Dynamics and Free-Body Diagrams — OCR A Level Physics
Module 3 · Forces and motion

Dynamics and Free-Body Diagrams

Specification: OCR A H556  |  Sections: 3.2.1 Newton’s laws, weight, tension, normal contact force, friction, free-body diagrams  |  Focus: Newton’s three laws, drawing correct free-body diagrams, equilibrium, and F = ma with multiple forces

By the end of this topic you should be able to…
  • state and apply Newton’s three laws of motion
  • identify and name the common forces acting on an object
  • draw correct free-body diagrams with force arrows originating from the object
  • label each force with magnitude and unit
  • resolve forces on inclined planes using perpendicular components
  • apply F = ma in situations with multiple forces and interpret equilibrium as resultant force = 0

Big idea: forces explain why motion changes. A free-body diagram is your map of what is pushing or pulling on the object and which way.

Part 1

Newton’s three laws of motion

OCR expects you to know and apply all three laws, not just recite them.

First law

An object stays at rest or moves with constant velocity unless a resultant force acts on it.

Second law

Force is equal to the rate of change of momentum in the direction of the force.

Third law

Body A exerts a force on body B, body B will exert a force back on body A that is equal in magnitude, opposite in direction and of the same type.

OCR habit

When you apply the second law, always identify the resultant force first. Do not plug just one force into F = ma unless that force is already the resultant.

Part 2

Common forces and how to name them

ForceSymbolDirectionNotes
WeightW or mgVertically downwardsAlways acts through the centre of mass.
Normal contact forceR or NPerpendicular to the surfaceOnly exists when surfaces touch.
TensionTAlong the string or rope, away from the objectStrings pull; they never push.
FrictionF or fOpposes relative motion or attempted motionActs parallel to the surface.
Applied forceP or FSet by the situationLabel clearly with magnitude and unit.
Common misconception

Weight and normal contact force are not always equal. They are only equal when there is no vertical acceleration and no other vertical forces.

Part 3

Drawing free-body diagrams correctly

A free-body diagram shows all the forces acting on a single object and no others.

Rules for correct free-body diagrams

  • draw the object as a simple box or dot
  • draw each force arrow starting from the object and pointing away in the direction the force acts
  • label each arrow with a symbol and, if given, its magnitude and unit
  • show only forces acting on the object, not forces exerted by the object
  • keep arrow lengths roughly to scale if magnitudes are given

Object on a horizontal surface

Horizontal surface free-body diagram W = mg R
Weight (W = mg) acts downwards from the centre; normal reaction (R) acts upwards from the surface.

Pulled across a rough surface

Rough surface free-body diagram W R P = 12 N F = 4 N
Applied force P to the right; friction F opposes motion; weight W down; normal R up.
Exam tip

Always draw the force arrows starting on the object and pointing outwards in the direction the force acts. This avoids the common mistake of drawing forces that act on other objects.

Part 4

Resolving forces on inclined planes

On a slope, weight acts vertically downwards but it is often useful to resolve it into components parallel and perpendicular to the plane.

Weight components on a slope at angle θ to the horizontalcomponent down the slope = mg sin θ
component into the slope = mg cos θ

Box on a slope

Normal force Friction Weight Horizontal
Weight acts vertically down. Normal force is perpendicular to the slope surface. Friction acts parallel to the slope, opposing motion.
Common trap

Do not assume sin is always down the slope and cos is always into the slope. Draw the triangle and check which side is opposite the labelled angle.

Part 5

Equilibrium and the condition resultant = 0

An object is in equilibrium if the resultant force on it is zero. This can mean it is stationary or moving with constant velocity.

Equilibrium conditionΣFx = 0 and ΣFy = 0

Three-force equilibrium

Three coplanar forces in equilibrium can be drawn as a closed triangle.

Resolving

Choose perpendicular axes and resolve each force. Set the sum in each direction to zero.

Check

If the object is in equilibrium and you know all but one force, you can find the missing force.

Worked example
A 5.0 kg block rests on a horizontal surface. State the magnitude of the normal contact force.
Weight = mg = 5.0 × 9.81 = 49.05 N
No vertical acceleration, so resultant vertical force = 0
R = weight = 49 N
Part 6

Applying F = ma with multiple forces

Newton’s second law relates the resultant force to mass and acceleration. In realistic situations, several forces act at once.

Key relationshipΣF = ma

Steps:

  1. draw a free-body diagram
  2. resolve forces into perpendicular components if needed
  3. find the resultant force in each direction
  4. apply F = ma in the direction of acceleration
Worked example
A 3.0 kg block is pulled across a rough horizontal surface by a horizontal force of 15 N. Friction is 3.0 N. Find the acceleration.
Resultant force = 15 − 3.0 = 12 N
F = ma → 12 = 3.0 × a
a = 12 ÷ 3.0 = 4.0 m s−2
Worked example
A 2.0 kg box slides down a frictionless slope at 30° to the horizontal. Find its acceleration.
Component of weight down slope = mg sin θ
= 2.0 × 9.81 × sin 30° = 9.81 N
F = ma → 9.81 = 2.0 × a
a = 4.9 m s−2
Interactive

Force builder and free-body checker

Build a free-body diagram and check equilibrium or acceleration
Resultant = 9.0 N
Acceleration = 3.0 m s⁻²
Motion: accelerating to the right

This is a horizontal-surface model. Try making resultant force zero and verify that the acceleration becomes zero too.

