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Apparent Weight

Key idea: Use Newton’s 2nd law to relate apparent weight (normal reaction) to acceleration in lifts and free fall (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain inertia and momentum, then apply Newton's laws using free-body diagrams.
  • Describe normal, frictional, buoyant and viscous forces qualitatively.
  • Apply moments, couples and force-and-torque equilibrium using free-body diagrams and vector triangles.

1. Definitions (Must Know)

A. Apparent weight, R

Apparent weight is the normal reaction force R measured by a scale (the force the scale exerts on you).

True weight is W = mg (gravitational force).

B. Weightlessness

Weightlessness means apparent weight is zero:

R = 0

This happens in free fall when the only force is weight.

2. Key Ideas (What Earns Marks)

  • Pick a sign convention (commonly upwards +).
  • For a person in a lift:
    • forces: R up, mg down
    • apply Newton’s 2nd law: ∑ F_y = ma
  • If lift accelerates upward: R > mg (you feel heavier).
  • If lift accelerates downward: R < mg (you feel lighter).
  • If the lift is in free fall (a = g downward): R = 0 (weightlessness).
Exam pitfall: acceleration sign in lift questions

Keep one sign convention from start to end. If up is positive, downward acceleration must be entered as a negative value in R - mg = ma.

3. Detailed Explanations

A. Core equation (upwards positive)

Take upwards as positive.

Forces on the person:

  • R (normal reaction) upwards
  • mg downwards

Newton’s 2nd law:

∑ F_y = ma ⇒ R - mg = ma

So:

R = m(g + a)

This single equation covers all lift cases as long as you use the correct sign for a.

B. Interpreting the cases

  • Lift accelerating upward: a > 0 ⇒ R > mg
  • Lift accelerating downward: a < 0 ⇒ R < mg
  • Lift moving at constant speed: a = 0 ⇒ R = mg
  • Lift in free fall: a = -g ⇒ R = 0

C. Visual: how R changes with acceleration

For a given person, R = m(g + a) is a straight-line relationship between apparent weight and lift acceleration.

Apparent weight vs lift acceleration

As acceleration increases upward, the scale reading increases. At free fall (a = -g), the scale reads zero.

Scroll across the graph to read all labels.

As acceleration increases upward, the scale reading increases. At free fall (a = -g), the scale reads zero.As acceleration increases upward, the scale reading increases. At free fall (a = -g), the scale reads zero.
Example for m = 60 kg and g = 9.81 m s⁻²: free fall (a = -g) gives R = 0, constant speed (a = 0) gives R = mg, upward acceleration increases R, and downward acceleration decreases R.
Open full-size graph
View figure data
Values and uncertainty for Apparent weight vs lift acceleration
SeriesLift acceleration, a (m s⁻²)Lift acceleration, a uncertaintyApparent weight, R (N)Apparent weight, R uncertainty
R = m(g + a) for m = 60 kg-9.810
R = m(g + a) for m = 60 kg101188.6
Key cases-9.810
Key cases0588.6
Key cases1.5678.6
Key cases-2468.6

4. Common Mistakes

  • Mixing up weight mg with apparent weight R (the scale reads R, not mg).
  • Using R = mg± ma without stating a sign convention (easy to flip signs).
  • Saying “no gravity” for weightlessness (gravity still acts; R is what becomes zero).
  • Forgetting that “slowing down” means acceleration is opposite the velocity direction.

5. Exam Tips

  • Write: “take upwards as positive” before your equations.
  • Always start from R - mg = ma (or the equivalent with your sign choice).
  • If the lift is slowing while moving upward, acceleration is downward, so a < 0.
  • Sanity check: if you are “heavier”, your scale reading must be larger.

6. Worked Examples

Modelled example 1

Lift accelerating upwards

Core

Problem

A 60 kg person stands on a scale in a lift accelerating upward at 1.5 m s⁻². Take g = 9.81 m s⁻². Find the scale reading.
Study the worked solution

  1. Identify the scale reading

    Method

    The scale reads the upward normal reaction R, not the person’s weight mg.

    Reason

    A scale measures the contact force it exerts on the person.

    Working

    R upward, mg downward

  2. Apply Newton's second law

    Method

    R = m(g + a).

    Reason

    Taking upward as positive, the resultant upward force produces the stated positive acceleration.

    Working

    R-mg = ma ⇒ R = m(g + a)

  3. Calculate the reading

    Method

    R ≈ 6.79 × 10² N.

    Reason

    The upward acceleration requires the normal reaction to exceed the person’s weight.

    Working

    R = 60(9.81 + 1.5) = 678.6 N ≈ 679 N

Guided practice 2

Lift accelerating downwards

About 4 min

Problem

The same 60 kg person is in a lift accelerating downward at 2.0 m s⁻². Find the scale reading.

