use a vector triangle to represent coplanar forces in equilibrium

Equilibrium of Coplanar Forces (Cambridge IGCSE/A‑Level 9702 – Topic 4.2)

1. Core Principles

• Resultant force zero – a body is in static equilibrium when the vector sum of all forces acting on it is zero: $$\displaystyle \sum\vec F = \vec 0$$
• Force polygon (vector triangle) – for any set of coplanar forces the arrows can be placed head‑to‑tail. The polygon closes iff the forces are in equilibrium.
• Moment equilibrium – the algebraic sum of the moments of all forces about any point must be zero: $$\displaystyle \sum \tau = 0,\qquad \tau = \vec r \times \vec F$$ (SI unit = N·m).
• Both conditions – Σ F = 0 (components) **and** Σ τ = 0 – are required for true static equilibrium.

2. Principle of Moments

• The moment of a force about a point O is the product of the force magnitude and the perpendicular distance from O to the line of action of the force: $$\tau = F\,r_{\perp}$$
• Geometrically, this is the magnitude of the cross‑product \(\vec r \times \vec F\). The direction (out of the plane) follows the right‑hand rule; in planar problems we treat clockwise moments as negative and anticlockwise as positive.
• Choosing a pivot that makes the perpendicular distance of one or more unknown forces zero removes those unknowns from the moment equation – a vital exam strategy.

3. Constructing a Force Polygon (General Method for ≥ 3 Forces)

1. Resolve every force into its components in the plane of interest.
2. Select a scale (e.g. 1 cm = 10 N) and a convenient starting point.
3. Draw the first vector to scale, using a ruler and protractor for the correct direction.
4. From the **head** of the previous vector, draw the next force to scale and with its proper direction.
5. Continue until all forces are placed. If the final head meets the original tail, the polygon is closed – the system is in equilibrium.
6. If the polygon does **not** close, the line joining the start point to the final head is the resultant \(\vec R\). Adding a force equal and opposite to \(\vec R\) restores equilibrium.

4. Choosing the Pivot – A Quick Tip

5. Three‑Force Example (Beam on Two Supports)

Problem: A uniform horizontal beam of length 4 m and weight 200 N rests on two supports A and B. Support A is at the left end, support B is 3 m from A. Find the reaction forces at A and B.

1. Take moments about A (eliminates the unknown reaction at A): $$\tau_B = W \times \frac{L}{2} \;\Longrightarrow\; R_B \times 3 = 200 \times 2$$ $$R_B = \frac{400}{3}\;{\rm N}=133.3\;{\rm N}$$
2. Apply ΣF_y=0: $$R_A + R_B - W = 0 \;\Longrightarrow\; R_A = 200 - 133.3 = 66.7\;{\rm N}$$
3. Force polygon: draw \(\vec R_A\) upward (66.7 N), then from its head draw \(\vec R_B\) upward (133.3 N) at the same line of action (vertical). Finally draw the weight \(\vec W\) downward (200 N). The three vectors close to a triangle, confirming equilibrium.

6. Graphical Check of Moment Equilibrium

• Select any point O (often a hinge, bolt or support).
• For each force, draw a perpendicular from O to the line of action; the length of this perpendicular is the moment arm \(r_{\perp}\).
• Calculate the algebraic moment: \(\tau = r_{\perp}F\) (CCW = +, CW = −).
• Verify that the sum of all moments is zero. In a clean diagram the clockwise and anticlockwise moments will balance visually.

7. Worked Example – Sign Supported by Two Cables

Problem statement

A uniform rectangular sign (weight \(W = 120\;{\rm N}\)) is attached to a vertical wall by two steel cables. Cable A is fixed at the top‑right corner and makes an angle of \(30^{\circ}\) above the horizontal. Cable B is fixed at the bottom‑left corner and makes an angle of \(45^{\circ}\) below the horizontal. The bolts at the corners are frictionless pivots. The sign is 2.0 m wide and 1.0 m high. Determine the tension in each cable.

