Equilibrium of a particle | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Translate equilibrium into two zero sums

Definition

What is meant by a particle in equilibrium?

Model answer: A particle for which the vector sum of all forces acting on it is zero.

\sum F_x=0,\qquad \sum F_y=0

Swipe left or right to see the whole diagram →

Equilibrium is checked one direction at a time: the horizontal components cancel and the vertical components cancel.

A particle at rest can be in equilibrium.
Zero resultant also means zero acceleration, so a particle moving with constant velocity can be in equilibrium. Constant velocity means that both speed and direction stay unchanged.
You may resolve in any two perpendicular directions. Choose the pair that produces the simplest equations.

Let each direction supply one equation

Worked example

Recover one unknown force vector

A particle is acted on by 12 N to the right, 5 N upward and one unknown force $P$ . The particle is in equilibrium.

The unknown force must cancel the combined known force, so its components are 12 N left and 5 N down.

P=\sqrt{12^2+5^2}=13\text{ N}

\theta=\tan^{-1}\!\left(\dfrac{5}{12}\right)=22.6^\circ

Answer

P=13\text{ N}

Direction: $22.6^\circ$ below the leftward horizontal.

Key idea

If an unknown force lies along a known line, try resolving perpendicular to that line first. Its component is then zero, so the first equation can be solved without that unknown.

Common mistake

Writing

\sum F=0

is not enough when the forces are not collinear. Show the horizontal and vertical equations separately.

Do not assume different strings have equal tension

Each light string has a constant tension along itself, but two different strings can have different tensions. Resolve the forces on the selected particle rather than assigning equality from the letter “T”.

Draw the particle and every string direction.
Label different strings with different unknowns.
Resolve horizontally and vertically.
Check that the calculated tensions are positive.

Examiner note

A 2024 mark scheme explicitly rejects setting all three string tensions equal in a suspended-particle problem. The equality belongs along one light string, not automatically across different strings. (9709/42/O/N/24 Q4)

Your turnIndependent check [4 marks]

A particle of weight 30 N is held in equilibrium by a horizontal string of tension $S$ and another string of tension $T$ acting at 60° above the horizontal. Find $S$ and $T$ .

Show worked answer

Resolve vertically first because only one tension has a vertical component.

T\sin60^\circ=30

T=34.6\text{ N}

S=T\cos60^\circ=17.3\text{ N}

Answer

S=17.3\text{ N}

T=34.6\text{ N}

Translate equilibrium into two zero sums

Definition

What is meant by a particle in equilibrium?

Model answer: A particle for which the vector sum of all forces acting on it is zero.

\sum F_x=0,\qquad \sum F_y=0

Swipe left or right to see the whole diagram →

Equilibrium is checked one direction at a time: the horizontal components cancel and the vertical components cancel.

A particle at rest can be in equilibrium.
Zero resultant also means zero acceleration, so a particle moving with constant velocity can be in equilibrium. Constant velocity means that both speed and direction stay unchanged.
You may resolve in any two perpendicular directions. Choose the pair that produces the simplest equations.

Let each direction supply one equation

Worked example

Recover one unknown force vector

A particle is acted on by 12 N to the right, 5 N upward and one unknown force $P$ . The particle is in equilibrium.

The unknown force must cancel the combined known force, so its components are 12 N left and 5 N down.

P=\sqrt{12^2+5^2}=13\text{ N}

\theta=\tan^{-1}\!\left(\dfrac{5}{12}\right)=22.6^\circ

Answer

P=13\text{ N}

Direction: $22.6^\circ$ below the leftward horizontal.

Key idea

If an unknown force lies along a known line, try resolving perpendicular to that line first. Its component is then zero, so the first equation can be solved without that unknown.

Common mistake

Writing

\sum F=0

is not enough when the forces are not collinear. Show the horizontal and vertical equations separately.

Do not assume different strings have equal tension

Draw the particle and every string direction.
Label different strings with different unknowns.
Resolve horizontally and vertically.
Check that the calculated tensions are positive.

Examiner note

Your turnIndependent check [4 marks]

A particle of weight 30 N is held in equilibrium by a horizontal string of tension $S$ and another string of tension $T$ acting at 60° above the horizontal. Find $S$ and $T$ .

Show worked answer

Resolve vertically first because only one tension has a vertical component.

T\sin60^\circ=30

T=34.6\text{ N}

S=T\cos60^\circ=17.3\text{ N}

Answer

S=17.3\text{ N}

T=34.6\text{ N}