1-4 Moment of a Force

Chapter 1 · Mechanics

❓ Lesson question

Push a door with the same force in two places — why does it rotate easily at the handle but barely move near the hinge?

You will learn to

1Calculate the moment of a force as the product of the force and its perpendicular distance from the point of rotation.
2Determine the perpendicular moment arm for a force acting at an angle.
3Assign a positive or negative sign to a moment for its rotation direction.

Before you start — this lesson changes the subject

Lessons 1-1 to 1-3 were one idea in three costumes: perpendicular motions do not interfere. That idea is finished. From here a force is no longer only something that pushes a body along — it is something that can turn it, and the turning depends on where you push as much as on how hard. Do not go looking for components of velocity here; there are none.

A door opens easily when pushed at the handle, but is much harder to open when pushed near the hinge. Likewise, a longer wrench loosens a tight wheel nut more easily than a shorter one.

In both cases the force is the same. The only thing that changed is the distance from the pivot.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) Does the magnitude of the force alone decide its turning effect?

(b) Does pushing the door close to the hinge make the door easier to open, because the force is closer to the pivot?

Write a line for each before this opens.

🔬 Explore — the ruler and the coins

What you’ll do: place identical coins at measured distances on both sides of a pivot and record the arrangements that balance the ruler.

⏱️ 12 min🧰 a straight ruler · a pencil for the pivot · a few identical coins · a flat table

1Rest the ruler across the pencil until it balances horizontally, with the pencil under the midpoint mark.
2Place one coin near the right end and watch the ruler tilt down on that side.
3Now add coins on the left side, close to the pivot, until the ruler balances again.
4Count how many coins near the pivot are needed to balance the single far coin, and measure both distances.
5Move the single coin closer to the pencil and repeat. Use the simulation below if you have no ruler to hand — it has the same two columns to compare.

🎛️ Simulation — turning effect

Coins on the left3

Left distance5 cm

Coins on the right1

Right distance15 cm

Left: coins × distance

3 × 5 = 15

Right: coins × distance

1 × 15 = 15

Result

Balanced

Try 1 coin at 15 cm against 3 coins at 5 cm. The side with fewer coins does not go down, because 1 × 15 equals 3 × 5. Now try to balance 8 coins at 2 cm against 1 coin — you will need it at 16 cm, and the ruler only reaches 20. Distance is doing as much work here as weight.

Analyse what you collected

(a) When the single coin was placed farther from the pivot, did you need more coins near the pivot to balance it, or fewer?

(b) Multiply the number of coins on each side by its distance from the pivot. How do the two products compare at balance?

✍️ Write your explanation in two lines before you read on.

Measurement note

Read each distance to the nearest scale division, and repeat the balancing twice — the coin count can vary by one near the balance point, because a ruler balances over a small range rather than at a single mathematical point.

In the Explore, a single coin far from the pivot was balanced by several coins close to it. At balance, the force multiplied by its distance matched on both sides — not the force alone. You may well have predicted the heavier group would tip the ruler down. It did not.

Moment of a force

A force does not just push a body — it can turn it about a point. The magnitude of that turning effect depends on two things together: the size of the force, and how far its line of action lies from the pivot. This turning effect is the moment of the force about the point:

M = F d

with $F$ in newtons and $d$ — the moment arm — in metres, giving the moment in newton metres.

Direction of rotation — the sign

A force can turn a body about a point in one of two senses. We separate them with a sign convention: counterclockwise is positive, clockwise is negative. So every moment carries both a magnitude and a sign.

When the force is inclined

Here is the part that costs marks. If a force $F$ acts at a point a distance $L$ along a bar from $O$ , but at an angle $θ$ to that bar, then $L$ is not the moment arm. The moment arm is the perpendicular distance from $O$ to the line of action:

d = L sin θ so M = F L sin θ

Switch the simulation to The moment arm and drag the angle. The purple segment is $d$ ; the bar is $L$ ; they are equal only at 90°.

Point

An ideal extended object that does not deform when a force is applied is a rigid body. The turning effect of a force about a point O is the product of the magnitude of the force and the perpendicular distance from O to the line of action. This is the moment of the force, measured in newton metres. That perpendicular distance is the moment arm.
With moment $M$ [N m], force $F$ [N] and moment arm $d$ [m]: $M = F d$ . For a force acting at an angle, $d = L sin θ$ , so $M = F L sin θ$ .
Counterclockwise is positive, clockwise is negative.
The moment of a force about a point on its own line of action is 0 N m, because there the perpendicular distance is zero.

📐 Equation box

Moment of a force

M = F d

Moment of an inclined force

M = F L sin θ

Sign convention

+ counterclockwise - clockwise

Key terms

Moment of a force: — The turning effect of a force about a point, measured in N m.
Moment arm: — The perpendicular distance from the point to the force’s line of action.
Line of action: — The infinite straight line along which the force acts, extended both ways.
Rigid body: — An ideal extended object that does not deform when a force is applied.

🛠️ Worked example

Situation

A force of 4.0 N acts at 60° to the line AB. Point A lies 4.0 m from where the force is applied and point B lies 3.0 m from it, on the other side. Determine the moment of the force about A and about B, taking counterclockwise as positive.

Step 1 of 6 — try it yourself before you open the next one.

The same 4.0 N force acts on the same body. Take a point C on the body, 2.0 m from the point where the force is applied, and it turns the body clockwise about C. Determine the moment of the force about C in N m.

