⚖ Statics — Forces, Moments & Equilibrium
Statics is the study of objects that are not accelerating. Engineers use it to design beams, brackets, structures and connections that stay still under load — calculating reaction forces, bending moments and checking nothing tips or slides.
2Topics
8Examples
4Calculators
12Practice Qs
📌 Level 3 context: Every beam, bracket, shelf support, crane jib and retaining wall must satisfy two equilibrium conditions: all forces sum to zero, and all moments sum to zero. Get either wrong and the structure moves, tips or collapses.
⚖Forces & MomentsM = F × d, resultants
⚖EquilibriumΣF = 0, ΣM = 0
⚙️Applications4 structural calculators
🎮Game12 randomised questions
📝Practice12 graded questions
⚖
Forces & Moments
A moment is the turning effect of a force about a pivot point. It depends on both the magnitude of the force and its perpendicular distance from the pivot.
🎮Live explorer — moment calculatorDrag sliders — diagram updates instantly
💡 Moment = how hard a force tries to rotate something. Double the distance and you double the turning effect.
📋Try it — Moment & Resultant CalculatorUpdates as you type
Select what to find and enter values above.
📚 Key vocabulary
- Moment (M)
- the turning effect of a force about a pivot — force multiplied by perpendicular distance. Unit: N m.
- Perpendicular distance
- the shortest distance from the line of action of the force to the pivot point.
- Resultant force
- the single force that has the same effect as all individual forces combined.
- Coplanar forces
- forces acting in the same plane — the common case for 2D statics problems.
- Couple
- two equal and opposite parallel forces that produce a pure turning effect with no resultant translation.
- Torque
- another name for moment, especially in rotating systems — units N m, same as moment.
👁️ Visual aid — resolving forces into components: a 500 N force at 30° splits into horizontal and vertical parts
✍️Worked example 1 — moment of a force step through it
QuestionA 300 N force acts vertically on a horizontal spanner at a perpendicular distance of 0.4 m from the nut. Calculate the moment.
Formula
M = F × dSubstitute
M = 300 × 0.4Multiply
300 × 0.4 = 120AnswerM = 120 N m (clockwise) ✓
Step 1 of 5
✍️Worked example 2 — resultant of parallel forces
QuestionTwo forces act at the same point: 400 N to the right and 150 N to the left. Find the resultant.
Convention
Right = positive, left = negativeResultant
R = +400 + (−150) = +250AnswerR = 250 N to the right
✍️Worked example 3 — resolving a force into components
QuestionA 600 N force acts at 40° above the horizontal. Find the horizontal and vertical components.
Horizontal
Fₓ = 600 × cos 40° = 600 × 0.766 = 459.6 NVertical
Fᵧ = 600 × sin 40° = 600 × 0.643 = 385.7 NAnswerFₓ = 459.6 N Fᵧ = 385.7 N
✍️Worked example 4 — finding force from a moment
QuestionA moment of 840 N m is produced about a pivot. The perpendicular distance is 1.2 m. Find the force.
Formula
F = M ÷ dSubstitute
F = 840 ÷ 1.2Divide
840 ÷ 1.2 = 700AnswerF = 700 N
⚖
Equilibrium & Reactions
An object is in equilibrium when the resultant force and the resultant moment are both zero. This lets us find unknown reaction forces in beams, frames and structures.
🎮Live explorer — simply supported beamDrag sliders to find reaction forces
R₁ (left)
625 N
R₂ (right)
375 N
Check R₁+R₂
1000 N ✓
💡 The closer the load to a support, the greater that support's reaction. Load at midspan → both reactions = W/2.
📋Try it — Beam Reaction CalculatorUpdates as you type
Enter load, span and position to find reactions.
📚 Key vocabulary
- Equilibrium
- the state where the resultant force and resultant moment on a body are both zero — it is not accelerating.
- Reaction force
- a force exerted by a support on a structure — equal and opposite to the load the support carries.
- Simply supported beam
- a beam resting on two supports that can only exert vertical (upward) reactions.
- ΣM = 0
- sum of all moments about any point equals zero — used to find one unknown reaction at a time.
- Taking moments about a point
- choosing a pivot to eliminate one unknown — the reaction at that point has zero moment arm.
- Free body diagram (FBD)
- a diagram showing all external forces on a body, used as the starting point for every statics problem.
👁️ Visual aid — simply supported beam with two point loads: take moments about R₁ to find R₂
✍️Worked example 5 — simply supported beam, one point load step through it
QuestionA simply supported beam of span 6 m carries a 900 N point load 2 m from the left support. Find R₁ and R₂.
Moments at R₁
Take moments about R₁ (eliminates R₁)ΣM = 0
R₂ × 6 = 900 × 2 = 1800R₂
R₂ = 1800 ÷ 6 = 300 NΣFᵧ = 0
R₁ + R₂ = 900 → R₁ = 900 − 300 = 600 NCheckR₁=600N, R₂=300N 600+300=900 ✓
Step 1 of 6
✍️Worked example 6 — two point loads on a simply supported beam
QuestionBeam span 4 m. Load 1: 400 N at 1 m from left. Load 2: 600 N at 3 m from left. Find R₁ and R₂.
ΣM at R₁
R₂ × 4 = (400 × 1) + (600 × 3) = 400 + 1800 = 2200R₂
R₂ = 2200 ÷ 4 = 550 NR₁
R₁ = (400 + 600) − 550 = 1000 − 550 = 450 NCheck450 + 550 = 1000 N ✓
✍️Worked example 7 — beam with UDL (uniformly distributed load)
QuestionA 5 m simply supported beam carries a UDL of 200 N/m over its full length. Find R₁ and R₂.
