Forces, moments and equilibrium

The two tests are independent. Equal opposite forces can have zero resultant while producing a couple, so checking vertical forces alone may miss rotation. Conversely, a zero moment about one chosen point does not guarantee that the object will not translate. In a two-dimensional problem, write horizontal and vertical force equations and a moment equation as needed. Static equilibrium also means the object is initially at rest; zero resultant force and torque can otherwise describe motion at constant linear and angular velocity. State the physical situation instead of equating balanced forces with rest in every context.

The moment of a couple does not depend on the point chosen for taking moments. If the pivot lies between the force lines, both turning effects act in the same rotational sense; their lever arms add to the full separation. If the pivot lies elsewhere, subtracting the opposite contributions gives the same total. Do not multiply by two again when the given distance is already the separation between forces. A steering wheel turned by opposite hand forces is a useful physical model, provided the hands’ forces are approximately equal and opposite.

This is a mathematical choice, not necessarily the point about which the object is physically hinged. Any point can be used when the object is in equilibrium. For a simply supported beam, taking moments about one support often gives the other support’s reaction directly. Then use force balance to find the reaction at the chosen support. Include the beam’s own weight at its centre of gravity, and check that all distances are measured from the same pivot. A force applied far away can still have zero moment if its line of action passes through that point.

The angle must be the one between the force and the pivot-to-application direction. An angle measured from a different reference line may require cosine after conversion. Drawing the perpendicular avoids memorising the wrong function. A force directed along the lever arm produces zero moment because its line of action passes through the pivot. This does not mean the force is absent: it can still contribute to translational force balance. Give the rotational direction separately from the magnitude, following any sign convention established in the problem.

For a body resting on a base, the support forces can combine to balance weight only while the required reaction lies within that base. Tilting shifts the vertical projection of the centre of gravity relative to the contact region. A lower centre of gravity or wider base allows a greater angle before the projection crosses the edge. Do not confuse the centre of gravity’s height with its vertical line of action; both geometry and orientation matter. If other external forces act, include their moments because tipping may occur before the weight alone reaches the simple threshold.

Friction is not automatically equal to a limiting value. In static equilibrium it takes the value needed for balance, up to the maximum available. Mark the normal perpendicular to the surface and determine friction’s direction from the tendency to slide. In a ladder problem, the wall and floor can have different assumptions, so read each description separately. Omitting the wall reaction or assuming both contacts are rough changes the moment equation and can make an otherwise neat calculation wrong. Keep the diagram tied to the specified surfaces rather than to a memorised example.