Moments, Centre of Gravity and Equilibrium

The moment of a force about an axis is force multiplied by the perpendicular distance from that axis to the force’s line of action. This distance is not necessarily the distance along a beam to the point where the force is applied. Choose a pivot, draw the line of action, then identify its shortest separation from the pivot.

Assign clockwise and anticlockwise senses consistently. A force whose line of action passes through the pivot has zero moment about that pivot, even if its magnitude is large. For an angled force, resolve perpendicular to the beam or use the correct perpendicular lever arm; these give the same result.

For a rigid body in static equilibrium the vector sum of forces is zero and the sum of clockwise moments equals the sum of anticlockwise moments about any point. Balancing moments alone cannot establish vertical or horizontal force balance.

A useful pivot eliminates an unknown reaction from the moment equation. After finding one support reaction, use force balance to find the other. Check that the two reactions add to the total supported weight. A negative calculated contact reaction can indicate that the assumed support contact cannot be maintained.

The centre of gravity is the point through which the resultant weight acts. For a uniform straight rod in a uniform gravitational field it is at the midpoint. An irregular object can be suspended from different points: each vertical plumb line passes through its centre of gravity, so their intersection locates it.

For a freely standing object, tipping begins when the vertical line through the centre of gravity passes outside the base of support. A wider base or lower centre of gravity usually increases the tilt needed to reach that condition. State the geometry rather than saying a heavy object must always be stable.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.