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Newton’s laws and resultant force

Key idea: Newton’s laws connect motion to the resultant external force. The key is to choose one body and keep force pairs on their correct bodies.

  • GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

  • State and apply all three Newton’s laws.
  • Use force as rate of change of momentum and F = ma for constant mass.
  • Identify third-law pairs without cancelling forces on different bodies.

Learn the idea

Big question: How do Newton’s three laws connect force diagrams to changes in motion without mixing forces on different bodies?

Start with the system and the resultant

Newton's first law says a body remains at rest or at constant velocity unless a resultant external force acts. Newton's second law says the rate of change of momentum is proportional to, and in the direction of, the resultant force; in SI units the proportionality becomes equality.

Newton’s first and second laws describe one chosen body or system. Add only forces acting on it, find their vector resultant, then connect that resultant to acceleration or rate of momentum change. Velocity may point elsewhere while the object slows or turns.

For constant mass, F = ma follows from F = dp/dt. This shortcut is powerful, but writing the resultant component equation first prevents a single force from being mistaken for the net force.

Check your understanding: A car moves east while its resultant force is west. What happens?

Its acceleration is west, so its eastward speed initially decreases; force need not point with velocity.

Third-law partners belong to different diagrams

A third-law pair comes from one interaction: A pushes B and B simultaneously pushes A with equal magnitude and opposite direction. Because the forces act on different bodies, they never cancel within the free-body diagram of either one.

Check your understanding: What is the partner of Earth’s gravitational force on a falling ball?

The gravitational force exerted by the ball on Earth, equal in magnitude and opposite in direction.

Resultant forces in one and two dimensionsThree labelled cases compare forces in the same direction, forces in opposite directions, and two perpendicular forces. Each case shows the direction and magnitude of the resultant.Resultant-force casesSame direction10 N20 Nresultant 30 NOpposite directions40 N20 Nresultant 20 NPerpendicular forces4 N east3 N northresultant 5 Nnorth-east
Scroll diagram horizontally to read all labels.
Add collinear forces with signs. For forces at an angle, preserve both magnitude and direction in a scale drawing.

Key ideas

  • Resultant force points with acceleration, not necessarily velocity.
  • Zero resultant force permits rest or constant-velocity motion.
  • A third-law pair cannot both appear on the free-body diagram of one object.

Relationships to know

  • F = Δp/Δt
  • for constant mass, F = ma

Follow the reasoning

Worked example

Apparent weight in an accelerating lift

Question: A 60 kg passenger accelerates upward at 2.0 m s⁻². Find the normal contact force from the lift and identify its third-law partner. Use g = 9.81 m s⁻².

  1. Step 1: Draw forces on the passenger

    Why: The equation must contain only forces acting on the chosen body.

    Working: Normal force N upward; weight mg downward.

  2. Step 2: Apply Newton’s second law

    Why: Upward acceleration means the upward force must exceed weight.

    Working: N − mg = ma, so N = m(g + a) = 60(9.81 + 2.0) = 708.6 N.

  3. Step 3: Name the third-law partner

    Why: The partner must be the same interaction acting on the other body.

    Working: The passenger exerts 708.6 N downward on the lift.

Answer: The lift exerts about 709 N upward on the passenger; the passenger exerts 709 N downward on the lift.

Check: N is greater than the 589 N weight, which is required for upward acceleration.

Now try it with support

Practise with support

A 1200 kg car slows from 20 m s⁻¹ to 8.0 m s⁻¹ in 4.0 s. Find the average resultant force, taking forward as positive.

Hints

  1. Find the signed acceleration or momentum change.
  2. A negative answer means the force is backward.
View the guided answer

a = (8.0 − 20)/4.0 = −3.0 m s⁻², so F = 1200(−3.0) = −3.6 × 10³ N.

Your turn

Practise independently

A crate is pulled right while accelerating left. Draw its free-body diagram and explain how this is possible without treating a third-law pair as cancelling forces.

Check your answer

This is possible whenever the horizontal resultant is left, for example if leftward kinetic friction is greater than a rightward pull. Acceleration follows the resultant force, while the crate can still have a rightward velocity as it slows. The reaction to friction acts on the floor, not on the crate.

Common mistakes and exam guidance

Watch out for

  • Saying a moving object must have a resultant force in the direction of motion.
  • Treating weight and normal contact on one body as a third-law pair.

In an exam

  • For a law question, state the law before applying it to the context.
  • Write F = ma only after identifying the resultant force and confirming constant mass.

Put the ideas together

Exam-style practice [7 marks]

A 20 kg crate is pulled right by 90 N while friction is 30 N left. (a) Find its acceleration. (b) The pull is removed while the crate is still moving right. Find its acceleration and explain the subsequent motion using Newton’s laws. (c) State the third-law partner of the friction on the crate.

Plan before you answer

  • Write a horizontal resultant equation for each stage.
  • Keep velocity and acceleration directions separate.
  • Name the other body in the third-law interaction.
View the marking points and model answer

Marking points

  1. First resultant = 60 N right.
  2. First acceleration = 3.0 m s⁻² right.
  3. After removal, resultant = 30 N left.
  4. Second acceleration = 1.5 m s⁻² left.
  5. States that the crate continues right initially but slows.
  6. Links the change to the leftward resultant/Newton’s second law.
  7. Identifies friction exerted by crate on floor to the right as the partner.

Model answer

Initially Fres = 90 − 30 = 60 N right, so a = 60/20 = 3.0 m s⁻² right. After the pull is removed, friction is the 30 N leftward resultant, so a = 30/20 = 1.5 m s⁻² left. The crate continues moving right at first but slows because acceleration is opposite velocity. The third-law partner is the friction force exerted by the crate on the floor to the right.

Finish from memory

Three-question recap

  1. State Newton’s first law in terms of resultant external force.

    Check

    With zero resultant external force, a body remains at rest or continues at constant velocity.

  2. When does F = ma follow directly from F = dp/dt?

    Check

    When mass is constant.

  3. Why do third-law forces not cancel on one free-body diagram?

    Check

    They act on different bodies.

Try this next

Practise a lift, a braking vehicle and a connected-body force diagram before combining Newton’s laws with energy.

Check the whole topicReturn to Motion and ForcesReview Mass, inertia and linear momentum

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H1 Physics
Edition
GCE A-Level H1 Physics 2027