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Forces and Newton's laws (SACE Stage 2 Physics)

Syllabus dot point

“Apply Newton's three laws of motion, drawing free-body diagrams and resolving forces to determine the acceleration of objects and the forces in interacting systems.”

9 min read

Quick answer

Newton's first law (inertia) says an object keeps its velocity unless a net force acts. The second law says net force equals mass times acceleration, F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a}. The third law says forces come in equal and opposite pairs acting on different objects. You apply these by drawing free-body diagrams and finding the net force.

Jump to a section
  1. What this dot point is asking
  2. The three laws
  3. Free-body diagrams
  4. Forces on an inclined plane
  5. Connected systems
  6. Equilibrium
  7. Exam-style questions

What this dot point is asking

You need to state and apply Newton's three laws, draw free-body diagrams, resolve forces (including on inclined planes), and calculate acceleration from the net force on an object or connected system.

Block on a frictionless incline

Acceleration down a 25 degree slope

A 4.0 kg4.0\ \text{kg} block slides down a frictionless incline at 25∘25^\circ. Find its acceleration and the normal force. Use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}.

Along the slope: Fnet=mgsin⁡θ=4.0×9.8×sin⁡25∘=16.6 NF_{\text{net}} = mg\sin\theta = 4.0 \times 9.8 \times \sin 25^\circ = 16.6\ \text{N}.

So a=gsin⁡θ=9.8×sin⁡25∘=4.1 m s−2a = g\sin\theta = 9.8 \times \sin 25^\circ = 4.1\ \text{m s}^{-2} down the slope.

Perpendicular: N=mgcos⁡θ=4.0×9.8×cos⁡25∘=35.5 NN = mg\cos\theta = 4.0 \times 9.8 \times \cos 25^\circ = 35.5\ \text{N}.

Note the mass cancels in the acceleration but not in the forces.

The three laws

Newton's three laws
  1. First law (inertia): an object continues at rest or constant velocity unless acted on by a non-zero net force.
  2. Second law: F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a}. The net force and the acceleration point the same way.
  3. Third law: if A exerts a force on B, then B exerts an equal and opposite force on A. The two forces act on different objects and never cancel.

The second law is the workhorse. "Net force" means the vector sum of all forces acting on the object. If the net force is zero the object is in equilibrium (first law); if not, it accelerates in the direction of the net force, with a=Fnet/ma = F_\text{net}/m.

Free-body diagrams

A free-body diagram shows one object as a point (or simple shape) with every force on it drawn as an arrow from that point. Typical forces are:

  • Weight W=mgW = mg, always vertically down.
  • Normal force NN, perpendicular to the surface.
  • Tension TT, along a rope or cable, pulling away from the object.
  • Friction ff, parallel to the surface, opposing relative motion.
  • Applied force FF, whatever pushes or pulls.

Once drawn, resolve the forces into perpendicular axes (usually horizontal and vertical, or along and across an incline) and apply F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a} to each axis separately.

Forces on an inclined plane

For an object on a frictionless incline at angle θ\theta, choose axes along and perpendicular to the slope. The weight resolves into:

W∥=mgsin⁡θ (down the slope)W⊥=mgcos⁡θ.W_{\parallel} = mg\sin\theta\ (\text{down the slope}) \qquad W_{\perp} = mg\cos\theta.

Perpendicular to the slope there is no acceleration, so N=mgcos⁡θN = mg\cos\theta. Along the slope, mgsin⁡θmg\sin\theta produces an acceleration a=gsin⁡θa = g\sin\theta (frictionless). If friction acts, subtract ff from mgsin⁡θmg\sin\theta before dividing by the mass.

Connected systems

For objects joined by a rope (e.g. a trolley pulled by a hanging mass over a pulley), treat the whole system to find the common acceleration, then isolate one object to find the tension. For the whole system:

a=net external forcetotal mass.a = \frac{\text{net external force}}{\text{total mass}}.

Then apply Newton's second law to a single object to get the internal tension. The rope tension is an internal force for the system but an external force for each block, which is why isolating one block reveals it.

Equilibrium

When F⃗net=0\vec{F}_{\text{net}} = 0 the object is in equilibrium (at rest or constant velocity). Then the components of all forces balance in each direction. This is how you find unknown tensions in a hanging sign or the friction holding a box still on a slope. Set the sum of components to zero along each axis and solve.

Common errors
Pairing forces on the same object
Newton's third-law pairs act on different objects. The weight of a book and the normal force from the table are not a third-law pair (they act on the same book and are only equal because the book is in equilibrium).
Forgetting the normal force changes on a slope
On an incline N=mgcos⁡θN = mg\cos\theta, not mgmg.
Using weight where you need mass
F⃗net=ma⃗\vec{F}_{\text{net}} = m\vec{a} uses mass in kg, not weight in newtons.
Leaving out a force
A missing friction or tension arrow on the free-body diagram is the most common reason an answer is wrong.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2024-style4 marks
A 3.0 kg3.0\ \text{kg} block on a frictionless bench is connected by a light string over a frictionless pulley to a hanging 2.0 kg2.0\ \text{kg} mass. (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. Use g=9.80 m s−2g = 9.80\ \text{m s}^{-2}.
Show worked answer →

Treat the two masses as one system to find the acceleration, then isolate one mass for the tension.

(a) The only external force driving the system is the weight of the hanging mass, W=(2.0)(9.80)=19.6 NW = (2.0)(9.80) = 19.6\ \text{N}. The total mass is 3.0+2.0=5.0 kg3.0 + 2.0 = 5.0\ \text{kg}.

a=Fnetmtotal=19.65.0=3.92 m s−2.a = \dfrac{F_\text{net}}{m_\text{total}} = \dfrac{19.6}{5.0} = 3.92\ \text{m s}^{-2}.
(2 marks)

(b) Isolate the 3.0 kg3.0\ \text{kg} block on the bench: the only horizontal force is the tension, so T=ma=(3.0)(3.92)=11.8 NT = m a = (3.0)(3.92) = 11.8\ \text{N}. (2 marks) Markers reward the whole-system step for aa and isolating one mass for TT.

SACE 2023-style3 marks
A 4.0 kg4.0\ \text{kg} block is held on a frictionless incline at 25.0∘25.0^\circ to the horizontal, then released. (a) Calculate the acceleration of the block down the slope. (b) Calculate the normal force from the surface on the block. Use g=9.80 m s−2g = 9.80\ \text{m s}^{-2}.
Show worked answer →

Choose axes along and perpendicular to the slope and resolve the weight W=mgW = mg.

(a) Along the slope the net force is mgsin⁡θmg\sin\theta, so a=gsin⁡θ=(9.80)sin⁡25.0∘=4.14 m s−2a = g\sin\theta = (9.80)\sin 25.0^\circ = 4.14\ \text{m s}^{-2} down the slope. The mass cancels. (2 marks)

(b) Perpendicular to the slope there is no acceleration, so N=mgcos⁡θ=(4.0)(9.80)cos⁡25.0∘=35.5 NN = mg\cos\theta = (4.0)(9.80)\cos 25.0^\circ = 35.5\ \text{N}. (1 mark) Markers reward resolving the weight into the two slope components and recognising N=mgcos⁡θN = mg\cos\theta, not mgmg.

Practise this

Sources & how we know this

ExamExplained