What are the conditions of equilibrium for parallel and general coplanar forces, and how are couples balanced?

In one line

A body under parallel or general coplanar forces is in equilibrium only when the net force is zero and the net moment about every point is zero.

Explained simply

A seesaw with two children is balanced only if the total weight is supported (no up-down motion) and the turning effects on both sides are equal (no rotation). Both conditions are needed.

Technically, for a coplanar general system: \( \Sigma F_x = 0, \Sigma F_y = 0, \Sigma M = 0 \) (three independent equations). For parallel forces, say vertical: \( \Sigma F_y = 0 \) and \( \Sigma M = 0 \) (two equations).

How it works

1. Draw the FBD of the whole body showing loads and reactions.
2. For parallel forces write \( \Sigma F_y = 0 \) and a moment equation about a point on one unknown.
3. For general forces, use \( \Sigma F_x = 0 \), \( \Sigma F_y = 0 \), \( \Sigma M_A = 0 \); choose A where two unknowns meet so the moment equation has only one unknown.
4. Couples: a couple has zero net force, so it cannot be balanced by a single force. It is balanced only by an equal and opposite couple (often formed by two reactions).
5. An applied couple appears only in the moment equation, with the same value about any point.
6. Max solvable unknowns: 3 for a general system, 2 for parallel forces.

Key formula

\( \Sigma F_x = 0 \), \( \Sigma F_y = 0 \), \( \Sigma M_O = 0 \) (forces in N, moments in N·m).

Quick example

A horizontal beam AB, 4 m long, is hinged at A and held by a cable at B inclined at 30° to the beam; a 10 kN load acts at the middle. \( \Sigma M_A = 0 \): \( T\sin30^\circ \times 4 = 10 \times 2 \), so T = 10 kN. \( \Sigma F_x = 0 \): \( H_A = T\cos30^\circ = 8.66 \) kN. \( \Sigma F_y = 0 \): \( V_A = 10 - 5 = 5 \) kN.

Common mistakes

• Including a couple in the force equations; it belongs only in the moment equation.
• Writing three moment equations about collinear points, which are not independent.
• Forgetting the moment of the horizontal component of an inclined reaction when it has a moment arm.

How MU asks it

Theory 3 marks: 'State the conditions of equilibrium for parallel and general coplanar force systems'. Mostly used in 6 to 8-mark numericals on beams, bent bars and hinged bodies. Write which equation you use at each step, e.g. '\( \Sigma M_A = 0 \) (clockwise +ve)'.