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
- 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.
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.
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.
Common forces and how to name them
| Force | Symbol | Direction | Notes |
|---|---|---|---|
| Weight | W or mg | Vertically downwards | Always acts through the centre of mass. |
| Normal contact force | R or N | Perpendicular to the surface | Only exists when surfaces touch. |
| Tension | T | Along the string or rope, away from the object | Strings pull; they never push. |
| Friction | F or f | Opposes relative motion or attempted motion | Acts parallel to the surface. |
| Applied force | P or F | Set by the situation | Label clearly with magnitude and unit. |
Weight and normal contact force are not always equal. They are only equal when there is no vertical acceleration and no other vertical forces.
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
Pulled across a rough surface
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.
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.
component into the slope = mg cos θ
Box on a slope
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.
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.
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.
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.
Steps:
- draw a free-body diagram
- resolve forces into perpendicular components if needed
- find the resultant force in each direction
- apply F = ma in the direction of acceleration
Force builder and free-body checker
This is a horizontal-surface model. Try making resultant force zero and verify that the acceleration becomes zero too.
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.
Always state the direction you are taking as positive when you write ΣF = ma. This avoids sign errors.
Worked examples
Knowledge Check
- Object remains at rest or continues with constant velocity unless a resultant external force acts on it.
- The magnitude of the force.
- Normal contact force / reaction force.
- Resultant force = 0.
- mg sin θ.
Exam-Style Questions
a) Draw a free-body diagram for the block. [3 marks]
b) Calculate the acceleration of the block. [2 marks]
generated exam-style
- 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.
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
- 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.
- Third-law pairs act on different objects.
- Both weight and normal act on the same object, so they cannot be a third-law pair.
a) Resolve forces horizontally and vertically. [2 marks]
b) Find the tension in each string. [3 marks]
generated exam-style
- 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.
Topic Summary
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.