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Newton's laws and the resultant force

Newton's laws say that a resultant force changes motion rather than sustaining it, and that only the resultant force enters F = ma. A free-body diagram drawn before any algebra is what makes that resultant possible to find. A third-law pair is one type of force shared between two bodies, which is why weight and the normal contact force are not a pair.

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The maths behind it: Forces and Newton's laws on InkMaths.

IN THIS TOPIC

  • State and apply all three laws of motion in appropriate situations.
  • Use ΣF = ma for constant mass, starting from a labelled free-body diagram.
  • Identify genuine third-law pairs, and explain why weight and the normal contact force are not one.

COMMON MISCONCEPTION

Weight and the normal force are an action-reaction pair.

Weight and the normal force act on the same body, so they are not a third-law pair. The partner of your weight is your gravitational pull on the Earth; third-law pairs act on different bodies.

All three laws, properly

First law. An object stays at rest, or keeps moving at constant velocity, unless a resultant force acts on it. The first law defines what a force does: forces change motion rather than sustain it, so no force is needed to keep an object moving at constant velocity.

Second law. The resultant force on an object equals the rate of change of its momentum, which for the constant-mass situations set at this level becomes

F=maF = maIN THE AQA DATA BOOKLET

with F meaning the resultant force, ΣF, and the acceleration always parallel to it.

Third law. If body A exerts a force on body B, then B exerts a force on A that is equal in magnitude, opposite in direction, of the same type, and acting on a different body.

The free-body diagram

Begin by drawing and labelling every force acting on the chosen object. If a force is omitted, the calculated resultant will be wrong. Isolate one object, draw each arrow from the object, and label it by type. Draw the diagram before starting the algebra.

A block on level ground with four labelled arrows: weight down and an equal normal contact force up, a long applied force to the right and a shorter friction force to the left, so the resultant points right.
FIG. 1A free-body diagram for a dragged block: weight, normal contact, applied force and friction, all drawn from the object. The imbalance of the horizontal pair is the resultant.

With the diagram drawn, resolve along and perpendicular to the motion, sum each direction, then feed the resultant into ΣF = ma. A resultant of zero returns you straight to the first law, meaning constant velocity, and standing still counts as constant velocity.

WORKED EXAMPLE

A person in an accelerating lift

A 70 kg person stands in a lift accelerating upward at 2.0 m s−2. Find the force the floor exerts on them.

Free-body diagram first. Two forces act on the person, weight mg down and the normal contact force N up. Nothing else.

Newton's second law, taking up as positive, gives N − mg = ma, so N = m(g + a) = 70 × (9.81 + 2.0) = 830 N.

Check the limiting cases. At a = 0 the floor pushes with exactly the weight, 687 N. Accelerating upward it has to push harder, and that extra is the heaviness you feel as a lift sets off.

GUIDED PRACTICE

The force between the blocks

A 30 N force pushes a 4.0 kg block, which in turn pushes a 2.0 kg block ahead of it on a frictionless surface. Find the acceleration, then the force between the blocks. Sketch the free-body diagrams first.

Show the working

Treat both blocks as one body first, giving a = 30/(4.0 + 2.0) = 5.0 m s−2.

Now isolate the front block on its own. The contact force is the only horizontal force acting on it, so F = 2.0 × 5.0 = 10 N. The choice of which object to isolate determines which forces appear in the equation.

INDEPENDENT PRACTICE

Car and trailer

A 1200 kg car tows a 400 kg trailer with a driving force of 3200 N; ignore resistance. Find the acceleration and the tension in the tow bar.

Show the working

Whole system first, so a = 3200/1600 = 2.0 m s−2.

Then the trailer alone, where the tow bar supplies the only horizontal force, so T = 400 × 2.0 = 800 N.

As a check, the car alone obeys 3200 − 800 = 1200 × 2.0. The same tension appears in both bodies' diagrams, pointing opposite ways: an action-reaction pair acting through the bar.

Weight and the normal contact force are not a third-law pair

The weight of a book and the normal contact force from the table beneath it are not an action-reaction pair. They act on the same object, the book, and they are different types of force, one gravitational and one contact. Their equality here is an accident of the book not accelerating. Put the book in an accelerating lift and the two stop matching, whereas a genuine third-law pair never does.

Two panels. In the first, the table pushes the book up while the book pushes the table down, equal and opposite on different bodies. In the second, the Earth pulls the book down while the book pulls the Earth up. Weight and normal contact never appear as a pair.
FIG. 2The genuine pairs. Table pushes book, book pushes table; Earth pulls book, book pulls Earth. Each pair is one type of force shared between two bodies.

The real pairs run like this. The Earth pulls the book down, so the book pulls the Earth up. The table pushes the book up, so the book pushes the table down. Same type, equal size, opposite direction, two different bodies, every single time.

ASSESSMENT FOCUS

  • Write ΣF, not F. The second law applies to the resultant force, so identify the resultant explicitly before substituting.
  • For third-law questions, name both bodies in both sentences: “the Earth pulls the book down; the book pulls the Earth up”. A pair described on a single object is not a third-law pair.
  • Draw the free-body diagram before any algebra and label each force by type: weight, normal contact, friction, tension.
  • Resolve along and perpendicular to the direction of motion; on a slope that means along the slope. The perpendicular direction usually gives you the normal force for free.
  • “Constant velocity” anywhere in the question means the resultant force is zero, so set ΣF = 0 rather than looking for an acceleration.

CHECK YOURSELF

A book rests on a table. A student claims the book's weight and the table's normal contact force on it are a Newton's third law pair. Give two reasons the claim is wrong, and state the two genuine pairs.

Show a hint

Check the third law's small print: same type of force, different bodies.

Show the answer

Both forces act on the same body, the book, and they are different types, gravitational and contact. Either count alone rules out a third-law pair; this claim fails on both.

The genuine pairs are gravitational and contact respectively. The Earth pulls the book down and the book pulls the Earth up. The table pushes the book up and the book pushes the table down.

The equality of weight and normal force here is equilibrium, not the third law, and it breaks the moment the book accelerates vertically.

Forces change motion, never sustain it.

Third-law pairs live on different bodies.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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  • State and apply all three laws of motion in appropriate situations.
  • Use ΣF = ma for constant mass, starting from a labelled free-body diagram.
  • Identify genuine third-law pairs, and explain why weight and the normal contact force are not one.

Open the full revision checklist to track your progress across the whole unit.