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Using F = ma: slopes, lifts and connected bodies

Key idea: Choose one body, draw its forces and resolve them along the acceleration before using F = ma. A normal force or string tension need not equal the body's weight.

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  • GCE A-Level H2 Physics 2027
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Why does a bathroom scale read more when a lift starts to rise? How fast does a block slide down a slope? These problems look different, but each is solved the same way: pick one body, find the resultant force on it and apply F = ma. This lesson sets out that method and uses it for slopes, lifts and bodies joined by a string.

Predict the scale reading

Check your understanding 1

A passenger stands on a scale in a lift moving downward but slowing. Predict the scale reading relative to the passenger’s weight.

Scale reading
Show answer

The acceleration is upward. For the passenger, R-mg = ma with upward positive, so R > mg. The direction of motion does not set the reading.

The method

  1. Choose one body and draw every force acting on it. Name what exerts each force.
  2. Choose a positive direction along the acceleration, if you know it.
  3. Resolve forces along that direction and, if needed, perpendicular to it. The components replace the force; do not draw both.
  4. Write resultant force = ma along each direction. If the body does not accelerate in a direction, the forces in that direction balance.
  5. Solve, then check that the signs and sizes make sense.

The resultant force is the sum of the forces on the body. It is never an extra force to add to the diagram.

Bodies on a slope

On a slope at an angle θ to the horizontal, use axes along the slope and perpendicular to it, because the body moves along the slope. Resolve the weight mg: its component down the slope is mg sin θ, and its component into the slope is mg cos θ. For a fixed straight slope with no other perpendicular forces, the body does not accelerate into the slope, so the normal force balances that component: N = mg cos θ, less than mg.

Forces and weight components on a smooth inclineThe left panel isolates a block on a smooth plane at angle theta to the horizontal, with normal force perpendicular away from the plane and weight vertically downward. The right panel separately resolves that same weight into mg sin theta down the plane and mg cos theta into the plane; these component vectors add to the vertical weight.Forces on the blockResolve the same weightNmgθmg sin θmgmg cos θθHead-to-tail component sum
Forces and weight components on a smooth inclineThe left panel isolates a block on a smooth plane at angle theta to the horizontal, with normal force perpendicular away from the plane and weight vertically downward. The right panel separately resolves that same weight into mg sin theta down the plane and mg cos theta into the plane; these component vectors add to the vertical weight.Forces on the block NmgθResolve the same weight mg sin θmgmg cos θθHead-to-tail component sum
Left: only N and mg act on the block in the smooth-plane model. Right: the two components represent the same mg, not extra forces. Arrows are schematic, and the slope drawing is representative.

Worked example 1

A block on a smooth slope

Problem

A block of mass m slides down a smooth slope inclined at θ to the horizontal. Find its acceleration and the normal force on it.

Show full solution

  1. Identify the forces

    Method

    Only weight and the normal force act.

    Reason

    The slope is smooth, so there is no friction.

    Working

    Answer

    Weight mg vertically down; normal force N perpendicular to the slope.

  2. Apply F = ma along the slope

    Reason

    Only the weight has a component along the slope.

    Working

    Answer

    mg sin θ = ma, so a = g sin θ.

  3. Balance across the slope

    Working

    Answer

    No acceleration perpendicular to the slope: N = mg cos θ.

Guided practice 2

A block on a rough slope

About 5 min

Problem

A 2.0 kg block slides down a slope inclined at 30° to the horizontal. A friction force of 4.0 N acts up the slope. Take g = 9.81 m s⁻². Find the block’s acceleration and the normal force on it.

Find the acceleration and the normal force

Take down the slope as positive. Choose the weight component, enter the resultant down the slope in newtons, then calculate the acceleration and normal force to two significant figures.

Weight component down the slope
Show the steps Optional working
Unit: N
Unit: m s⁻²
Unit: N

Hints

Hint 1

along the slope

With down the slope positive, which forces contribute along the slope, and which way does each act?

Show solution step by step

  1. Along the slope

    Reason

    The resultant along the slope gives the acceleration.

    Working

    Answer

    mg sin 30° = (2.0)(9.81)(0.50) = 9.81 N, so the resultant is 9.81-4.0 = 5.81 N and a = 5.81/2.0 = 2.9 m s⁻² down the slope.

