Forces and equilibrium

📚The mathematics

The force diagram

Almost every lost mark in mechanics traces back to a missing or invented force. Draw the body, then ask what touches it and what acts at a distance:

• Weight \( W = mg \), always vertically downwards, always present. Take \( g = 10 \) m s\(^{-2}\) unless told otherwise, and check what the question specifies.
• Normal contact force \(R\), perpendicular to the surface. On a slope it is perpendicular to the slope, not vertical — and it is not equal to the weight there.
• Tension in a string, always pulling away from the body along the string. A string can pull but never push.
• Friction \(F\), along the surface, opposing the motion or the tendency to move.
• Applied forces, as described in the question.

Mark each force with a labelled arrow. Do not put an arrow for “the force of motion” — motion is not a force, and adding one is a classic error.

Resolving

A force \(F\) at angle \( \theta \) to a chosen direction has component \( F\cos\theta \) along it and \( F\sin\theta \) perpendicular to it.

The reliable rule: the component adjacent to the marked angle takes \( \cos \), the opposite one takes \( \sin \). Never memorise “horizontal is cos” — it is false as soon as the angle is measured from the vertical, which questions do deliberately.

Resultants

To combine forces, resolve them all into two perpendicular directions, add the components separately, then recombine:

Give both the magnitude and the direction, and say what the direction is measured from — “at 34° above the horizontal” is an answer; “34°” is not. Sketch the resultant to check the quadrant, exactly as with the argument of a complex number.

Equilibrium

A body in equilibrium has zero resultant force, so the components balance in every direction. In practice, choose two perpendicular directions and write:

Choosing the directions well is most of the skill. On a slope, resolve along and perpendicular to the slope rather than horizontally and vertically — that way the normal force and any motion each lie along one of your axes, and only the weight needs splitting.

For exactly three forces in equilibrium, they form a closed triangle of forces, and the sine or cosine rule then solves it in one step. That is often much faster than resolving, and is worth spotting.

Friction

The inequality is the whole point. Friction is a response: it takes whatever value is needed to prevent motion, up to a maximum of \( \mu R \).

• If the body is in equilibrium and not on the point of moving, \( F \) is whatever the equilibrium equations say, and \( F < \mu R \).
• If the body is on the point of moving (“limiting equilibrium”, “about to slip”, “the least force required”), then and only then \( F = \mu R \).
• If the body is moving, friction is at its maximum, \( F = \mu R \), directed against the motion.

The inclined plane

For a body of mass \(m\) on a slope at angle \( \theta \) to the horizontal, resolving along and perpendicular to the slope:

Note \( R \ne mg \) on a slope — it is reduced by the \( \cos\theta \). Getting \( \sin \) and \( \cos \) the wrong way round here is the most frequent error on Paper 4; check by extremes. If \( \theta = 0 \) (flat), \( R \) should be the full \( mg \), and \( \cos 0 = 1 \) delivers that, while the down-slope component should vanish, and \( \sin 0 = 0 \) delivers that too.