Try this before viewing the solution

Unit: N

Hints

Hint 1: give acceleration its sign
Keep upward positive, so the downward acceleration is a = -2.0 m s⁻².
View solution step by step

  1. Use the signed acceleration

    Method

    a = -2.0 m s⁻².

    Reason

    The chosen positive direction remains upward throughout the calculation.

    Working

    a = -2.0 m s⁻²

  2. Calculate the normal reaction

    Method

    R ≈ 4.69 × 10² N.

    Reason

    A downward resultant requires the upward scale force to be less than weight.

    Working

    R = m(g + a) = 60(9.81-2.0) = 468.6 N

Common misconception 3

Free fall (weightlessness)

Find and correct the mistake

Learner claim

A lift cable snaps and the lift enters free fall. A learner says the scale reads zero because gravity no longer acts. Decide what the scale reads and diagnose the explanation.

Try this before viewing the solution

Forces during free fall

View solution step by step

  1. Retain the gravitational force

    Method

    The person’s weight mg still acts downward.

    Reason

    Free fall is acceleration caused by gravity, not absence of gravity.

    Working

    ∑ F_y = -mg when upward is positive.

  2. Use free-fall acceleration

    Method

    The normal reaction is zero.

    Reason

    Both person and scale accelerate downward at g, so the scale does not need to push on the person.

    Working

    R-mg = m(-g) ⇒ R = 0

  3. Interpret the reading

    Method

    The scale reads 0 N.

    Reason

    Apparent weight is the normal contact force.

    Working

    R = 0 N while W = mg remains non-zero.

Examiner practice 4

Find the lift’s acceleration from a scale reading

4 marks

Examination question

A 60 kg student stands on a scale in a lift. The scale reads 450 N. Find the lift’s acceleration, giving its magnitude and direction. Take g = 9.81 m s⁻². [4 marks]

Try this before viewing the solution

Unit: m s^-2
Acceleration direction

View solution step by step

  1. Write the force equation

    1 mark

    Method

    R-mg = ma with upward positive.

    Reason

    The scale reaction is upward and weight is downward.

    Working

    R-mg = ma

  2. Substitute the reading

    1 mark

    Method

    450-588.6 = 60a.

    Reason

    The stated scale reading is R.

    Working

    450-60(9.81) = 60a

  3. Calculate signed acceleration

    1 mark

    Method

    a = -2.31 m s⁻².

    Reason

    The negative result is relative to the upward-positive convention.

    Working

    a = (-138.6)/60 = -2.31 m s⁻²

  4. State magnitude and direction

    1 mark

    Method

    The lift accelerates downward at 2.31 m s⁻².

    Reason

    A complete vector answer translates the sign into a direction.

    Working

    |a| = 2.31 m s⁻², downward.

Challenge 5

Moving downward but slowing down

Minimal support

Independent transfer

A 70 kg person is in a lift moving downward but slowing at 1.2 m s⁻². Find the scale reading. Take g = 9.81 m s⁻².

Try this before viewing the solution

Unit: N

Hints

Hint 1: separate velocity from acceleration
An object slowing down accelerates opposite to its direction of motion.
View solution step by step

  1. Infer the acceleration direction

    Method

    The acceleration is upward.

    Reason

    The downward velocity is decreasing in magnitude, so acceleration opposes it.

    Working

    a = +1.2 m s⁻² for upward positive.

  2. Calculate the scale reading

    Method

    R ≈ 771 N.

    Reason

    The upward acceleration requires R to exceed mg even though the lift is moving downward.

    Working

    R = m(g + a) = 70(9.81 + 1.2) = 770.7 N

7. Mind Stretchers

Mind stretcher 1: “Slowing down” sign trapExtension

A lift is moving upward but slowing down.

Is R greater than, equal to, or less than mg? Explain.

Show Answer

Moving upward but slowing down means acceleration is downward (a < 0 if upwards is positive).

So R = m(g + a) < mg. You feel lighter.

Mind stretcher 2: Optional (Enrichment)Extension

A. Impulse reminders (belongs to Momentum)

If you see force–time graphs and “average force”, that is impulse/momentum content:

  • impulse is area under F–t,
  • F_avg = Δ p/Δ t.

See: Momentum & Impulse.

Mind stretcher 3: Can the scale reading be “negative”?Extension

Some calculations can produce R < 0 (e.g. if you assume the lift accelerates downward faster than g). What does that mean physically?

Show Answer

The scale reading is the normal reaction R, and a contact force cannot pull you downward (it can only push up), so R cannot be negative.

If your equation gives R < 0, it means you would lose contact with the scale. In that case, the correct physical situation is R = 0 (weightlessness) and your acceleration would be g downward in free fall (until contact is regained).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027