MathsMechanics › Statics of a particle

Statics of a particle

Nothing moves, so everything cancels. The resultant force on a particle in equilibrium is zero in every direction you care to resolve. Two directions, two equations, and the unknowns fall out.

Builds on Friction and inclined planes.

IN THIS TOPIC

  • Resolve a set of coplanar forces in two perpendicular directions and set each sum to zero.
  • Solve string-and-weight equilibrium problems for unknown tensions.
  • Handle equilibrium on rough surfaces, including the least and greatest force that keeps an object still.

COMMON MISCONCEPTION

If a particle is in equilibrium, there are no forces acting on it.

Zero, twice

A particle in equilibrium has zero resultant force. In a plane that is two scalar statements, the components summing to zero horizontally and vertically, or along and perpendicular to a slope, whichever pair of directions makes the forces simplest. Choose the axes before writing anything down. The best pair usually lines up with the most awkward force.

A 10 newton weight held by strings at 30 and 60 degrees: tensions 5 and about 8.7 newtons balance it exactly10 NT₁ = 5 N at 30°T₂ ≈ 8.7 N at 60°
FIG. 1Two tensions, one weight, no motion: resolve twice and both unknowns surrender.

WORKED EXAMPLE

A weight on two strings

A 10 N weight hangs from two strings making 30° and 60° with the horizontal. Find both tensions.

Horizontally: T₁ cos 30° = T₂ cos 60°, so T₂ = √3 T₁.

Vertically: T₁ sin 30° + T₂ sin 60° = 10, so 0.5T₁ + 1.5T₁ = 10.

T₁ = 5 N and T₂ = 5√3 ≈ 8.7 N.

The steeper string carries more of the weight, which is what should happen and is worth checking every time.

Equilibrium with friction

On a rough surface, friction joins the balance sheet as the force that makes equilibrium possible. It is sized by whatever the other forces demand and capped at μR. A block resting on a slope is held there by friction acting up the slope with magnitude mg sin θ. Ask “what is the largest force before it slips?” and you are asking about limiting equilibrium, F = μR, alongside the same two resolutions as ever.

Because friction can act either way, a block on a rough slope usually sits still over a range of holding forces, and questions ask for both ends of that range.

The range of the holding force P on the rough slope: equilibrium for P between 5.25 and 28.3 newtons, with friction reversing direction between the two endsP (N)5.2528.3equilibrium: 5.25 ≤ P ≤ 28.3about to slip downP = 16.8 − 11.5 = 5.25about to slip upP = 16.8 + 11.5 = 28.3friction 11.5 N16.8 N down the slope throughout; only friction reverses
FIG. 2Friction reverses direction between the two ends of the range, so the same block needs a different holding force depending on which way it is about to slip: equilibrium holds for 5.25 ≤ P ≤ 28.3 N.

WORKED EXAMPLE

The range of a holding force

A 5 kg block rests on a rough slope at 20°, μ = 0.25, held by a force P acting up the slope. Find the least and greatest values of P for equilibrium.

Perpendicular: R = 5g cos 20° = 46.0 N, so μR = 11.5 N. Down the slope, 5g sin 20° = 16.8 N.

About to slip down, friction acts up the slope at full strength: P + 11.5 = 16.8, so P = 5.25 N.

About to slip up, friction reverses: P = 16.8 + 11.5 = 28.3 N.

So equilibrium holds for 5.25 ≤ P ≤ 28.3 N. Sketch the friction arrow both ways before writing either equation.

The wording deserves care. Show equilibrium is possible by showing the friction needed lies within the ceiling. Find the friction force by resolving, never by quoting μR.

ASSESSMENT FOCUS

  • Two resolutions, each set to zero, each labelled with its direction. That layout earns the method marks on its own.
  • Pick axes that simplify. Along and perpendicular to a slope is easier than horizontal and vertical whenever a slope is involved.
  • In “on the point of” questions, add F = μR to the two equilibrium equations and solve the trio.
  • Tension pulls away from the particle along the string. One arrow drawn backwards poisons both equations.

CHECK YOURSELF

A particle of weight 20 N on a smooth slope inclined at 30° is held by a string parallel to the slope. Find the tension and the normal reaction.

Show a hint

Resolve along the slope, then perpendicular to it.

Show the answer

Along: T = 20 sin 30° = 10 N.

Perpendicular: R = 20 cos 30° ≈ 17.3 N.

Equilibrium means the components sum to zero in two chosen directions.

Choose axes that flatten the geometry, and let friction fill the gap up to μR.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

5 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

  • Resolve a set of coplanar forces in two perpendicular directions and set each sum to zero.
  • Solve string-and-weight equilibrium problems for unknown tensions.
  • Handle equilibrium on rough surfaces, including the least and greatest force that keeps an object still.

Open the full revision checklist to see every objective in the course in one place.