  2. Across the slope

    Working

    Answer

    N = mg cos 30° = (2.0)(9.81)(0.866…) = 16.99… N ≈ 17 N (two significant figures).

People in lifts

A bathroom scale measures the normal force R it exerts on the person standing on it. Take upward as positive. The forces on the person are R upward and the weight mg downward, so

R-mg = ma ⇒ R = m(g + a)

The scale reads more than mg when the acceleration is upward and less when it is downward. What matters is the direction of the acceleration, not of the motion: a lift moving down and slowing to a stop accelerates upward, so the reading is high.

Forces on the passenger and the contact partner on the liftThe passenger's free-body diagram has upward normal contact force N from the lift and downward gravitational weight mg from Earth. On the other body, the lift, the passenger exerts the equal contact force N downward. The lift panel shows only this contact force, not its cable tension or weight.On the passengerContact force on the liftpassengerNlift on passengermgEarth on passengerliftNpassenger on liftOther forces on the lift are omitted.
Forces on the passenger and the contact partner on the liftThe passenger's free-body diagram has upward normal contact force N from the lift and downward gravitational weight mg from Earth. On the other body, the lift, the passenger exerts the equal contact force N downward. The lift panel shows only this contact force, not its cable tension or weight.On the passengerContact force on the liftpassengerNlift on passengermgEarth on passengerliftNpassenger on liftOther forces on the lift are omitted.
The two N forces are a third-law pair on different bodies. N and mg both act on the passenger and are not a third-law pair. Arrows are schematic; lengths are not a quantitative force scale.

Spot the mistake 3

Does the normal force always equal the weight?

About 4 min

Learner claim

A 60.0 kg student stands on a scale in a lift that is accelerating upward at 1.50 m s⁻². A learner says the scale reads the student’s weight, because the student stays on the floor. Take g = 9.81 m s⁻². Find the actual reading to three significant figures.

Find the scale reading

Show the steps Optional working
Unit: N
Unit: N
Show solution step by step

  1. Use the resultant force

    Method

    Apply F = ma to the student, upward positive.

    Reason

    The student accelerates with the lift, so the vertical forces do not balance.

    Working

    Answer

    R-mg = ma, so R = m(g + a).

  2. Calculate

    Working

    Answer

    R = 60.0(9.81 + 1.50) = 679 N, more than the student’s weight of 589 N.

Try it yourself 4

A lift’s acceleration from a scale reading

Minimal support

Problem

The same 60.0 kg student now finds that the scale reads 450 N. Find the lift’s acceleration to three significant figures, giving its size and direction. Can you tell which way the lift is moving?

Find the acceleration

Show the steps Optional working
Unit: N
Unit: m s⁻²
Acceleration direction
Direction of motion

Hints

Hint 1

compare first

How does the reading compare with the student’s weight?

Show full solution

  1. Solve for the acceleration

    Working

    Answer

    450-588.6 = 60.0a, so a = -2.31 m s⁻²: 2.31 m s⁻² downward. The reading does not show the direction of motion: the lift could be speeding up on the way down or slowing down on the way up.

Bodies joined by a string

When two bodies are joined by a light, inextensible string over a light, smooth pulley, the tension is the same all along the string, and both bodies have the same size of acceleration. Apply F = ma to each body separately, each with its own positive direction along its motion, then solve the two equations together.

Exam-style question 1

A table block and a hanging mass

5 marks

Examination question

A 2.00 kg block on a smooth horizontal table is joined by a light inextensible string, passing over a light, smooth pulley, to a hanging 1.00 kg mass. The system is released from rest. Take g = 9.81 m s⁻². Find the acceleration of the system and the tension in the string. [5 marks]

Two connected masses and separate force diagramsA block on a smooth horizontal table is connected by a light taut inextensible string over a fixed ideal pulley to a hanging mass. The isolated table block has tension right, normal force up and weight down. The hanging mass has tension up and weight down. Positive directions are right for the table block and down for the hanging mass.The setupOn the table blockOn the hangingmassm₁m₂Fixed ideal pulleyTNm₁gPositive: rightTm₂gPositive: down
Two connected masses and separate force diagramsA block on a smooth horizontal table is connected by a light taut inextensible string over a fixed ideal pulley to a hanging mass. The isolated table block has tension right, normal force up and weight down. The hanging mass has tension up and weight down. Positive directions are right for the table block and down for the hanging mass.The setupm₁m₂Fixed ideal pulleyOn the table blockTNm₁gPositive: rightOn the hanging massTm₂gPositive: down
The same T acts on both masses in the light-string, ideal-pulley model. Their acceleration magnitudes are equal for this taut inextensible string and fixed pulley. Force-arrow lengths are schematic, not a force scale.

