Free-Body Diagrams and Force Analysis

Overview

Many Dynamics questions are solved not by memorising formulas, but by correctly identifying forces and applying:

$$\sum \vec{F}=m\vec{a}$$

A free-body diagram (FBD) is one of the most powerful tools in H2 Physics. It helps convert a physical situation into clear mathematical equations.

This page focuses on:

• choosing the system
• identifying forces correctly
• drawing clean free-body diagrams
• resolving forces
• connected-body problems
• lifts and vertical motion
• inclined planes
• exam methods

Related hub: Dynamics

Why It Matters

Most mechanics errors begin before the algebra, when the wrong forces are identified or the wrong body is analysed.

Definition

A free-body diagram isolates one chosen object or system and shows all the external forces acting on it.

Key Representations

$$\sum \vec{F}=m\vec{a}$$ $$W=mg$$

What Is a Free-Body Diagram?

A free-body diagram isolates one chosen body or system and shows all external forces acting on it.

It does not show:

• surrounding objects
• forces the body exerts on others
• motion arrows unless useful
• resultant force as a separate force

Step 1: Choose the System

Before drawing anything, decide what object you are analysing.

Common choices:

• one block
• one particle
• one trolley
• one person
• two connected bodies as a system

Why It Matters

Different system choices give different equations.

Example:

Two blocks connected by string:

• choose one block → tension appears
• choose both blocks together → internal tension cancels

Step 2: Identify External Forces

Common Forces

Weight

$$\vec{W}=m\vec{g}$$

• acts vertically downward
• through centre of mass

---

Normal Contact Force

$$\vec{N}$$

• exerted by surface
• perpendicular to surface

Not always equal to weight.

---

Tension

$$\vec{T}$$

• acts along string/cable
• pulls away from object

For light inextensible string over smooth pulley, tension is same throughout.

---

Friction

$$\vec{f}$$

Acts parallel to contact surface and opposes relative motion or attempted motion.

---

Resistive Force / Drag

Acts opposite direction of motion through fluid or air.

---

Applied Force / Thrust

Pushing or pulling force supplied externally.

Step 3: Draw a Clean Diagram

Rules

• Use a simple box / dot / block representation
• Draw forces from object
• Label clearly
• Use correct directions
• Include only external forces

Example: Block on Rough Floor Pulled Right

Forces:

• weight downward
• normal upward
• pull right
• friction left

Common Mistakes

Wrong: Drawing Motion as Force

Velocity arrow is not a force.

Wrong: Drawing Resultant Force Separately

The resultant is the vector sum of real forces.

Wrong: Missing Weight

Almost every Earth-based mechanics question includes weight.

Wrong: Tension Pushing

Tension pulls, never pushes.

Applying Newton’s Second Law

After drawing FBD:

$$\sum \vec{F}=m\vec{a}$$

Usually apply separately in perpendicular directions.

Example:

Horizontal:

$$\sum F_x=m{a}_x$$

Vertical:

$$\sum F_y=m{a}_y$$

If no vertical acceleration:

$$\sum F_y=0$$

Force Resolution

When force acts at angle $\theta$, resolve into components.

For force $F$:

Horizontal:

$$F\cos \theta$$

Vertical:

$$F\sin \theta$$

Use Vectors carefully.

Example 1: Pulled Block on Horizontal Surface

A 5.0 kg block is pulled with force 20 N on smooth floor.

FBD

• $W=mg$ downward
• $N$ upward
• 20 N right

Vertical

$$N=mg$$

Horizontal

$$20=5a$$ $$a=4.0{\text{m s}}^{-2}$$

Example 2: Rough Surface

Same block, friction = 6 N.

Horizontal:

$$20-6=5a$$ $$a=2.8{\text{m s}}^{-2}$$

Connected Bodies

Key Ideas

If two bodies are connected by taut light string:

• same magnitude of acceleration
• tension usually same (ideal string/pulley)

Example 3: Two Blocks on Smooth Surface

Masses 2 kg and 3 kg pulled by 10 N.

Treat whole system:

$$a=\frac{10}{5}=2.0$$

Then analyse one block:

For 2 kg block:

$$T=2(2)=4\text{ N}$$

Hanging Mass / Atwood-Type Systems

For two hanging masses:

Choose direction of motion as positive.

For heavier mass $m_1$:

$$m_{1}g-T=m_{1}a$$

For lighter mass $m_2$:

$$T-m_{2}g=m_{2}a$$

Solve simultaneously.

Vertical Motion / Lifts

A person in lift experiences:

• weight $W$ downward
• normal reaction $R$ upward

Scale reading = $R$

Lift Accelerating Upward

$$R-W=ma$$

Thus:

$$R=m(g+a)$$

Heavier feeling.

Lift Accelerating Downward

$$W-R=ma$$

Thus:

$$R=m(g-a)$$

Lighter feeling.

Free Fall

If lift accelerates downward at $g$:

$$R=0$$

Apparent weightlessness.

Inclined Plane Analysis

A block on slope angle $\theta$.

Resolve Weight

Parallel to slope:

$$mg\sin \theta$$

Down slope.

Perpendicular:

$$mg\cos \theta$$

Into slope.

If Smooth Plane

Normal reaction:

$$N=mg\cos \theta$$

Acceleration down slope:

$$mg\sin \theta =ma$$ $$a=g\sin \theta$$

If Rough Plane

Include friction opposite motion.

If moving down slope:

$$mg\sin \theta -f=ma$$

Example 4: Inclined Plane

2.0 kg block on smooth 30° slope.

$$a=g\sin 30^{\circ }$$ $$a=9.81(0.5)=4.9{\text{m s}}^{-2}$$

Choosing Positive Direction

Choose direction that simplifies equations.

Usually:

• direction of acceleration
• direction of expected motion

If answer is negative, actual direction is opposite.

Multi-Step Strategy for Exams

For Any Dynamics Question

1. Identify object/system.
2. Draw FBD.
3. Choose axes.
4. Resolve angled forces.
5. Apply:

$$\sum \vec{F}=m\vec{a}$$

6. Solve algebra carefully.
7. Check units and sign.

Common Exam Pitfalls

1. Assuming $N=mg$ Always

False on slopes or accelerating systems.

2. Friction Direction Wrong

Friction opposes relative motion.

3. Missing Tension on Connected Bodies

Tension acts on each connected object.

4. Using Different Accelerations for Connected Bodies

If string taut and inextensible, magnitudes match.

5. Mixing Forces from Different Bodies

Each FBD must represent one chosen object.

6. Forgetting Weight Components on Slopes

Use:

$$mg\sin \theta ,mg\cos \theta$$

Summary Table

Situation	Key Equation
Horizontal motion	$\sum F=ma$
Vertical equilibrium	$\sum F=0$
Lift upward	$R-W=ma$
Lift downward	$W-R=ma$
Smooth slope	$mg\sin \theta =ma$
Normal on slope	$N=mg\cos \theta$

Formula Summary

$$\sum \vec{F}=m\vec{a}$$ $$W=mg$$ $$F_x=F\cos \theta$$ $$F_y=F\sin \theta$$ $$mg\sin \theta ,mg\cos \theta$$

Summary

Good force analysis comes from identifying the correct system, the correct external forces, and the correct directions before writing equations.

Links

Related Links

• Dynamics
• Momentum and Impulse
• Momentum Conservation and Collisions
• Forces
• Kinematics
• Vectors

Final Takeaway

Students often lose marks not because the physics is hard, but because the diagram is wrong. A correct free-body diagram usually makes the mathematics straightforward.

Draw first. Calculate second.