HSC Exam Practice

Sample response. The resultant force is the single force that has exactly the same effect on an object as all individual forces acting together; it is found by vector addition of all forces. Equilibrium is the state of an object when the net force is zero in every direction, meaning it is either stationary or moving at constant velocity. For two-dimensional equilibrium, both ΣF_x = 0 and ΣF_y = 0 must hold simultaneously.

Marking notes. 1 mark for a correct definition of resultant force (single equivalent force, vector addition); 1 mark for a correct definition of equilibrium (net force = 0, stationary or constant velocity); 1 mark for explicitly stating the two simultaneous 2D conditions (ΣF_x = 0 AND ΣF_y = 0).

Sample response. $F_x = F\cos\theta$ (horizontal component) and $F_y = F\sin\theta$ (vertical component), where F is the magnitude of the force in newtons (N), θ is the angle measured from the horizontal reference axis in degrees, F_x is the component acting along the horizontal (x) axis, and F_y is the component acting along the vertical (y) axis. The angle θ must be measured from the reference direction (usually horizontal); if the angle is measured from the vertical, cos and sin are swapped.

Marking notes. 1 mark for F_x = F cos θ with correct identification as horizontal component; 1 mark for F_y = F sin θ with correct identification as vertical component; 1 mark for defining all symbols (F, θ, F_x, F_y) correctly.

Sample response. The tip-to-tail method gives the same resultant regardless of order because vector addition is commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$. Each vector has a fixed magnitude and direction (it is not changed by being placed in a different position). Because the resultant arrow always runs from the tail of the first vector to the tip of the last, and the total displacement of the chain is the same regardless of order, the resultant magnitude and direction are unchanged. Mathematically, the total x-component and total y-component of the sum are the same no matter what order the individual x and y contributions are summed.

Marking notes. 1 mark for identifying that vector addition is commutative (order does not change the result); 1 mark for explaining that each vector retains its magnitude and direction regardless of where it is placed in the chain; 1 mark for linking to the resultant arrow definition (from first tail to last tip, total displacement unchanged by order).

Sample response. Static equilibrium: the object is at rest (velocity = 0) and the net force is zero in every direction. Example: a traffic light hanging stationary from two symmetric cables over a Sydney intersection — the weight is balanced by the upward components of cable tension. Dynamic equilibrium: the object is moving at constant velocity (non-zero but unchanging) and the net force is still zero in every direction. Example: a V8 Supercar moving at constant speed along a straight track — the driving force exactly balances air resistance and friction, so net force = 0 and velocity is constant.

Marking notes. 1 mark for correct definition of static equilibrium (rest, net force = 0); 1 mark for a valid real-world static example; 1 mark for correct definition of dynamic equilibrium (constant velocity, net force = 0); 1 mark for a valid real-world dynamic example.

Sample response. The student is incorrect because the two forces are perpendicular — they act at 90° to each other (one north, one east) — so their magnitudes cannot simply be added. Scalar addition only applies when forces act in exactly the same direction. The correct resultant magnitude is found using Pythagoras’ theorem: $F = \sqrt{60^2 + 60^2} = \sqrt{7200} = 84.9\;\text{N}$. The direction is 45° north of east (northeast), since both components are equal.

Marking notes. 1 mark for identifying the error (perpendicular forces cannot be added as scalars / Pythagoras must be used); 1 mark for the correct magnitude (84.9 N or $60\sqrt{2}$ N); 1 mark for the correct direction (45° north of east or northeast).

Sample response. Tension increases significantly when the angle decreases. Using $T = W/(2\sin\theta)$: at 60°, $T = W/(2\sin 60°) = W/1.732 \approx 0.577W$; at 15°, $T = W/(2\sin 15°) = W/(2\times 0.259) = W/0.518 \approx 1.93W$. The tension more than triples [1]. The reason is that a shallower cable has a smaller vertical component per unit of tension (F_y = T sin 15° = 0.259T vs 0.866T at 60°); to provide the same total vertical force needed to support the sign’s weight, a larger tension is required [1]. At a very shallow angle, even a small weight requires enormous tension, which is why suspension cables are never made nearly horizontal in practice [1].

Marking notes. 1 mark for correctly identifying that tension increases; 1 mark for a valid quantitative comparison using $T = W/(2\sin\theta)$ at the two stated angles; 1 mark for correctly explaining the physics (smaller vertical component per unit tension at shallower angles requires greater tension to maintain equilibrium).

