockwise moments

Step 2 — Calculate the clockwise moment

Since the beam is uniform, its own weight (40 N) acts at its centre of gravity — the exact midpoint of the beam, i.e. 5 ÷ 2 = 2.5 m from either end.

The pivot is 2 m from the end where W hangs, so the beam's centre of gravity is 2.5 m − 2 m = 0.5 m from the pivot (on the opposite side to W).

Clockwise moment = 40 N × 0.5 m = 20 N m

Step 3 — Calculate the anticlockwise moment

The load W acts 2 m from the pivot.
Anticlockwise moment = W × 2 m

Step 4 — Equate the moments

20 N m = W × 2 m

Step 5 — Solve for W

W = 20 ÷ 2 = 10 N → Answer A

📌 Examiner Tip Always double-check whether the beam or rule is described as "uniform." If it is, its weight acts at its exact geometric centre and you MUST include it as one of the moments — a hugely common mistake is forgetting the weight of the beam itself, treating it as if it were massless.

Practice Question 4

A uniform see-saw of length 4 m and weight 200 N is pivoted at its centre. A child of weight 300 N sits 1.2 m from the pivot on the left side. How far from the pivot on the right side must a second child of weight 250 N sit for the see-saw to balance?

Practice Question 5

A shop sign of weight 60 N hangs from the end of a uniform horizontal bracket of length 0.8 m and weight 15 N. The bracket is fixed to a wall at one end (this is the pivot) and supported by a wire attached to the same end as the sign, pulling vertically upward. What tension must the wire provide to keep the bracket in equilibrium?

What to Memorise

Term / Formula	Meaning
Moment	The turning effect of a force about a pivot.
Moment = F × x	Force perpendicular to the pivot arm: multiply force by perpendicular distance.
Moment = F × x cos(θ)	Force at an angle θ to the arm: use the perpendicular component of the distance.
Unit	Newton metre (N m), sometimes newton centimetre (N cm) — keep units consistent.
Centre of gravity	The single point where an object's entire weight can be considered to act.
Stability rule	An object stays upright as long as its centre of gravity remains above its base.
Principle of Moments	For equilibrium: sum of clockwise moments = sum of anticlockwise moments (about the same point).
Uniform object's weight	Acts at its exact geometric centre — never forget to include it if the beam isn't massless.

Concepts Checklist

I can state the definition of a moment in my own words.
I can calculate a moment when the force is perpendicular to the pivot arm (Moment = F × x).
I can calculate a moment when the force acts at an angle, using Moment = F × x cos(θ).
I understand why door handles are placed far from the hinge.
I can define centre of gravity and find it for symmetrical shapes.
I can explain why a wide base and low centre of gravity make an object more stable.
I can state the Principle of Moments accurately.
I remember to include the weight of a uniform beam/rule as a moment, acting at its centre.
I can identify which moments are clockwise and which are anticlockwise from a diagram.
I can solve for an unknown force or distance using the Principle of Moments.

Exam Tips & Common Mistakes

Forgetting the weight of a "uniform" beam

If a question says the rod/beam/rule is uniform, it has weight, and that weight acts at its exact centre. This is one of the most commonly missed moments in exam answers — always check for this word.

Using the wrong distance for angled forces

Don't just multiply force by the raw distance to the pivot if the force isn't perpendicular. Sketch the right-angle triangle and use x cos(θ) (or occasionally x sin(θ), depending on how the angle is defined) to find the true perpendicular distance.

Mixing up clockwise and anticlockwise

Before writing any equation, physically trace with your finger which way each force would rotate the object. Label each moment "CW" or "ACW" on your diagram before doing any calculation — it prevents sign errors.

Including forces that create no moment

Exam diagrams often show extra forces deliberately, such as forces acting exactly at the pivot (distance = 0, so no moment) or forces that are decoys with no real turning effect. Only include forces that genuinely contribute a moment about your chosen pivot.

Not keeping units consistent

If some distances are given in cm and others in m, convert everything to the same unit before substituting into the formula. Mixing units is a very easy way to lose marks on an otherwise correct method.

Choosing a "bad" pivot point

You can technically take moments about any point, but choosing the point where an unknown force acts is a smart strategy — that unknown force then has zero moment (distance = 0) and disappears from the equation entirely, leaving you with one less unknown to solve for.