How To Draw Free Body Diagrams In Singapore: An O-Level Physics Tutorial

If you’re taking O-Level Physics or Combined Science (Physics) in Singapore, you cannot run away from free body diagrams (FBDs).

They appear in forces questions, Newton’s laws, friction, moments, even dynamics questions in Paper 2 and Paper 3. And the annoying thing is: sometimes you understand the topic, but you still lose marks because your free body diagram isn’t clear or is missing a force.

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This guide is written for Singapore secondary students like you, following the MOE O-Level syllabus. I’ll walk you through:

• How to draw free body diagrams step by step
• What O-Level examiners actually look for
• Practice-style questions (including hard variants)
• Common mistakes Singapore students make

And along the way, I’ll show you how you can use Tutorly.sg, a 24/7 AI tutor built just for Singapore students, to practise and check your understanding. Tutorly.sg has already been used by thousands of students in Singapore and has even been mentioned on Channel NewsAsia (CNA), so you’re in good company.

You can try it here any time:

• Main AI tutor page: https://tutorly.sg/ai-tutor-singapore
• Web app with practice Q&A: https://tutorly.sg/app

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Step-by-step tutorial

Let’s start with the basics, then move into trickier O-Level-style situations.

1. What exactly is a free body diagram?

A free body diagram (FBD) is a simplified diagram that shows only one object and all the forces acting on that object, using arrows.

Key ideas:

• One object at a time (not the whole system)
• Forces are arrows starting on the object and pointing outwards
• You label each force clearly (e.g. weight, normal reaction, tension, friction, air resistance)
• You don’t draw the surroundings (table, rope, wall etc.) in detail – just the object and forces

In O-Level marking schemes, if your FBD is wrong, your whole forces calculation can fall apart. So let’s get this solid.

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2. The 5-step method for any FBD

You can use this same method for almost every O-Level Physics forces question.

Step 1: Isolate the object

Decide which object you’re drawing the FBD for.

Examples:

• A block on a table
• A lift passenger
• A car on a slope
• A trolley being pulled by a string

Don’t mix multiple objects in one FBD.

Step 2: Mark the object as a simple shape

Draw a simple box or dot to represent the object. Don’t waste time drawing a realistic car or block.

Example: just a rectangle labelled “block” or “car”.

Step 3: Identify all forces acting on it

Ask yourself these questions:

• Is there weight? (Almost always yes)
• Is there a surface in contact? (Then there’s a normal reaction perpendicular to the surface)
• Is there friction? (Opposes motion or intended motion, along the surface)
• Is there a tension in a string/rope/cable?
• Is there air resistance / drag?
• Is there any applied force $e.g. a person pulling/pushing$?

For O-Level, the most common forces are:

• Weight ($W = mg$)
• Normal reaction ($R$ or $N$)
• Friction ($f$)
• Tension ($T$)
• Resultant force ($F_\text{net}$, sometimes drawn separately in explanations)

Step 4: Draw arrows from the object, in the correct directions

Rules:

• Arrows start at the object (or its centre) and point outwards
• Arrow direction shows direction of the force
• Arrow length is sometimes drawn to indicate relative size, but at O-Level, direction and correct type of force are more important

Examples:

• Weight: always vertically downwards, towards the centre of the Earth
• Normal reaction: always perpendicular to the surface, away from the surface
• Friction: along the surface, opposing motion or intended motion
• Tension: along the string, away from the object

Step 5: Label every force clearly

Write labels like:

• $W$ or $mg$ for weight
• $R$ or $N$ for normal reaction
• $T$ for tension
• $f$ for friction
• $F$ for applied force

In O-Level marking schemes, unlabeled arrows can cost marks.

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3. Basic examples (O-Level style)

Let’s go through some classic Singapore exam-style situations.

Example 1: Block resting on a horizontal table

Scenario: A 2 kg block is resting on a horizontal table at rest.

Forces on the block:

• Weight $W = mg$ acting vertically down
• Normal reaction $R$ from the table acting vertically up

So your FBD:

• Draw a box
• Arrow down labelled $W$ or $2 g$ N
• Arrow up labelled $R$

Since the block is at rest and on a horizontal surface:
$R = W = mg$

Example 2: Block being pulled horizontally

Scenario: A block on a horizontal surface is pulled to the right by a horizontal force $F$. There is friction.

Forces:

• Weight $W$ (down)
• Normal reaction $R$ (up)
• Pulling force $F$ (to the right)
• Friction $f$ (to the left, opposing motion to the right)

FBD: one box, four arrows, all labelled.