O is at the left end of a horizontal bar. A horizontal arm 0.20 m long reaches out from O, and a force of 3.0 N acts perpendicular at the far end. Determine the moment about O.

An inclined bar has a point O at its lower end. A force of 4.0 N is applied at the upper end, 0.30 m from O along the bar, at an angle of 45° to the bar. Determine the moment about O.

An inclined bar has O at its left end. A force of 2.0 N is directed leftward at the far end, 0.40 m from O, at an angle of 30° to the bar. Determine the moment about O.

Point O is at the top left end of a bar. A force of 4.0 N acts 0.20 m from O at 60° to the bar, rotating the bar clockwise. Determine the moment about O.

In a New Context

In a car or aircraft workshop a wheel nut must be tightened to produce a specific turning effect. Too little and it can loosen in service; too much and the bolt may be damaged. So mechanics use a torque wrench — a tool that clicks audibly when the required moment is reached.

As the mechanic pulls, the turning effect depends on the force and the moment arm. A longer wrench gives a larger arm, so a smaller force produces the same moment. The click is triggered by the set moment — often called torque in engineering — not by the force alone.

➡️ Your turn: Explain why a longer torque wrench reaches the same preset moment with a smaller force, using force × perpendicular distance.

The answer, if you want to check yourself

The click is set by a fixed product

M = F d

. If

d

is doubled, then

F

must halve for the product to stay the same. A 0.20 m wrench needing 200 N becomes a 0.40 m wrench needing 100 N — the same 40 N m either way. This is exactly the coins on the ruler, with the balance replaced by a preset target.

🧪 Nature of Science — Archimedes and the lever — the early law of mechanics

More than two thousand years ago, Archimedes formulated the law of the lever: a small weight far from the pivot can balance a large weight close to it, when weight × distance is equal on both sides. He is said to have declared, “Give me a place to stand and I will move the Earth.”

It was one of the earliest precise, quantitative rules for the rotational effect of a force — and it is the same product, force × moment arm, that you used throughout this lesson. (Source: Archimedes, On the Equilibrium of Planes, c. 250 BCE.) Your ruler and coins reproduced a result that is older than almost everything else in this book.

Your turn: Suggest one everyday tool that uses the law of the lever — a large moment produced from a small force applied far from the pivot.

✅ Lesson question, answered

Push a door with the same force in two places — why does it rotate easily at the handle but barely move near the hinge?

Because the moment depends on both the force and its perpendicular distance from the hinge — not on the force alone. At the handle the arm is long, so the moment is large; near the hinge the arm is short, so the moment is small.

Assign a sign to that moment — counterclockwise positive, clockwise negative — and the turning effect is fully described for this planar convention.

The moment of a force is the product of the force and the perpendicular distance from the point to its line of action. Apply the force farther from the pivot and the same force produces a much larger moment.

A force F of magnitude 4.0 N is applied to a body. Determine the moment of the force about each of the points A (4.0 m), B (3.0 m) and C (2.0 m), taking counterclockwise as positive.

For each of the following, determine the moment of the force about point O:
(i) 15 N perpendicular, 0.20 m from O. (ii) 30 N at 45°, 0.40 m from O. (iii) 40 N at 30°, 0.60 m from O. (iv) 15 N acting horizontally along the line through O, 0.20 m away.

📝 Exam-style question

A force of 6.0 N acts at the end of a lever 0.50 m long, at 30° to the lever. Determine the moment of the force about the other end of the lever.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Part 1 — multiple choice (50% of the paper)

Watch for the option built from using $L$ instead of $L sin θ$ . It appears in almost every real paper.

A force of 10 N acts perpendicular to a bar, 0.50 m from the pivot. The moment about the pivot is:

A force of 10 N acts at 30° to a bar, at a point 0.80 m from the pivot. The moment is:

A force whose line of action passes through the pivot produces a moment of:

A 4.0 N weight sits 0.60 m from a pivot. To balance it with a weight placed 0.30 m from the pivot on the other side, that weight must be:

The correct unit for the moment of a force is:

A longer spanner turns a stiff nut more easily because it:

Part 2 — written answer (the other 50%)

Three steps every time: find the moment arm, multiply by the force, assign the sign. If the force is perpendicular, step one is free. Each question has its own notebook page, so your working survives to be compared with the model answer.

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

A force of 12 N acts perpendicular to a spanner, 0.25 m from the nut. Determine the moment about the nut.

1.2

A moment of 18 N m is required. The spanner is 0.30 m long and the force is applied perpendicular to it. What force is needed?

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

A force of 25 N acts at 40° to a bar, 0.60 m from the pivot, turning it clockwise. Determine the moment with its sign.

2.2

A door handle is 0.80 m from the hinge. A child pushes perpendicular at the handle with 20 N. An adult pushes perpendicular only 0.10 m from the hinge. What force must the adult use to produce the same moment?

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

A uniform beam is pivoted at its centre. A 4.0 N weight hangs 0.30 m to the left of the pivot.
(i) Where must a 6.0 N weight hang on the right to balance it? (ii) A 2.0 N weight is then added 0.45 m to the left. Where must the 6.0 N weight go now?

3.2

A mechanic must apply a moment of 45 N m to a wheel nut.
(i) What perpendicular force is needed with a 0.30 m wrench? (ii) If the mechanic can pull only 120 N, how long must the wrench be? (iii) With a 0.40 m wrench pulled at 70° to the handle instead of perpendicular, what force is needed?

Keep reading — this lesson marks itself done automatically.