Find the acceleration and tension

Use a for the common acceleration magnitude. Take towards the pulley as positive for the table block and downward as positive for the hanging mass. Choose both equations, then enter acceleration and tension to three significant figures.

Table block equation
Hanging mass equation
Show the steps Optional working
Unit: N
Unit: kg
Unit: m s⁻²
Show the steps Optional working
Unit: m s⁻²
Unit: N
Show solution step by step

  1. One equation per body

    2 marks

    Method

    Apply F = ma to each mass along its own direction of motion.

    Reason

    Tension is the only horizontal force on the block; weight and tension act on the hanging mass.

    Working

    Answer

    Block (towards the pulley): T = 2.00a. Hanging mass (downward): 1.00(9.81)-T = 1.00a.

  2. Solve for the acceleration

    2 marks

    Method

    Add the two equations.

    Reason

    Both masses have the same acceleration, so the tension cancels.

    Working

    Answer

    9.81 = 3.00a, so a = 3.27 m s⁻².

  3. Find the tension

    1 mark

    Working

    Answer

    T = 2.00 × 3.27 = 6.54 N

The tension, 6.54 N, is less than the hanging mass’s weight of 9.81 N. It has to be, or the hanging mass would not accelerate downward.

Check your understanding 2

For the table block and hanging mass above, draw one force diagram per body on paper. Explain why the accelerations have equal magnitudes and why the tensions have equal magnitudes. Use both equations to justify the acceleration and tension.

Show answer

For the block draw weight down, normal force up and tension towards the pulley; the vertical pair balances. For the hanging mass draw weight down and tension up. A taut inextensible string constrains the acceleration magnitudes to match. A light string over the ideal smooth pulley has uniform tension. The equations are T = 2.00a and 9.81-T = 1.00a. Adding gives a = 3.27 m s⁻² and then T = 6.54 N. Compare your diagrams, string constraints, separate force balances and substitution.

Check your understanding 3

The table is now rough. The 2.00 kg block is moving towards the pulley while a constant 3.00 N friction force opposes it. The hanging mass remains 1.00 kg. Find the acceleration magnitude and tension, and explain whether the block speeds up. Use g = 9.81 m s⁻². Show your equations; work on paper if you prefer.

Show answer

The table block now has T-3.00 = 2.00a. The hanging mass still has 9.81-T = 1.00a. Adding gives 6.81 = 3.00a, hence a = 2.27 m s⁻² and T = 2.00(2.27) + 3.00 = 7.54 N. The acceleration of the block is towards the pulley, in its direction of motion, so it speeds up. Friction lowers the acceleration but raises the tension needed to pull the table block. Compare both changed force balances, the direction and the three significant figure results.

Common mistakes

  • Assuming the normal force always equals mg. It equals mg only on a level surface with no vertical acceleration and no other vertical forces.
  • Resolving the normal force instead of the weight on a slope.
  • Mixing forces from two bodies in one equation.
  • Deciding a lift’s scale reading from its direction of motion instead of its acceleration.
  • Taking the tension in the string as equal to the weight of the hanging mass.

Before you move on

Check your understanding 4

Without looking back: a lift moves upward but is slowing down. Is the scale reading greater than, equal to or less than the passenger’s weight? On a fixed slope at angle θ, a block has no perpendicular acceleration and no other perpendicular forces. Choose its normal force.

Scale reading in the slowing upward lift
Slope normal force
Show answer

Less: slowing on the way up means the acceleration is downward, so R = m(g + a) with a negative. On the slope, N = mg cos θ, because the block does not accelerate perpendicular to the slope.

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