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4. Slopes and inclined planes (very common in O-Levels)

This is where many students in Singapore start losing marks, especially when components of weight are involved.

Example 3: Block on a rough slope

Scenario: A block is on a rough slope inclined at angle $\theta$ to the horizontal. It is at rest.

Forces on the block:

• Weight $W = mg$ vertically downwards
• Normal reaction $R$ perpendicular to the slope
• Friction $f$ along the slope, opposing motion (or possible motion)

If the block is at rest and tends to slide down the slope, friction acts up the slope.

On your FBD:

• Draw the block on a slanted line (the slope)
• Arrow down: $W$
• Arrow perpendicular to slope: $R$
• Arrow along slope, up: $f$

Important: For O-Level, you may need to resolve weight into components:

• Perpendicular to slope: $mg \cos\theta$
• Parallel to slope (down the slope): $mg \sin\theta$

You can either:

• Draw just $W$ and resolve in your workings, or
• Draw the components $mg\cos\theta$ and $mg\sin\theta$ on a separate diagram

Just don’t mix both in one FBD unless the question explicitly guides you.

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5. Tension and connected objects

Example 4: Two blocks connected by a string on a horizontal surface

Scenario: Two blocks A and B are on a smooth horizontal surface connected by a light string. A is pulled to the right by a force $F$.

If you’re drawing the FBD for block B:

Forces on B:

• Weight $W_B$ (down)
• Normal reaction $R_B$ (up)
• Tension $T$ from the string (to the right or left, depending which side A is on)

You do not draw the pulling force $F$ on B if $F$ is applied only to A. That’s a common mistake.

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6. Vertical motion: lifts and passengers

Very Singapore-flavoured question: lifts in HDBs or malls.

Example 5: Passenger in a lift

Scenario: A 60 kg student stands in a lift. The lift accelerates upwards.

Forces on the student:

• Weight $W = mg$ (down)
• Normal reaction $R$ from the floor (up)

If the lift accelerates upwards, then $R > W$.
If the lift accelerates downwards, then $R < W$.
If the lift moves at constant speed, $R = W$.

Your FBD is just two vertical forces, opposite directions, labelled clearly.

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Exam strategy guide

Now that you know how to draw FBDs, let’s talk about how to use them strategically in O-Level exams.

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1. Read the command words carefully

Look out for phrases like:

• “Draw a labelled diagram showing the forces acting on the block.”
• “On the diagram, draw and label all the forces acting on the trolley.”
• “State and explain the forces acting on…”

When you see this, you must think: “Time to draw a free body diagram.”

2. Always start with the FBD before using $F = ma$

For any forces/dynamics question:

1. Draw FBD
2. Decide which direction is positive (usually direction of motion or acceleration)
3. Write $F_\text{net} = ma$ along that direction
4. Substitute values and solve

Example (slope question):

• Choose “down the slope” as positive
• FBD shows $mg \sin\theta$ down, friction $f$ up
• Then:
  $mg\sin\theta - f = ma$

Examiners want to see that you understand where the equation comes from, not just plug numbers randomly.

3. Use FBD to decide if forces are balanced or unbalanced

For questions involving:

• “at rest”
• “moving at constant speed”
• “terminal velocity”
• “equilibrium”

Your FBD should show that resultant force is zero.

Example: A lorry moving at constant speed on a level road.

Forces:

• Driving force (forward)
• Friction/air resistance (backward)
• Weight (down)
• Normal reaction (up)

Since speed is constant, resultant force is zero:

• Forward force = backward force
• Upward force = downward force

If you can see that from your FBD, the explanation becomes easy.

4. How many marks can FBDs carry?

In O-Level Physics, a forces question might be 4–8 marks. The FBD itself can be worth 1–2 marks, but it also affects method marks for $F = ma$.

If your FBD is wrong:

• You may choose wrong forces in your equation
• You may use wrong directions (signs)
• You may end up with a wrong final answer and lose both method and accuracy marks

So drawing the FBD properly is not “extra work”. It’s insurance for your marks.

5. Time management: how long to spend on the diagram?

For a typical 6-mark forces question:

• Spend about 30–60 seconds on the FBD
• Then move on quickly to equations and calculations

Don’t over-decorate the diagram. Simple and clear is enough.

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6. Using Tutorly.sg to practise exam-style FBD questions

You don’t improve at FBDs just by reading. You improve by doing many questions and seeing the patterns.

