Revise Equilibrium of a rigid body for Cambridge 9231 (3.2)

For any rigid body to be in static equilibrium, two conditions must be met simultaneously. Firstly, there must be no net force, meaning the body is not accelerating linearly. Secondly, there must be no net turning effect, meaning the body is not accelerating angularly. These two simple rules form the basis for solving all static equilibrium problems.

$\sum \mathbf{F} = \mathbf{0} \quad \text{and} \quad \sum \mathbf{M}_P = \mathbf{0} \quad (\text{for any point P})$

While the moments equation holds true for any point you choose as a pivot, a strategic choice can drastically simplify your calculations. The most common and effective strategy is to choose a pivot point through which one or more unknown forces act. Since the moment of a force about a point on its own line of action is zero (as the perpendicular distance is zero), these forces will not appear in your moment equation, leaving you with fewer variables to solve for.

Problems may specify that a rigid body is 'on the point of tilting' or 'about to topple'. This provides a crucial piece of information: the body is about to rotate about a specific pivot point, and in doing so, it loses contact with another support. This means the normal reaction force at the point where contact is lost becomes zero. You can then take moments about the pivot point of the tilt to solve the problem.