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QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Recall, describe and apply Newton's three laws of motion, including the use of free-body diagrams to identify forces acting on an object and solve problems involving weight, normal force, friction and tension

A focused answer to the QCE Physics Unit 2 dot point on Newton's three laws and force analysis. States each law, walks through free-body diagrams for the standard QCAA problem types (level surface with friction, inclined plane, connected bodies, hanging tension), and works a force-on-an-incline example that recurs in IA1 stimulus and EA Paper 2.

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  1. What this dot point is asking
  2. Newton's three laws
  3. Free-body diagrams
  4. Inclined planes
  5. Connected bodies
  6. Examples in context
  7. Try this

What this dot point is asking

QCAA expects you to state and apply Newton's three laws of motion, and to construct free-body diagrams to analyse the forces on a single body. The standard problem types are level-surface motion with friction, motion on an inclined plane, connected bodies (pulleys and trains of carts), and tension in a hanging string or cable.

Newton's three laws

First law (inertia). An object continues in a state of rest or uniform motion in a straight line unless a net external force acts on it. A net force of zero means constant velocity (which includes being at rest).

Second law. The net force on an object equals its mass times acceleration:

Fnet=ma\vec{F}_{\text{net}} = m \vec{a}

Force has SI unit newton (N): 11 N is the force that accelerates a 11 kg mass at 11 m s2^{-2}. Force is a vector. Add forces vectorially.

Third law. For every action there is an equal and opposite reaction. If body A exerts a force on body B, then body B exerts a force of equal magnitude and opposite direction on body A. The forces act on different bodies and never on the same body, which is why a third-law pair does not "cancel out" the way two opposing forces on one body do.

Free-body diagrams

Isolate one object and draw every external force acting on it as an arrow from the object's centre. Standard forces:

  • Weight (W=mgW = mg). Straight down. Always present unless explicitly in deep space.
  • Normal force (NN). Perpendicular to the contact surface, pointing away from the surface.
  • Friction (ff). Parallel to the surface, opposing relative motion or relative motion tendency. Kinetic: fk=μkNf_k = \mu_k N. Static: fsμsNf_s \le \mu_s N.
  • Tension (TT). Along a string or cable, pulling away from the body.
  • Applied force. As stated.

Sum forces vectorially in two perpendicular directions and apply F=maF = ma in each.

Inclined planes

On a frictionless ramp at angle θ\theta:

  • Weight component along the slope (down the slope): W=mgsinθW_\parallel = mg \sin\theta.
  • Weight component perpendicular to slope: W=mgcosθW_\perp = mg \cos\theta.
  • Normal force balances WW_\perp: N=mgcosθN = mg \cos\theta.
  • If friction acts up the slope (object sliding down): Fnet=mgsinθμkmgcosθF_{\text{net}} = mg \sin\theta - \mu_k mg \cos\theta.

Choose the xx-axis along the slope and the yy-axis perpendicular. This eliminates the need to resolve the normal force or friction.

Connected bodies

For two masses joined by a light, inextensible string over a frictionless pulley, both masses have the same magnitude of acceleration and the tension is the same throughout the string. Write F=maF = ma for each mass, then solve simultaneously.

Examples in context

Example 1. A Gladstone port crane lifts a 4000 kg4000 \text{ kg} container at constant velocity. The free-body diagram shows weight mg=39200 Nmg = 39200 \text{ N} downward balanced by cable tension T=39200 NT = 39200 \text{ N} upward (Newton's first law). Accelerating the same container upward at 1.2 m s21.2 \text{ m s}^{-2} requires Tmg=maT - mg = ma, so T=m(g+a)=44000 NT = m(g+a) = 44000 \text{ N} - an extra 4800 N4800 \text{ N} from Newton's second law that is the design basis for the crane's safe working load margin.

Example 2. A Cairns light-rail tram traverses a 55^\circ incline approaching a stop. The tram (mass 45000 kg45000 \text{ kg}) experiences a component of weight along the incline mgsinθ=38500 Nmg\sin\theta = 38500 \text{ N} plus rolling resistance 1800 N1800 \text{ N}. The traction motor must apply 40300 N40300 \text{ N} to hold the tram stationary, by Newton's third law equally pushing the rail backward. QCAA Unit 2 EA Paper 2 incline-stem decomposition follows this template.

Try this

Q1. State Newton's third law and give one example. [2 marks]

  • Cue. Forces on two interacting bodies are equal in magnitude and opposite in direction; e.g. rocket pushes gas down, gas pushes rocket up.

Q2. A 3.0 kg3.0 \text{ kg} block on a horizontal surface (μk=0.20\mu_k = 0.20) is pulled with 15 N15 \text{ N} horizontal. Calculate the friction force, the net force and the acceleration. [4 marks]

  • Cue. fk=μkmg=5.88 Nf_k = \mu_k mg = 5.88 \text{ N}; Fnet=9.12 NF_{net} = 9.12 \text{ N}; a=3.04 m s2a = 3.04 \text{ m s}^{-2}.

Q3. A 1500 kg1500 \text{ kg} car ascends a 66^\circ incline at constant velocity. (a) Draw the free-body diagram. (b) Calculate the driving force and the normal reaction. (c) If the car then accelerates at 1.5 m s21.5 \text{ m s}^{-2} up the slope, determine the new driving force. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

  • Cue. (a) weight, normal, drive, opposed by gravity-component (and friction if specified); (b) F=mgsin6=1538 NF = mg\sin 6 = 1538 \text{ N}, N=mgcos6=14620 NN = mg\cos 6 = 14620 \text{ N}; (c) F=1538+2250=3788 NF = 1538 + 2250 = 3788 \text{ N}.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC5 marksA 5.05.0 kg block is pulled along a horizontal surface by a horizontal force of 3030 N. The coefficient of kinetic friction is μk=0.30\mu_k = 0.30. Using g=9.8g = 9.8 m s2^{-2}, calculate (a) the friction force and (b) the acceleration of the block.
Show worked answer →

(a) Friction force. Normal force on a horizontal surface equals weight.

N=mg=(5.0)(9.8)=49N = mg = (5.0)(9.8) = 49 N.

Kinetic friction: fk=μkN=(0.30)(49)=14.7f_k = \mu_k N = (0.30)(49) = 14.7 N, opposing motion.

(b) Acceleration. Net horizontal force divided by mass.

Fnet=Fappliedfk=3014.7=15.3F_{\text{net}} = F_{\text{applied}} - f_k = 30 - 14.7 = 15.3 N.

a=Fnet/m=15.3/5.0=3.06a = F_{\text{net}} / m = 15.3 / 5.0 = 3.06 m s2^{-2}.

Markers reward an explicit free-body diagram, separate horizontal and vertical force balances, and the substitution into F=maF = ma rather than guessing.

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