Total load
W = 200 × 5 = 1000 N, acting at midspan (2.5 m from each end)Symmetry
R₁ = R₂ = 1000 ÷ 2 = 500 NCheck500 + 500 = 1000 N ✓
✍️Worked example 8 — checking moment equilibrium for a bracket
QuestionA wall bracket carries a 2400 N load at 0.6 m from the wall. A tie rod at 0.8 m from the wall resists tipping. Find the tension T in the tie rod.
ΣM at wall
T × 0.8 = 2400 × 0.6 = 1440T
T = 1440 ÷ 0.8 = 1800AnswerT = 1800 N
⚙️
Engineering Applications
Four structural calculators — results update as you type.
⚖ Beam reactions — single point load
Find the support reactions for a simply supported beam with a single point load anywhere along its span.
R₂ = W × a ÷ L R₁ = W − R₂
🧮 Calculator
Enter load, span and load position.
🔨 Spanner / torque — moment
Calculate the torque applied to a fastener — force on spanner multiplied by effective length.
M = F × d
🧮 Calculator
Enter force and spanner length.
📈 Force components — resolve at angle
Resolve a force at any angle into horizontal and vertical components for structural analysis.
Fₓ = F cosθ Fᵧ = F sinθ
🧮 Calculator
Enter force and angle.
🔋 Two-load beam reactions
Find reactions for a simply supported beam carrying two separate point loads at different positions.
R₂×L = W₁×a₁ + W₂×a₂ R₁ = W₁+W₂−R₂
🧮 Calculator
Enter both loads, positions and span.
🎮
Statics Challenge
12 randomised questions per round — moments, reactions and components.
⚖ Quick-fire round
Question 1 / 12
Score: 0
Streak: 0 🔥
Press Start to begin!
📝
Practice Questions
12 graded questions — type your answer to check it, then reveal full working.
⚖ Show your free body diagram, list all forces, apply equilibrium conditions. Answers to 1 d.p. where appropriate.
Q1 · Moment1 mark
Calculate the moment produced by a 250 N force acting at a perpendicular distance of 0.8 m from a pivot.
M = F × d = 250 × 0.8 = 200 N m
Q2 · Moment2 marks
A torque of 600 N m is required to tighten a bolt. The spanner is 0.4 m long. What force must be applied?
F = M ÷ d = 600 ÷ 0.4 = 1500 N
Q3 · Moment2 marks
A moment of 360 N m is produced by a force of 450 N. Find the perpendicular distance from the pivot.
d = M ÷ F = 360 ÷ 450 = 0.8 m
Q4 · Resultant2 marks
Three collinear forces act on a bracket: 500 N right, 200 N right, 350 N left. Find the resultant and its direction.
R = 500 + 200 − 350 = 350 N to the right
Q5 · Components2 marks
A 700 N force acts at 50° to the horizontal. Find the horizontal component. (cos 50° = 0.643)
Fₓ = F cosθ = 700 × 0.643 = 450.1 N
Q6 · Components2 marks
Find the vertical component of a 450 N force at 25° above horizontal. (sin 25° = 0.423)
Fᵧ = F sinθ = 450 × 0.423 = 190.35 N
Q7 · Beam reactions3 marks
A simply supported beam of span 8 m carries a 1200 N point load 3 m from the left support. Find R₁ and R₂.
ΣM at R₁: R₂×8 = 1200×3 = 3600 → R₂ = 450 N
R₁ = 1200−450 = 750 N
Check: 750+450=1200 ✓
R₁ = 1200−450 = 750 N
Check: 750+450=1200 ✓
Q8 · Beam reactions3 marks
A 6 m beam carries 800 N at 2 m from the left and 400 N at 5 m from the left. Find R₁ and R₂.
ΣM at R₁: R₂×6 = 800×2 + 400×5 = 1600+2000 = 3600
R₂ = 3600÷6 = 600 N
R₁ = 1200−600 = 600 N
R₂ = 3600÷6 = 600 N
R₁ = 1200−600 = 600 N
Q9 · UDL2 marks
A 4 m beam carries a UDL of 300 N/m across its full length. Find R₁ and R₂.
Total W = 300×4 = 1200 N at midspan
By symmetry: R₁ = R₂ = 1200÷2 = 600 N
By symmetry: R₁ = R₂ = 1200÷2 = 600 N
Q10 · Bracket / tie rod3 marks
A bracket fixed at a wall carries 3000 N at 0.5 m from the wall. A tie rod connects 1.0 m from the wall to prevent rotation. Find the tie rod tension T.
ΣM at wall: T×1.0 = 3000×0.5 = 1500
T = 1500 N
T = 1500 N
Q11 · Combined3 marks
A 10 m beam carries 2000 N at 4 m from the left and a UDL of 100 N/m over its full length. Find R₁ and R₂.
UDL total = 100×10=1000N acting at 5m
ΣM at R₁: R₂×10 = 2000×4 + 1000×5 = 8000+5000=13000
R₂ = 1300 N
R₁ = 3000−1300 = 1700 N
ΣM at R₁: R₂×10 = 2000×4 + 1000×5 = 8000+5000=13000
R₂ = 1300 N
R₁ = 3000−1300 = 1700 N
Q12 · Combined3 marks
A simply supported beam span 5 m carries 600 N at 1 m and 900 N at 4 m from the left support. Find R₁, R₂ and verify equilibrium.
ΣM at R₁: R₂×5 = 600×1 + 900×4 = 600+3600=4200
R₂ = 4200÷5 = 840 N
R₁ = 1500−840 = 660 N
Check: 660+840=1500 ✓
R₂ = 4200÷5 = 840 N
R₁ = 1500−840 = 660 N
Check: 660+840=1500 ✓