About this chapter
Statics is dynamics with the acceleration set to zero: every force balances. The skill is resolving in two directions and choosing those directions well, then handling friction, which is not automatically μR but only as large as it needs to be until the particle is on the point of slipping.
This chapter covers equilibrium, choosing directions to resolve, strings and pulleys, the triangle of forces, forces in components, friction at rest, limiting cases on slopes, inequalities for μ, and the normal reaction on a ring threaded on a rod. The longer questions find the least and greatest force that holds a particle on a slope.
The ten questions
- Q1 (6 marks): a 4 kg particle hanging from two strings at 30° and 45°: both tensions.
- Q2 (6 marks): three forces in equilibrium as vectors, then the acceleration when one is removed.
- Q3 (6 marks): a force at an angle on a particle at rest: the normal reaction and show μ ≥ 4/7.
- Q4 (6 marks): three horizontal forces in equilibrium, given by bearings: find the unknown sizes.
- Q5 (6 marks): a particle on the point of slipping down a slope, then the least force to move it up.
- Q6 (10 marks): a box on a slope held by a string at 30° to it, on the point of slipping.
- Q7 (10 marks): a particle held by two strings, one over a pulley to a hanging mass: the angle and tension.
- Q8 (11 marks): a horizontal force holding a particle on a rough slope: its least and greatest values.
- Q9 (10 marks): a ring on a rough rod pulled at 30°: whether it stays at rest, and the limiting force.
- Q10 (11 marks): a particle on a slope connected over a pulley to a hanging mass M: the range of M for equilibrium.
Key skills tested
Equilibrium. The resultant force is zero, so resolve in two perpendicular directions and set each total to zero. There is no ma term.
Choosing directions. Resolve perpendicular to a force you do not need, so it drops out. On a slope, resolve parallel and perpendicular to it.
Strings. Tension acts along the string, away from the particle. Over a smooth pulley the tension is the same on both sides.
Triangle of forces. Three forces in equilibrium form a closed triangle, so the sine rule links each force to the angle opposite it.
Forces in components. In equilibrium the i parts add to zero and the j parts add to zero. Removing one force leaves a resultant equal and opposite to it.
Friction at rest. Friction takes whatever value is needed, up to μR. It equals μR only when the particle is on the point of slipping.
Limiting cases. About to slip down, friction acts up the slope; about to slip up, it acts down. Each case gives one end of a range.
Inequalities. Staying at rest means friction is at most μR, so μ is at least friction ÷ R. Keep the inequality the right way round.
Normal reaction. R acts perpendicular to the surface, and a pull may reduce it, or even reverse it for a ring on a rod.
Worked example
Question 1 from this chapter. A 4 kg particle hangs in equilibrium from two strings, inclined to the horizontal at 30° and 45°. Find the tension in each string.
Resolving horizontally: T1 cos 30° = T2 cos 45°. Resolving vertically: T1 sin 30° + T2 sin 45° = 4g = 39.2.
From the first equation, T2 = T1 cos 30° ÷ cos 45°, which is about 1.225 T1. Substituting into the second: T1 (0.5 + 0.866) = 39.2, so T1 = 28.7 N and T2 = 35.1 N.
The weight is 4g, not 4. A triangle of forces, using the sine rule, gives the same answers.
Seen on real papers
Every question in our booklets is original. These are the recent papers where each question type has appeared.
- January 2025, Q6: a box on a slope held by a string, on the point of slipping, with the friction found from limiting equilibrium, as in Q6.
Where marks are lost
- Leaving friction out. In January 2025 a few did not write down an expression for the friction at all, although part (b) then needed it.
- μ not replaced. Some used F = μR but never substituted the value of μ, or gave R when F was asked for.
- A resultant not set to zero. A few wrote the difference of the forces as ma, then did not set a = 0 for equilibrium.
- Not understanding limiting friction. Students who thought friction could push the box up the slope lost the final mark.
Common questions
Which directions should I resolve in?
Pick one perpendicular to a force you do not need to find, so that force does not appear in the equation.
Is friction always equal to μR?
No. At rest it is only as large as needed. It reaches μR when the particle is on the point of moving.
How do I find a range of values for a force?
Work out the two limiting cases, one with friction up the slope and one with it down, and each gives one end of the range.
Where this chapter leads
- Mechanics 1 Chapter 8: Moments: equilibrium of rods, where forces must balance and have no turning effect.
- Mechanics 1 Chapter 5: Forces and Friction: the same slopes with the particle moving.