On Tutorly.sg (built specifically for MOE syllabus from Sec 1 to JC 2), you can:

• Ask O-Level Physics questions anytime $even at 1am before your test$
• Get Singapore-style explanations that match what your teacher expects
• See step-by-step reasoning from the FBD to the final answer
• Try variations of the same type of question to really drill the concept

Start from here:

• Overview & info: https://tutorly.sg/ai-tutor-singapore
• Jump straight into asking Physics questions: https://tutorly.sg/app

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Worksheet practice

Let’s run through some practice-style questions similar to what you might see in school tests or O-Levels. Try to sketch the FBDs and think about the forces before reading the explanations.

Practice 1: Basic horizontal forces

Question 1
A 5.0 kg box is pulled along a horizontal floor by a horizontal force of 20 N. The frictional force is 8.0 N.

1. Draw a free body diagram of the box.
2. Calculate the acceleration of the box.

FBD (description):

Forces on the box:

• Weight $W = mg$ down
• Normal reaction $R$ up
• Pulling force 20 N to the right
• Friction 8.0 N to the left

Working:

Take right as positive.

Net horizontal force:
$F_\text{net} = 20 - 8.0 = 12\ \text{N}$

Using $F = ma$:
$12 = 5.0 \times a \Rightarrow a = 2.4\ \text{m s}^{-2}$

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Practice 2: Block on a rough slope (classic O-Level type)

Question 2
A 3.0 kg block rests on a rough slope inclined at $30^\circ$ to the horizontal. The block is at rest.

1. Draw a free body diagram of the block.
2. Resolve the weight into components parallel and perpendicular to the slope.
3. If the frictional force is 5.0 N, find the normal reaction of the slope on the block.

FBD (description):

Forces on the block:

• Weight $W = mg$ vertically down
• Normal reaction $R$ perpendicular to the slope
• Friction $f$ acting up the slope (since the block would slide down without friction)

Resolving weight:

• Component perpendicular to slope:
  $W_\perp = mg\cos 30^\circ = 3.0 \times 9.8 \times \cos 30^\circ$
• Component parallel to slope (down the slope):
  $W_\parallel = mg\sin 30^\circ = 3.0 \times 9.8 \times \frac{1}{2} = 14.7\ \text{N}$

Equilibrium conditions (block at rest):

Perpendicular to slope:
$R = W_\perp = 3.0 \times 9.8 \times \cos 30^\circ \approx 25.5\ \text{N}$

Parallel to slope:
$W_\parallel = f = 14.7\ \text{N}$

$Here, the 5.0 N friction given in the question could be part of a variant where the block is just starting to move, or where you compare maximum static friction vs required friction. Your teacher might tweak the numbers.$

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Practice 3: Lift question (vertical forces)

Question 3
A 70 kg student stands in a lift. The lift accelerates downwards at $2.0\ \text{m s}^{-2}$.

1. Draw a free body diagram of the student.
2. Calculate the normal reaction force of the lift floor on the student.

FBD (description):

Forces on the student:

• Weight $W = mg = 70 \times 9.8 = 686\ \text{N}$ (down)
• Normal reaction $R$ from the floor (up)

Take upward as positive. Acceleration is downward, so $a = -2.0\ \text{m s}^{-2}$.

Using $F_\text{net} = ma$:

Upward forces – downward forces = $ma$

$R - W = ma$
$R - 686 = 70 \times (-2.0)$
$R - 686 = -140$
$R = 546\ \text{N}$

So the normal reaction is less than the weight, which makes sense since the lift is accelerating downwards (you feel lighter).

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Practice 4: Two-block system with tension (moderate-hard)

Question 4
Two blocks A $2.0 kg$ and B $3.0 kg$ are connected by a light string and pulled along a smooth horizontal surface by a horizontal force of 15 N applied to block A.

1. Draw a free body diagram for block B only.
2. Find the tension in the string.

FBD for block B (description):

Forces on B:

• Weight $W_B = 3.0 \times 9.8$ (down)
• Normal reaction $R_B$ (up)
• Tension $T$ from the string (to the right if A is on the left, or to the left if A is on the right – just be consistent)

No friction (surface is smooth).

Since both blocks move together, they have the same acceleration $a$.

Total mass: $2.0 + 3.0 = 5.0$ kg
Net horizontal force on system: 15 N

For the whole system:

$15 = 5.0 \times a \Rightarrow a = 3.0\ \text{m s}^{-2}$

Now focus on block B only:

$F_\text{net} = ma$

Only horizontal force on B is tension $T$:

$T = 3.0 \times 3.0 = 9.0\ \text{N}$

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Practice 5 (Hard variant): Block on slope with acceleration

This is closer to the level of harder school prelim questions.

