Stepwise
Start practisingPractise

Statics: forces, equilibrium & moments · Lesson 2 of 3

Equilibrium of a particle

A particle is in equilibrium when the forces on it balance: resolve in two directions, use Lami's theorem or the triangle of forces, and handle friction on rough planes.

20 minYou should already know: Vectors
  1. 1
  2. 2
  3. 3

A particle is in equilibrium when the resultant of all the forces on it is zero. It then stays at rest (or keeps moving at a steady speed). In components, that means two equations:

  • the parts in one direction add up to 0 (up = down);
  • the parts in the direction at right angles add up to 0 (left = right).

So one force that keeps a body in equilibrium is the resultant of the others, reversed.

Strings and resolving

A weight held by strings is the commonest case. Resolve across and up at the knot:

αβT₁T₂W
A weight on two stringsAcross: T₁ cos α = T₂ cos β. Up: T₁ sin α + T₂ sin β = W
Tensions in two stringsSet the weight and the angles
T₁T₂W
23.03 NT₁ (string at 40°)30.76 NT₂ (string at 55°)
Across: T₁ cos 40° = T₂ cos 55°. Up: T₁ sin 40° + T₂ sin 55° = 40. Solving, T₁ = 40 cos 55° ÷ sin 95° = 23.03 N and T₂ = 40 cos 40° ÷ sin 95° = 30.76 N. The steeper string (at 55°) carries more of the load.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q8

In the diagram, a mass of 12 kg12\text{ kg} hanging from a light inextensible string is pulled aside by a horizontal force RR, such that the string is inclined at 45∘45^\circ to the vertical. If the system is in equilibrium, calculate the: [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

TR12 kg45°P

tension in the string;

value of RR.

  1. The forces at the knot

    • The weight is 12×10=120{12 \times 10 = 120} N.
    • The string is at 45∘45^\circ to the vertical.

    Think first. Three forces: the tension T along the string, R across, and the weight down.

  2. Resolve up

    • Tcos⁡45∘=120{T\cos 45^\circ = 120}, so T=120cos⁡45∘=1202≈169.71{T = \frac{120}{\cos 45^\circ} = 120\sqrt2 \approx 169.71} N.

    Think first. Only T has an upward part.

  3. Resolve across

    • R=Tsin⁡45∘{R = T\sin 45^\circ}.
    • =1202×12=120{= 120\sqrt2 \times \frac{1}{\sqrt2} = 120} N.

More: strings

Lami’s theorem

When exactly three forces keep a particle in equilibrium, each force is proportional to the sine of the angle between the other two:

γβαPQW
Lami's theoremP ÷ sin α = Q ÷ sin β = W ÷ sin γ
Psin⁡α=Qsin⁡β=Wsin⁡γ\frac{P}{\sin\alpha} = \frac{Q}{\sin\beta} = \frac{W}{\sin\gamma}

Each angle is the one between the other two forces, opposite the force on top.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q7

A body of mass 1.5 kg1.5\text{ kg} is suspended by two light inextensible ropes inclined at 30∘30^\circ and 60∘60^\circ to the horizontal. Calculate the tensions in the ropes. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

  1. The angles

    • The weight is 1.5×10=15{1.5 \times 10 = 15} N, straight down.
    • The ropes meet at 180∘−30∘−60∘=90∘{180^\circ - 30^\circ - 60^\circ = 90^\circ}.
    • The rope at 30∘30^\circ makes 90∘+30∘=120∘{90^\circ + 30^\circ = 120^\circ} with the weight; the rope at 60∘60^\circ makes 150∘{150^\circ}.

    Think first. What angle do the two ropes make with each other, and with the weight?

  2. Lami

    • T1sin⁡150∘=T2sin⁡120∘=15sin⁡90∘{\frac{T_1}{\sin 150^\circ} = \frac{T_2}{\sin 120^\circ} = \frac{15}{\sin 90^\circ}}.
    • T1=15sin⁡150∘=7.5{T_1 = 15\sin 150^\circ = 7.5} N (the rope at 30∘30^\circ).
    • T2=15sin⁡120∘≈12.99{T_2 = 15\sin 120^\circ \approx 12.99} N (the rope at 60∘60^\circ).

    Think first. Each tension over the sine of the angle between the other two forces.

The triangle of forces

Three forces in equilibrium, drawn head to tail, close up into a triangle. The cosine rule then links them. Take care: the angle inside the triangle is not the angle between the forces.

60°120°10 N16 N14 N
Three forces in equilibriumInside the triangle 60°; between the forces 180° − 60° = 120°

More: forces in equilibrium

Rough planes and friction

On a slope, resolve along the plane and at right angles to it. The weight mgmg splits into mgsin⁡θmg\sin\theta down the slope and mgcos⁡θmg\cos\theta into the slope:

θmgRFmg sin θmg cos θ
A block on a rough slopeR = mg cos θ; limiting friction F = μR, against the motion

Friction acts against the way the body would move. When the body is just about to slide (limiting friction), F=μRF = \mu R.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q14

A body of mass 10 kg10\text{ kg} rests on a rough plane inclined at an angle of tan⁡−1(512)\tan^{-1}\left(\frac{5}{12}\right) to the horizontal. The coefficient of friction between the body and the plane is 34\frac34. A force of magnitude P NP\text{ N} acts on the body along the inclined plane. Find the value of PP if the body is at the point of moving: [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

down the plane;

up the plane.

  1. The slope

    • The triangle is 5, 12, 13, so sin⁡θ=513{\sin\theta = \frac{5}{13}} and cos⁡θ=1213{\cos\theta = \frac{12}{13}}.
    • The weight is 10×10=100{10 \times 10 = 100} N.

    Think first. tan θ = 5/12. What are sin θ and cos θ?

  2. Reaction and friction

    • R=100×1213≈92.31{R = 100 \times \frac{12}{13} \approx 92.31} N.
    • Limiting friction: μR=34×92.31≈69.23{\mu R = \frac34 \times 92.31 \approx 69.23} N.
    • Weight down the slope: 100×513≈38.46{100 \times \frac{5}{13} \approx 38.46} N.

    Think first. R = mg cos θ.

  3. About to move down

    • P+38.46=69.23{P + 38.46 = 69.23}, so P≈30.77{P \approx 30.77} N.

    Think first. Friction acts up the slope. P pushes down.

  4. About to move up

    • P=38.46+69.23≈107.69{P = 38.46 + 69.23 \approx 107.69} N.

    Think first. Now friction acts down the slope.

More: planes and friction

Your turn

WAEC 2022 · Paper 2 · Q7

A body of mass 18 kg18\text{ kg} is suspended by an inextensible string from a rigid support and is pulled by a horizontal force FF until the angle of inclination of the string to the vertical is 35∘35^\circ. If the system is in equilibrium, calculate the:

  1. (a)

    value of FF;

  2. (b)

    tension in the string.

Worked solution (try it first)
  1. The weight is 18×10=180 N18 \times 10 = 180\text{ N}.
  2. The string is at 35∘35^\circ to the vertical.
  3. Up: Tcos⁡35∘=180T\cos35^\circ = 180.
  4. Across: Tsin⁡35∘=FT\sin35^\circ = F.

(a)

  1. Divide: F=180tan⁡35∘F = 180\tan35^\circ
    ≈126.04 N\approx 126.04\text{ N}.

(b)

  1. T=180cos⁡35∘T = \dfrac{180}{\cos35^\circ}
    ≈219.74 N\approx 219.74\text{ N}.

Report a problem with this question