Equilibrium and centre of gravity

Hold a steering wheel with both hands and apply equal forces in opposite directions. Their sum may be zero, yet the wheel turns. What is missing when we represent a body as a single point particle?

For an extended body, where forces act matters as well as their sum. Equal opposite forces at different edges of a steering wheel can turn it without accelerating its centre. An initially stationary rigid body remains in equilibrium only if both the net force and net moment are zero. The first condition prevents translational acceleration; the second prevents rotation from starting. A point-particle model is no longer sufficient.

Try opening a door with the same force near its hinges and near its handle. The handle gives a greater effect because the force's line of action is farther from the axis. This perpendicular distance is the moment arm l. It is not always the door's width. Pushing from the handle directly towards the hinges puts the line of action through the axis, making the moment arm zero. Thus both the application point and force direction matter.

How can we compare two attempts to turn a door? In a planar problem, the moment magnitude is M = Fl: force F in newtons times moment arm l in metres, giving N·m. Doubling either force or moment arm doubles the moment. If l = 0, the moment is zero even for a large force. Moment describes the turning effect about a chosen axis. Having the same dimensions as work does not make it work or energy.

Choose one sign for clockwise turning and the opposite sign for anticlockwise turning. If only two significant moments act in opposite senses, equilibrium requires F₁l₁ = F₂l₂. Doubling one moment arm allows its force to be halved. The reaction at the pivot creates no moment about that pivot because its arm is zero. It still contributes to the net force.

Earth attracts every part of an object. In an almost uniform field, their combined effect can be replaced by one gravitational force whose line of action passes through the centre of gravity. For a uniform symmetric shape, symmetry locates this centre. A ring's centre lies in empty space: it is a point used to describe distributed forces, not necessarily a piece of matter. Uneven density can shift it away from the geometric centre.

Cut out an arbitrary flat shape and suspend it by an edge. Why does it settle in a particular orientation? If its centre of gravity is to one side of the suspension's vertical line, gravity has a moment arm and turns the shape. In the stable position, the centre lies below the suspension point. Draw this vertical line, then suspend the shape from another point and draw a second line. Their intersection gives the centre of gravity. The shape must turn freely without catching on the support.

Slowly tilt a box about an edge of its base. While the vertical through its centre of gravity lies inside the base, gravity tends to restore the box. Once the line passes outside the edge, gravity's moment continues the toppling. The edge itself is the limiting case. This reasoning assumes horizontal support and no additional overturning effects; base width and centre-of-gravity height together determine the allowable tilt.

Test stability by a small displacement. A ball in a bowl moves back after displacement: its equilibrium is stable. At a hilltop it moves farther away: unstable. On a horizontal plane it stays in its new position: neutral. 'Returns' means there is a restoring effect; without friction the ball may oscillate around the original point for a long time rather than stop there immediately. Rest alone does not distinguish these cases.

A force of 30 N acts on a lever with a moment arm of 0.4 m. What opposing moment should a force with a 0.6 m moment arm produce? Neglect the lever's weight and pivot friction.

For equilibrium, the second moment must have the same magnitude and opposite turning sense as the first. Choose the pivot as the axis: its reaction force has zero moment arm.

The first moment is M₁ = F₁l₁ = 30 · 0,4 = 12 Н·м.

For a 0.6 m moment arm, F₂l₂ = M₁ gives the required force: F₂ = 12/0,6 = 20 Н.

The second moment arm is 1.5 times longer, so its force is 1.5 times smaller. Checking the net force requires including the pivot reaction; equal moments alone do not replace this check.

Equilibrium requires a 20 N force with a 0.6 m moment arm, producing an opposing moment of 12 N·m.

An extended body can rotate with zero net force; equilibrium also requires checking moments.

The moment arm is the perpendicular distance to the line of action: a force through the axis does not turn the body about it.

Stability depends on what happens after a small displacement, rather than whether the body is currently at rest.

Our next way to describe interactions will track the momentum of the whole system.