Question 5
A 4.0 kg block slides down a rough slope inclined at $25^\circ$ to the horizontal. The frictional force acting on the block is 6.0 N.

1. Draw a free body diagram of the block.
2. Resolve the weight into components parallel and perpendicular to the slope.
3. Calculate the acceleration of the block down the slope.

FBD (description):

Forces on the block:

• Weight $W = mg$ vertically down
• Normal reaction $R$ perpendicular to slope
• Friction $f = 6.0$ N up the slope (opposes motion, which is down)

Resolving weight:

• Parallel to slope (down):
  $W_\parallel = mg\sin 25^\circ = 4.0 \times 9.8 \times \sin 25^\circ$
• Perpendicular to slope:
  $W_\perp = mg\cos 25^\circ$

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Numerical approximate:

$\sin 25^\circ \approx 0.4226$

So:
$W_\parallel \approx 4.0 \times 9.8 \times 0.4226 \approx 16.6\ \text{N}$

Net force down the slope:

Downwards along slope is positive.

$F_\text{net} = W_\parallel - f = 16.6 - 6.0 = 10.6\ \text{N}$

Using $F = ma$:

$10.6 = 4.0 \times a \Rightarrow a \approx 2.65\ \text{m s}^{-2}$

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Practice 6 (Hard variant): Lift with tension in cable

Question 6
A lift of mass 500 kg is suspended by a vertical cable. The lift is accelerating upwards at $1.5\ \text{m s}^{-2}$.

1. Draw a free body diagram of the lift.
2. Calculate the tension in the cable.

FBD (description):

Forces on the lift:

• Weight $W = mg = 500 \times 9.8 = 4900\ \text{N}$ (down)
• Tension $T$ in the cable (up)

Take upwards as positive.

Using $F_\text{net} = ma$:

Upward forces – downward forces = $ma$

$T - W = ma$
$T - 4900 = 500 \times 1.5$
$T - 4900 = 750$
$T = 5650\ \text{N}$

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If you want more practice like this, you can throw similar questions at Tutorly.sg and ask it to:

• Change the numbers
• Make it steeper / less steep
• Add friction / remove friction

Then you can compare your own working with the step-by-step explanation shown.

Try it here: https://tutorly.sg/app

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Common mistakes

These are mistakes I keep seeing from Singapore students in school tests, prelims, and O-Levels.

1. Mixing multiple objects in one FBD

Example: Drawing the car and the trailer in one diagram with all forces mixed together.

Fix:
Always ask, “Which object am I focusing on?” and draw only that object.

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2. Forgetting friction or drawing it in the wrong direction

Friction always opposes motion or intended motion.

Common mistakes:

• Drawing friction in the same direction as motion
• Forgetting friction entirely on “rough” surfaces

Fix:
Ask yourself: “If there was no friction, which way would it move?” Then put friction in the opposite direction.

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3. Wrong direction for normal reaction on a slope

Some students still draw the normal reaction vertically up even on a slope.

Fix:
Normal reaction is perpendicular to the surface. If the surface tilts, the normal tilts too.

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4. Confusing weight with normal reaction

Weight is always vertically down, regardless of slope. Normal reaction is perpendicular to surface.

If you draw weight perpendicular to the slope, you’ll mess up your components and equations.

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5. Drawing resultant force as an extra physical force

Resultant force is not a separate physical force. It’s the vector sum of all forces.

At O-Level, if the question asks for “forces acting on the object”, don’t add a random arrow labelled “resultant force” unless you’re specifically explaining the net effect. The main FBD should show actual forces only: weight, normal reaction, friction, tension, etc.

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6. Not labelling the forces

Unlabelled arrows are risky. Examiners don’t want to guess.

Fix:
Always label: $W$, $R$, $T$, $f$, $F$, etc. Even short labels are fine.

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7. Getting signs wrong in $F = ma$

Even if your diagram is correct, your equation can be wrong if you’re careless with directions.

Fix:

1. Choose a positive direction (e.g. “down the slope is positive”)
2. Write forces in that direction as positive, opposite direction as negative
3. Stick to that sign convention consistently

For example, if you choose “upwards positive” in a lift question, and acceleration is downwards, then $a$ should be negative.

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8. Overcomplicating the diagram

Some students draw:

• The whole building
• The entire slope with decorations
• Multiple copies of the same force

You’re not drawing art. You just need a clean, simple diagram that shows all forces and directions.

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9. Not practising enough variants

Your school might show

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