Sample response (a) — Resolve F₂ (2 marks). Positive direction: rightward = +x, upward = +y. $F_{2x} = 95\cos 55° = 95 \times 0.574 = 54.5\;\text{N}$ (positive, rightward) [1]. $F_{2y} = 95\sin 55° = 95 \times 0.819 = 77.8\;\text{N}$ (positive, upward) [1].

Sample response (b) — Components of F₃ using equilibrium (2 marks). $\Sigma F_x = 0$: $F_{3x} + 54.5 = 0$, so $F_{3x} = -54.5\;\text{N}$ (leftward) [1]. $\Sigma F_y = 0$: $F_{3y} + 77.8 - 80 = 0$, so $F_{3y} = 80 - 77.8 = +2.2\;\text{N}$ (upward) [1].

Sample response (c) — Magnitude and direction of F₃ (3 marks). $|F_3| = \sqrt{(-54.5)^2 + (2.2)^2} = \sqrt{2970.25 + 4.84} = \sqrt{2975.09} \approx 54.5\;\text{N}$ [1]. The angle with the negative x-axis: $\theta = \tan^{-1}(2.2/54.5) = \tan^{-1}(0.0404) \approx 2.3°$ above the negative x-axis [1]. Direction: F₃ acts at 54.5 N directed 2.3° above the horizontal, to the left (up-left in the second quadrant as stated in the problem) [1].

Marking notes. Part (a): 1 mark for correct F_2x with positive sign; 1 mark for correct F_2y with positive sign. Part (b): 1 mark for correct F_3x = −54.5 N (allow ±0.5 N); 1 mark for correct F_3y = +2.2 N (allow ±0.5 N). Part (c): 1 mark for correct magnitude using Pythagoras; 1 mark for correct angle using inverse tangent; 1 mark for a complete direction statement (“up-left” / second quadrant / angle above negative x-axis clearly stated).

Sample response. Component resolution is the primary analytical method for 2D force problems because it converts any force at an arbitrary angle into two perpendicular scalar quantities that can be handled algebraically. Its key strength is precision: once forces are resolved into x and y components, the conditions $\Sigma F_x = 0$ and $\Sigma F_y = 0$ can be solved exactly using arithmetic, regardless of the number of forces or their angles. For example, in the design of the suspension cables for the Sydney Harbour Bridge, engineers must calculate the tension in each diagonal cable precisely — a scale diagram would introduce drawing errors (mm-scale inaccuracies translating to significant force errors at engineering scale), whereas component resolution gives an exact algebraic answer. A second strength is that the method generalises seamlessly to three dimensions (adding a z-axis), which is essential for structural analysis of any real-world 3D structure. However, component resolution also has limitations. It requires knowledge of trigonometry and the ability to correctly assign signs based on direction; a sign error on a single component propagates through the entire calculation and produces a completely wrong answer. Students frequently confuse which component uses cos and which uses sin when the angle is measured from the vertical rather than the horizontal. Scale vector diagrams, by contrast, are more intuitive for qualitative understanding: seeing the tip-to-tail chain drawn to scale makes it physically obvious why the resultant changes direction when forces are added, and is accessible to students who have not yet mastered trigonometry. A scale diagram of the V8 Supercar pit-crew push (lesson context) can immediately show that the car moves at an angle between the two crew members — a result that requires several steps to obtain algebraically. The limitation of scale diagrams is that their accuracy depends entirely on drawing precision (typically ±2%) and they become impractical when many forces are involved. In summary, component resolution is superior whenever precise quantitative answers are required (structural engineering, exam calculation questions), while scale diagrams are superior for building conceptual understanding and quickly estimating direction. In professional engineering practice, both approaches are used: a quick sketch to confirm the qualitative picture, then component resolution for the final calculation.

Marking criteria (8 marks). 1 = identifies and explains at least two strengths of component resolution (precision, algebraic exactness, generalisability to 3D, or similar). 1 = identifies and explains at least two limitations of component resolution (sign errors, requires trig knowledge, angle measurement ambiguity). 1 = identifies and explains at least one strength of scale vector diagrams (intuitive, qualitative insight, accessible). 1 = identifies at least one limitation of scale diagrams (drawing precision, impractical for many forces). 1 = first named Australian/NSW engineering or real-world example used correctly to illustrate a strength of component resolution (Sydney Harbour Bridge cables, pit crew, traffic lights). 1 = second named example (different context). 1 = explicitly compares when each method is most appropriate (component resolution for precision; scale diagrams for conceptual clarity or estimation). 1 = reaches an explicit evaluative judgement that integrates both methods as complementary tools, not one-winner ranking, using precise physics terminology throughout (resultant, component, equilibrium, ΣF_x = 0, ΣF_y = 0, tip-to-tail).