State that, when there is no resultant force and no resultant moment, an object is in equilibrium

1.5.2 Turning Effect of Forces

Objective (Cambridge IGCSE 0625)

State that an object is in equilibrium when both of the following are true:

• the resultant (net) force acting on the object is zero, and
• the resultant (net) moment (torque) about any axis is zero.

Syllabus‑required statement

“When there is no resultant force and no resultant moment, the object is in equilibrium.”

Key Definitions

Term	Definition
Resultant Force (∑ \mathbf{F})	A single force that has the same linear effect as all the individual forces acting on the body. It is a vector quantity.
Resultant Moment / Torque (∑ M)	A single turning effect that reproduces the rotational effect of all the individual moments about a chosen pivot. Torque is a (pseudo‑)vector; its magnitude is M = F\,d\sin\theta and its sign follows a chosen sense of rotation (commonly anticlockwise = + , clockwise = –).
Moment arm (lever arm)	The perpendicular distance d from the line of action of the force to the pivot. Only the perpendicular component contributes to the moment.
Couple	A pair of equal and opposite forces whose lines of action do not coincide. A couple produces a pure rotation: resultant force = 0, resultant moment = F × distance between the forces.

Equilibrium Conditions (2‑D)

For a rigid body to be in static equilibrium the following three independent equations must be satisfied simultaneously:

1. Resultant (net) force is zero (vector form):
   $$\sum \mathbf{F}=0$$
2. Resultant moment about any chosen point O is zero:
   $$\sum M_O =0$$
3. In component form (useful for solving problems):
   $$\sum F_x =0,\qquad \sum F_y =0$$

Principle of Moments (General Form)

For any set of forces acting on a rigid body, the algebraic sum of the moments about a pivot O is zero when the body is in equilibrium:

$$\sum_{i} F_i d_i \sin\theta_i =0$$

• F_i – magnitude of the i‑th force.
• d_i – distance from O to the line of action of the i‑th force.
• θ_i – angle between the force direction and the line joining O to any point on the line of action (the sine term extracts the perpendicular component).

When all forces are perpendicular to their respective lever arms, \sin\theta =1 and the simpler form M = Fd applies.

Special case – one force each side of the pivot

If there is exactly one clockwise‑acting force and one anticlockwise‑acting force about the same pivot, equilibrium reduces to:

$$F_{\text{cw}} d_{\text{cw}} = F_{\text{acw}} d_{\text{acw}}$$

Couple

A couple of magnitude M_{\text{c}} = F \times s (where s is the separation of the two parallel forces) can rotate a body even though the resultant force is zero.

Everyday Illustrations of Moments

Worked Example 1 – One Force Each Side of the Pivot (Core 1)

Problem: A uniform beam 4 m long rests on two supports A (left) and B (right). A load of 200 N hangs 1 m from A. Find the reactions at A and B.

1. Choose a pivot – take moments about A.
2. Moment balance (clockwise = –, anticlockwise = +):
   $$\sum M_A = 0 \;\Rightarrow\; R_B (4\;\text{m}) - 200\;(1\;\text{m}) =0$$ $$\Rightarrow\; R_B = \frac{200\times1}{4}=50\;\text{N}$$
3. Vertical‑force balance:
   $$\sum F_y =0 \;\Rightarrow\; R_A + R_B -200 =0$$ $$\Rightarrow\; R_A =200-50 =150\;\text{N}$$

Both reactions are upward, the sum of forces and the sum of moments are zero – the beam is in equilibrium.

Worked Example 2 – Multiple Forces on Both Sides of the Pivot (Core 2)

Problem: A 6 m beam is hinged at O. Forces act as shown:

• 120 N downward at 1 m to the right of O.
• 80 N upward at 2 m to the left of O.
• 50 N downward at 4 m to the left of O, acting at an angle of 30° to the vertical (the line of action makes 30° with the beam).

Determine the hinge reaction (ignore its moment) and state whether the beam is in equilibrium.

1. Resolve the angled force into a component perpendicular to the beam:
   $$F_{\perp}=50\sin30^{\circ}=25\;\text{N}$$
2. Take moments about O (anticlockwise = +):
   $$\sum M_O = 0 \;\Rightarrow\; (+80)(2) - (120)(1) - (25)(4) =0$$ $$\Rightarrow\; 160 -120 -100 = -60\;\text{N·m}$$

   The sum is not zero, so the beam would rotate. To achieve equilibrium an additional external moment of +60 N·m (e.g., a supporting cable) would be required.

3. Because the moment condition fails, the beam is **not** in equilibrium, even though the vertical forces could be balanced by the hinge reaction.

Supplementary Concepts (Optional – for extended‑syllabus classes)

Concept	Key Formula / Idea
Resultant moment for a non‑perpendicular force	$$M = F\,d\sin\theta$$ where θ is the angle between the force and the lever arm.
Couple	Pure rotation: M_{\text{c}} = F\,s (s = separation of the two parallel forces). Resultant force = 0.
Static friction in equilibrium	Maximum frictional force F_f = \mu_s N can act to prevent slipping; it is included in the ∑F = 0 equation.
Equilibrium of a rigid body in two dimensions	Three independent equations: ∑F_x=0, ∑F_y=0, ∑M=0. Solving any two of the force components together with the moment equation gives the complete solution.

Summary Table – Conditions for Equilibrium

Condition	Mathematical Form	Physical Meaning
Resultant (net) force	$$\sum \mathbf{F}=0$$	No linear (translational) acceleration.
Resultant (net) moment (torque)	$$\sum M_O =0$$ (for any point O)	No angular (rotational) acceleration.
Component form (2‑D)	$$\sum F_x =0,\;\; \sum F_y =0$$	Ensures equilibrium in both horizontal and vertical directions.

Key Take‑away

An object (or rigid body) is in static equilibrium only when both the vector sum of all forces and the algebraic sum of all moments about any chosen point are zero. This dual condition underpins every problem involving static structures, levers, seesaws, bolts, and everyday tools.

Suggested Diagram for the Worked Example