Solution (AO2 – application)

1. Choose a pivot. Take moments about the bottom‑left corner (the point where cable B is attached). This removes the unknown tension in cable B from the moment equation.
2. Resolve forces. \[ \begin{aligned} T_A &: \; T_{Ax}=T_A\cos30^{\circ},\qquad T_{Ay}=T_A\sin30^{\circ}\\ W &: \; \text{acts vertically downwards at the centre }(1.0\;{\rm m},\,0.5\;{\rm m})\text{ from the pivot.} \end{aligned} \]
3. Moment arms. The vertical component of \(T_A\) acts at the top‑right corner \((2.0,\,1.0)\) m, giving a perpendicular (horizontal) distance of \(2.0\) m to the pivot. The weight acts at a horizontal distance of \(1.0\) m from the pivot.
4. Write the moment equation (Στ = 0). Counter‑clockwise moments are taken as positive. \[ T_A\sin30^{\circ}\;(2.0) \;-\; W\;(1.0) = 0 \] \[ T_A(0.5)(2.0) = 120 \;\Longrightarrow\; T_A = 120\;{\rm N} \]
5. Apply ΣF_x = 0 and ΣF_y = 0. \[ \begin{aligned} \Sigma F_x &: \; T_{Ax} - T_{Bx}=0 \;\Longrightarrow\; T_{Bx}=T_A\cos30^{\circ}=120\cos30^{\circ}=103.9\;{\rm N}\\[4pt] \Sigma F_y &: \; T_{Ay}+T_{By}-W=0 \;\Longrightarrow\; T_{By}=W-T_{Ay}=120-120\sin30^{\circ}=60\;{\rm N} \end{aligned} \] The tension in cable B is the resultant of its components: \[ T_B=\sqrt{T_{Bx}^{2}+T_{By}^{2}} =\sqrt{(103.9)^{2}+(60)^{2}} \approx 1.20\times10^{2}\;{\rm N} \]

Force Polygon for the Example

• Scale: 1 cm = 10 N.
• Draw \(\vec T_A\) (12 cm at \(30^{\circ}\) above the horizontal), then from its head draw \(\vec W\) (12 cm straight down), and finally \(\vec T_B\) (≈ 12 cm at the angle that closes the triangle). The three vectors meet, confirming equilibrium.

8. Summary Table – Procedure Checklist

9. Key Points to Remember

• Static equilibrium ⇔ Σ F = 0 **and** Σ τ = 0.
• The force polygon works for any number of coplanar forces; for three forces it is a triangle, for more it becomes a closed polygon.
• Always draw the polygon to scale; otherwise the visual check is unreliable.
• Moment balance can be performed about any point – choose the point that eliminates the most unknowns.
• In exam answers state the pivot, write the moment equation, then finish with the component equations.
• Interpretation matters: direction tells where the required force must act; magnitude indicates whether the supporting members are realistic.

10. Common Mistakes & How to Avoid Them

11. Related Concept – Couples (Turning Effect of Two Parallel Forces)

A pair of equal and opposite forces whose lines of action do not coincide forms a couple. The resultant of a couple is a pure moment: \[ \tau_{\text{couple}} = F\,d \] where \(d\) is the perpendicular distance between the two forces. In a force polygon a couple appears as a closed two‑force “triangle” (a line segment traced forward and then back). Recognising couples helps to simplify problems where a resultant moment is known but the individual forces are not required.

12. Assessment Objective (AO) Links

• AO1 – Knowledge: state the equilibrium conditions Σ F = 0 and Σ τ = 0, describe the force‑polygon method, and explain the moment‑arm concept.
• AO2 – Application: solve the sign‑support problem, construct the force polygon, and interpret the direction and magnitude of the required tensions.
• AO3 – Analysis (A‑Level optional): discuss how moving the cable attachment points would change the tension values and the shape of the force polygon.