Field, contact and elastic forces
Key idea: A complete H2 Physics lesson on field and contact forces, Hooke’s law, moments, couples, centre of gravity and equilibrium.
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The core idea
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Big question: Which real interactions belong on a force diagram, and when does a spring obey Hooke's law?
Isolate one body and draw only forces acting on it. Field interactions give gravitational, electric or magnetic forces; contact can give normal force, friction, upthrust, drag or tension. Upthrust is the resultant force from fluid pressure, usually upward because pressure is greater lower down. For a spring, extension is loaded length minus original length and F = kx applies only while force remains proportional to extension.
Choose the body before naming forces
A force diagram answers one question: what forces act on this chosen body? Weight is exerted by a gravitational field on mass; electric force by an electric field on charge; magnetic force by a magnetic field on a current-carrying conductor. Contact forces arise only where the body touches a surface, spring or fluid.
Naming the agent prevents invented forces. A flying ball has weight and perhaps drag, but it does not retain a forward force from the hand after contact ends.
Check your understanding: Why is a normal force not automatically equal to weight?
Its size is set by the complete force balance perpendicular to the surface; other forces or acceleration can make it different from weight.
Separate motion from force direction
Drag and friction oppose relative motion between contacting bodies or fluid layers. A spring force instead opposes deformation: a stretched spring pulls and a compressed spring pushes towards its natural length. Neither rule says that force must oppose the object’s instantaneous velocity.
Check your understanding: A compressed spring launches a block forward. Can spring force and velocity point the same way?
Yes. While the spring expands, its restoring force can accelerate the already forward-moving block in the same direction.
Key ideas to keep
- Force arrows are not velocity arrows.
- Friction and viscous force oppose relative motion, while normal force is perpendicular to contact.
- Use extension, not total spring length, in Hooke's law.
See the reasoning
Worked example
Use Hooke’s law inside an equilibrium argument
Question: A 0.60 kg block rests on a rough horizontal table. A spring of force constant 80 N m⁻¹ attached on the right is stretched by 0.050 m. Find the friction force and describe all other forces.
Step 1: Calculate the spring interaction
Why: The extension is within the stated proportional regime, so Hooke’s law applies.
Working: Fspring = kx = 80(0.050) = 4.0 N to the right.
Step 2: Apply horizontal equilibrium
Why: The resting block has zero resultant horizontal force.
Working: Ffriction + 4.0 = 0, so friction is 4.0 N to the left.
Step 3: Complete the vertical model
Why: A free-body account must include every external force, even those not needed numerically.
Working: Weight = 0.60g downward and normal contact = 0.60g upward.
Answer: Static friction is 4.0 N left. Weight and normal contact are equal and opposite vertically; spring force and friction are equal and opposite horizontally.
Check: Both component resultants are zero, consistent with the block remaining at rest.
Another worked model
Model 1
A mass, a positive charge and a current-carrying wire are placed separately in gravitational, electric and magnetic fields. State each force direction and the information needed to determine it.
Check the worked solution
Weight acts along the gravitational field. Force on a positive charge acts along the electric field. Force on the wire is perpendicular to both current and magnetic field, with its direction found using Fleming's left-hand rule. A negative charge would reverse the electric-force direction.
Model 2
A spring of force constant 250 N m⁻¹ has original length 0.180 m and loaded length 0.212 m. Find the elastic force.
Check the worked solution
Extension x = 0.212 − 0.180 = 0.032 m. F = kx = 250 × 0.032 = 8.0 N.
Use a hint if needed
Practise with support
Question 1
A horizontal current-carrying wire lies in a vertical magnetic field. Explain how to determine the direction of the force without calculating its magnitude.
Hint: Identify the current and field directions, then use Fleming's left-hand rule.
Check your answer
Use the first finger for magnetic field and the second finger for conventional current; the thumb then gives the force. The force must be perpendicular to both current and field.
Question 2
A spring extends from 4.0 cm to 7.5 cm under 14 N. Find k.
Hint: Convert the extension—not the final length—to metres.
Check your answer
x = 3.5 cm = 0.035 m, so k = 14/0.035 = 400 N m⁻¹.
Now work without the hint
Practise independently
Question 1
Draw separate force diagrams for (i) a mass in a gravitational field, (ii) a negative charge in an electric field and (iii) a wire carrying current perpendicular to a magnetic field. State every direction.
Check your answer
Weight acts along the gravitational field. Force on the negative charge is opposite to the electric field. Force on the wire is perpendicular to current and magnetic field as given by Fleming's left-hand rule.
Question 2
Describe normal force, friction, upthrust and viscous force by their source and typical direction. Do not use coefficients.
Check your answer
Normal force is a contact force perpendicular to a surface; friction opposes relative sliding at contact; upthrust is the resultant fluid-pressure force, typically upward; viscous force opposes motion through a fluid.
Avoid these traps
Common mistakes
Common mistake
A field name is itself a complete force description.
What is wrong with this reasoning?
Show better thinking
Name the body and interaction: gravitational force on mass, electric force on charge, or magnetic force on a current. Direction follows the relevant field and sign or current rule.
Common mistake
Motion always has a force in the direction of motion.
What is wrong with this reasoning?
Show better thinking
Forces come from interactions, not from velocity. Friction or viscous force opposes relative motion; a normal force is perpendicular to contact; upthrust results from fluid pressure.
Common mistake
Hooke’s law means every force–extension graph stays linear.
What is wrong with this reasoning?
Show better thinking
F = kx applies only over the Hooke-law region. Confirm proportionality and use extension, not total spring length.
Write for the examiner
Exam guidance
Name the source of every force and state whether the Hooke-law region is justified.
Exam-style practice [6 marks]
For each situation, name the force and state its direction: (i) a positive charge in an eastward electric field; (ii) a current-carrying wire that experiences an upward magnetic force; (iii) a sphere falling through oil below terminal speed. For the sphere, explain how the forces change as speed rises.
Plan before you answer
- Identify the object responding to each field or contact interaction.
- Use the field definition or relative-motion rule for direction.
- For the sphere, connect rising speed to viscous force and resultant force.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The positive charge experiences electric force east, along the electric field. The wire experiences magnetic force upward. The falling sphere has weight downward and viscous drag upward. As its speed increases, drag increases while weight is nearly constant, so the downward resultant and acceleration decrease. At terminal speed the two forces become equal.
Come back in three days
Check what stayed with you
Recall question 1
What interaction produces a normal force?
Check the answer
Contact between a body and a surface.
Recall question 2
When is F = kx valid?
Check the answer
Within the region where force is proportional to extension or compression.
Recall question 3
Does friction always oppose velocity?
Check the answer
It opposes relative slipping or the tendency to slip at the contact, not necessarily the body’s overall velocity.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. For friction and viscous force, qualitative treatment is required; coefficients of friction and viscosity are excluded.
- GCE A-Level H2 PhysicsTopic 2(a) / Topic 2(b) / Topic 2(c) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027