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M 4

Newton’s laws of motion

Mechanics · Paper 4

🎯What you need to be able to do

  • State and apply \( F = ma \) to a single body.
  • Combine \( F = ma \) with the constant-acceleration formulae.
  • Handle motion on a rough or smooth inclined plane.
  • Solve problems with connected particles joined by a string over a pulley.
  • Analyse a body in a lift, and explain apparent weight.
  • Choose between treating a system as a whole and treating each part separately.

📚The mathematics

The equation of motion

\[ F = ma \]

where \(F\) is the resultant force in the direction of the acceleration. Almost every error on this topic is a failure to find the resultant correctly, so the routine matters:

  1. Draw the body and mark every force.
  2. Choose the positive direction — sensibly, the direction of the acceleration.
  3. Write “resultant force = \( ma \)” along that direction, with each force signed.
  4. In the perpendicular direction there is no acceleration, so the forces balance there and you get a second equation, usually giving the normal contact force.

Newton’s third law is what lets you connect two bodies: the force each exerts on the other is equal in magnitude and opposite in direction. The tension in a light inextensible string is the same throughout, and pulls both bodies towards the pulley.

Modelling assumptions and what they mean

  • Smooth — no friction, so \( F = 0 \).
  • Light — the string or pulley has no mass, so the tension is unchanged along the string and the pulley needs no force to turn it.
  • Inextensible — the string does not stretch, so both connected bodies have the same magnitude of acceleration. This is the fact that makes pulley problems solvable.
  • Particle — the body has no size, so all forces act at one point and there is no rotation to consider.

The inclined plane

Resolve along and perpendicular to the slope, as in M 1. For a body sliding down a rough slope:

perpendicular: \( R = mg\cos\theta \)
along: \( mg\sin\theta - \mu R = ma \)

Friction opposes the motion, so it acts up the slope for a body sliding down, and down the slope for a body being pushed up. Redraw for each phase of the motion — a body that goes up and then comes back down has friction pointing one way and then the other, and therefore different accelerations on the two legs.

Connected particles

Two masses joined by a string over a smooth pulley. Because the string is inextensible, both have the same magnitude of acceleration \(a\); because it is light, the tension \(T\) is the same throughout.

Write \( F = ma \) for each particle separately, taking the direction of that particle’s motion as positive in each case. For masses \( m_1 > m_2 \) hanging vertically:

heavier, moving down: \( m_1g - T = m_1a \)
lighter, moving up: \( T - m_2g = m_2a \)

Adding these eliminates \(T\) immediately — which is why writing them in that orientation pays off. Then substitute back for \(T\).

Two equations, two unknowns — do not try to write one. The tension cannot be found without treating the particles separately, because \(T\) is internal to the system. You can treat the system as a whole to find the acceleration quickly — total driving force divided by total mass — but that method makes \(T\) invisible. Use whole-system for \(a\) if you like, then one particle’s equation for \(T\).

Lifts and apparent weight

For a person of mass \(m\) in a lift accelerating upwards at \(a\), the forces are the weight \(mg\) down and the normal force \(R\) from the floor up:

\[ R - mg = ma \;\Longrightarrow\; R = m(g + a) \]

\(R\) is what a set of bathroom scales would read — the apparent weight. Accelerating upwards makes you feel heavier; accelerating downwards (\(a\) negative) makes you feel lighter; and in free fall, \( a = -g \) gives \( R = 0 \), which is weightlessness. Note that moving at constant velocity, up or down, gives \( a = 0 \) and \( R = mg \): it is acceleration, not motion, that changes the reading.

✏️Worked example

Take \( g = 10 \) m s\(^{-2}\). Particles \(A\) of mass 5 kg and \(B\) of mass 3 kg are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest with both particles hanging freely. (a) Find the acceleration of the system and the tension in the string. (b) Find the speed of \(A\) after it has descended 1.2 m.

(a) \(A\) is heavier so it descends and \(B\) rises, both with acceleration \(a\) and a common tension \(T\).

For \(A\), taking downwards as positive:

\[ 5g - T = 5a \;\Longrightarrow\; 50 - T = 5a \]

For \(B\), taking upwards as positive:

\[ T - 3g = 3a \;\Longrightarrow\; T - 30 = 3a \]

Adding the two equations eliminates \(T\): \( 20 = 8a \), so \( a = 2.5 \) m s\(^{-2}\). Substituting into the second equation: \( T = 30 + 3(2.5) = 37.5 \) N.

