Angle of inclination of suspended bodies — A/L Combined Mathematics

When a rigid body is suspended freely from a fixed pivot point O, it attains static equilibrium under two forces: the tension/reaction at O and the body's total weight W acting vertically downwards through its centre of mass G. For rotational equilibrium, the line of action of the weight must pass through O, which mandates that in equilibrium, the centre of mass G lies vertically below the suspension point O (line OG is strictly vertical). The angle of inclination θ that a specified boundary edge makes with the horizontal or vertical is evaluated using right-angled trigonometry on the geometry of triangle OGX.

Check before you answer (A/L Applied Maths Paper 2)

1. Vertical Line Alignment Principle: Stating explicitly that the suspension point O and centre of mass G lie on the same vertical line.
2. Center of Mass Coordinates: Calculating the exact coordinates of G from the suspension point along orthogonal geometric axes.
3. Tangent Angle Formula: Deriving tan θ = x_G / y_G from the vertical triangle and evaluating the required angle of inclination.