Part 7

Common misconceptions and exam traps

“Weight and normal are always equal”

No. They are equal only when vertical acceleration is zero and no other vertical forces act.

“If the object moves, there must be a force in that direction”

No. Newton’s first law says motion can continue without a resultant force.

“Third-law pairs act on the same object”

No. Third-law pairs always act on different objects.

“Friction always equals the applied force”

Only up to the point of slipping. After that, kinetic friction is usually roughly constant.

High-value OCR habit

Always state the direction you are taking as positive when you write ΣF = ma. This avoids sign errors.

Worked Examples

Worked examples

Worked example 1
State Newton’s three laws of motion in your own words.
First law: an object stays at rest or at constant velocity unless a resultant force acts on it
Second law: Force is equal to the rate of change of momentum in the direction of the force
Third law: Body A exerts a force on body B, body B will exert a force back on body A that is equal in magnitude, opposite in direction and of the same type
Worked example 2
A 4.0 kg crate is pulled across a floor by a horizontal rope. The tension is 20 N and friction is 8.0 N. Find the acceleration.
Resultant force = 20 − 8.0 = 12 N
F = ma → 12 = 4.0 × a
a = 3.0 m s−2
Worked example 3
A 6.0 kg box rests on a slope at 25° to the horizontal. The surface is smooth. Find the acceleration down the slope.
Component of weight down slope = mg sin θ
= 6.0 × 9.81 × sin 25° = 24.9 N
F = ma → 24.9 = 6.0 × a
a = 4.15 m s−2
Worked example 4
For the box in example 3, find the normal contact force.
Perpendicular component of weight = mg cos θ
= 6.0 × 9.81 × cos 25° = 53.3 N
No acceleration perpendicular to the plane, so R = 53 N
Worked example 5
A 2.0 kg object is suspended from a string and pulled sideways by a horizontal force of 8.0 N. Find the tension in the string and the angle it makes with the vertical.
Vertical equilibrium: T cos θ = mg = 19.6 N
Horizontal equilibrium: T sin θ = 8.0 N
tan θ = 8.0 ÷ 19.6 → θ = 22.2°
T = 19.6 ÷ cos 22.2° = 21.2 N
Knowledge

Knowledge Check

1
State Newton’s first law.
2 marks
  • Object remains at rest or continues with constant velocity unless a resultant external force acts on it.
2
What does the length of a force arrow represent in a free-body diagram?
1 mark
  • The magnitude of the force.
3
Name the force that acts perpendicular to a surface.
1 mark
  • Normal contact force / reaction force.
4
State the equilibrium condition for forces.
1 mark
  • Resultant force = 0.
5
Write the component of weight down a slope at angle θ to the horizontal.
1 mark
  • mg sin θ.
Exam Practice

Exam-Style Questions

1
A 5.0 kg block is pulled across a horizontal surface by a horizontal force of 25 N. Friction is 10 N.

a) Draw a free-body diagram for the block. [3 marks]
b) Calculate the acceleration of the block. [2 marks]
generated exam-style
5 marks
  • a) Box with: weight downwards (labelled W or mg), normal upwards (R), applied force right (25 N), friction left (10 N). Arrows must start on the box.
  • b) Resultant = 25 − 10 = 15 N. F = ma → 15 = 5.0 × a → a = 3.0 m s−2.
2
A box of mass 4.0 kg rests on a smooth slope at 30° to the horizontal.

a) Draw a free-body diagram showing weight, normal reaction and the component of weight down the slope. [3 marks]
b) Calculate the acceleration down the slope. [2 marks]
c) Calculate the normal contact force. [2 marks]
generated exam-style
7 marks
  • a) Box on slope: weight vertically downwards (mg), normal perpendicular to slope (R), component down slope (mg sin 30°) or clearly resolved.
  • b) mg sin 30° = 4.0 × 9.81 × 0.5 = 19.62 N. a = 19.62 ÷ 4.0 = 4.9 m s−2.
  • c) mg cos 30° = 4.0 × 9.81 × 0.866 = 34.0 N.
3
Explain why the weight of an object and the normal contact force from a table are not a Newton’s third-law pair. generated exam-style
2 marks
  • Third-law pairs act on different objects.
  • Both weight and normal act on the same object, so they cannot be a third-law pair.
4
A 2.0 kg object hangs from two strings. One string is at 40° to the vertical and the other is horizontal. The object is in equilibrium.

a) Resolve forces horizontally and vertically. [2 marks]
b) Find the tension in each string. [3 marks]
generated exam-style
5 marks
  • a) Horizontal: T1 sin 40° = T2. Vertical: T1 cos 40° = mg.
  • b) mg = 19.6 N. T1 = 19.6 ÷ cos 40° = 25.6 N. T2 = 25.6 × sin 40° = 16.5 N.
Summary

Topic Summary

Dynamics and Free-Body Diagrams

Newton’s laws

First: no resultant force means constant velocity. Second: F = ma. Third: equal and opposite forces on different objects.

Free-body diagrams

Draw only forces on the object, with arrows starting on the object and pointing outwards.

Slopes

Resolve weight into mg sin θ down the slope and mg cos θ into the slope.

Equilibrium

Resultant force = 0 in each direction.

3.2.1(a) F = ma 3.2.1(e) free-body diagrams 3.2.3(f) triangle of forces 3.5.1(a) Newton’s laws