(b) The acceleration is constant, so use the constant-acceleration formulae with \( u = 0 \), \( a = 2.5 \), \( s = 1.2 \):

\[ v^{2} = u^{2} + 2as = 0 + 2(2.5)(1.2) = 6 \;\Longrightarrow\; v = 2.45\ \text{m s}^{-1} \]
Check it. The tension must lie between the two weights, \( 30 \) N and \( 50 \) N — and 37.5 N does. If it came out above 50 or below 30, the system could not behave as described. Also check the acceleration against the whole-system shortcut: the net driving force is \( 50 - 30 = 20 \) N on a total mass of 8 kg, giving \( a = \dfrac{20}{8} = 2.5 \) ✓.
The tension is not \( 5g \), and it is not the average of the weights either. A very common wrong answer is \( T = 40 \) N, the midpoint of 30 and 50 — plausible, and incorrect. Another is to use \( a = g \) for the falling particle, forgetting that the string holds it back. Always write both equations; the algebra is three lines and it removes the guesswork.

📝Practise

Work through these, then reveal the answer. Take \( g = 10 \) m s\(^{-2}\) throughout.

1. A box of mass 8 kg is pulled along a rough horizontal floor by a horizontal force of 30 N. The coefficient of friction is 0.25. Find the acceleration.
Vertically: \( R = mg = 80 \) N. Friction: \( F = \mu R = 0.25(80) = 20 \) N, opposing the motion. Horizontally: \( 30 - 20 = 8a \), so \( 10 = 8a \) and \( a = 1.25 \) m s\(^{-2}\).
2. A particle of mass 2 kg slides down a smooth plane inclined at \( 30^{\circ} \). Find its acceleration and the normal contact force.
Smooth, so no friction. Along the slope: \( mg\sin 30^{\circ} = ma \), so \( a = g\sin 30^{\circ} = 10(0.5) = 5 \) m s\(^{-2}\). Perpendicular: \( R = mg\cos 30^{\circ} = 20(0.8660) = 17.3 \) N. Note the acceleration is independent of mass — the \(m\) cancels.
3. A person of mass 70 kg stands in a lift. Find the force exerted by the floor when the lift (a) accelerates upwards at 2 m s\(^{-2}\), (b) descends at a constant 3 m s\(^{-1}\).
(a) \( R - mg = ma \), so \( R = m(g+a) = 70(12) = 840 \) N — heavier than the true weight of 700 N. (b) Constant velocity means \( a = 0 \), so \( R = mg = 700 \) N. The lift's speed and direction are irrelevant; only its acceleration changes the reading.
4. A particle of mass 4 kg is pulled up a rough plane inclined at \( 25^{\circ} \) by a force of 40 N acting up the slope. The coefficient of friction is 0.2. Find the acceleration.
Perpendicular: \( R = mg\cos 25^{\circ} = 40(0.9063) = 36.25 \) N. Friction opposes the upward motion, so acts down the slope: \( F = 0.2(36.25) = 7.25 \) N. Along the slope: \( 40 - mg\sin 25^{\circ} - F = ma \), that is \( 40 - 40(0.4226) - 7.25 = 4a \), so \( 40 - 16.90 - 7.25 = 15.85 = 4a \) and \( a = 3.96 \) m s\(^{-2}\).
5. Particles of mass 6 kg and 2 kg are connected over a smooth pulley and released from rest. Find the acceleration and the tension.
For the 6 kg (down): \( 60 - T = 6a \). For the 2 kg (up): \( T - 20 = 2a \). Adding: \( 40 = 8a \), so \( a = 5 \) m s\(^{-2}\). Then \( T = 20 + 2(5) = 30 \) N. Check: \(T\) lies between 20 N and 60 N ✓, and the whole-system shortcut gives \( \dfrac{60-20}{8} = 5 \) ✓.
6. A car of mass 1200 kg tows a trailer of mass 400 kg. The engine provides a driving force of 3200 N and resistances are negligible. Find the acceleration and the tension in the tow bar.
Whole system: \( 3200 = 1600a \), so \( a = 2 \) m s\(^{-2}\). For the tension, consider the trailer alone, since the only horizontal force on it is the tow bar: \( T = 400(2) = 800 \) N. Using the trailer rather than the car is the shortcut here — the car's equation would be \( 3200 - T = 1200(2) \), giving the same 800 N but with more arithmetic.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — Forces and Motion, for building intuition about \( F = ma \)
  • The Physics Classroom — Newton’s laws and free-body diagrams
  • Khan Academy — connected particles and pulley systems