Resolving vectors and equilibrium
Equilibrium means no acceleration and no rotation. This sheet focuses on the conditions for equilibrium, with interactive simulations you can use to explore tension, moments and force balance.
Work through it in three passes
Know the conditions
Learn the two conditions for equilibrium and how to draw a free-body diagram.
Play with the simulations
Use the tension, moment and force-polygon tools to build intuition for balanced forces.
Solve problems
Work through the examples and practice questions using the equilibrium method.
- State the two conditions for a body to be in equilibrium.
- Draw correct free-body diagrams for objects at rest.
- Resolve forces and set the sum of components equal to zero.
- Take moments about a pivot and set the sum of moments equal to zero.
- Solve cable-tension, beam and simple statics problems.
When an object is in equilibrium
An object is in equilibrium when it is not accelerating. For a rigid body this means two separate things must be true.
1. No resultant force
The vector sum of all forces is zero. In component form: ΣFx = 0 and ΣFy = 0.
2. No resultant moment
The sum of moments about any point is zero. Clockwise moments balance anticlockwise moments.
ΣF = 0 stops the object accelerating sideways or up/down. Στ = 0 stops it starting to rotate. Both are needed for a rigid body to remain completely at rest.
Two forces: equal and opposite, same line of action.
Three forces: lines of action meet at one point; can use Lami's theorem or triangle of forces.
Coplanar forces: resolve into two perpendicular directions and take moments.
How to draw a useful diagram
A free-body diagram shows only the forces acting on one object. It is the single most important step in solving equilibrium problems.
Do not draw forces acting on other objects. If a ladder rests against a wall, the forces on the ladder are its weight, the wall's normal reaction on the ladder, and the ground's reaction on the ladder. The force the ladder exerts on the wall belongs on the wall's diagram, not the ladder's.
Tension in two cables
A sign hangs from two cables at angles θ1 and θ2 to the horizontal. Change the angles and the weight to see how the tensions change.
Values update live as you type. The coloured arrows are the forces at the knot, scaled to size — together they close a triangle.
The tension in each cable increases as its angle to the horizontal gets smaller. A cable pulling almost horizontally must provide a very large tension to support the same vertical load.
Moments about a pivot
A uniform beam rests on a pivot. Add weights on either side and adjust their distances to see when the beam balances.
The beam is 4.0 m long and uniform. Distances are measured from the pivot; they snap to the beam if you go past the end. The beam tilts toward the larger moment.
Moment = force × perpendicular distance from pivot. For equilibrium, total clockwise moment equals total anticlockwise moment about any chosen point.
Three forces in equilibrium
Three forces are in equilibrium if their vector sum is zero, which means their head-to-tail arrows form a closed triangle. Enter three force vectors and see whether they close.
The three forces are drawn head-to-tail and auto-scaled to fit. If the arrows return to the start (a closed triangle), the point is in equilibrium.
For three forces in equilibrium, each force is proportional to the sine of the angle between the other two:
Very useful for problems with three forces, such as a mass on a slope held by a string.
Equilibrium problems
Example 1: sign hanging from two cables
A 60 N sign hangs from two cables at 40° and 50° to the horizontal. Find the tension in each cable.
Resolve vertically and horizontally. Let T₁ be the left cable at 40° and T₂ the right cable at 50°.
From the horizontal equation: T₁ = T₂ cos 50° / cos 40° = 0.839 T₂.
Substitute into the vertical equation:
Example 2: beam on a pivot
A uniform 4.0 m beam of weight 80 N is pivoted at its centre. A 30 N weight is placed 1.2 m from the pivot on the left. Where must a 50 N weight be placed on the right to balance the beam?
The beam is uniform, so its weight acts through the pivot and creates no moment. Take moments about the pivot.
Example 3: ladder against a wall
A 5.0 m ladder of weight 150 N rests against a smooth vertical wall with its base on rough ground. The ladder makes 60° with the ground. Find the normal reaction at the wall and the friction force at the ground.
Forces on the ladder: weight 150 N at the centre, wall reaction R (horizontal), ground reaction with vertical component N and horizontal friction F.
Take moments about the base to eliminate N and F:
Resolve horizontally: F = R = 43.3 N.
Resolve vertically: N = 150 N.
Example 4: mass on a slope held by a string
A 2.0 kg mass rests on a frictionless slope inclined at 30° to the horizontal. It is held by a string parallel to the slope. Find the tension.
Resolve parallel to the slope. The component of weight down the slope is mg sin θ.
Perpendicular to the slope, the normal reaction equals mg cos 30° = 17.0 N.
Common mistakes and quick fixes
Using sine instead of cosine
For a force at angle θ to the horizontal, the horizontal component is F cos θ and the vertical component is F sin θ.
Forgetting weight
Every object has weight. For uniform objects it acts at the geometrical centre.
Taking moments from the wrong point
Choose a point where unknown forces act — their moments are then zero, simplifying the equation.
Confusing distance with displacement
Moment = force × perpendicular distance from pivot. If the force is angled, use the perpendicular distance or resolve the force.
Self-check questions
1. A 10 kg box rests on a horizontal surface. What is the normal reaction?
Method: Resolve vertically. Weight = mg.
R = 10 × 9.81 = 98.1 N
2. A 20 N force acts at 30° above the horizontal. Find its horizontal and vertical components.
Method: Horizontal = F cos θ, vertical = F sin θ.
Fx = 20 cos 30° = 17.3 N; Fy = 20 sin 30° = 10 N
3. A uniform plank of weight 120 N rests on two supports at its ends. A 60 N person stands 1.5 m from the left end of the 4.0 m plank. Find the reaction forces at each support.
Method: Take moments about one support, then resolve vertically.
Moments about left support: RR × 4.0 = 120 × 2.0 + 60 × 1.5 → RR = 82.5 N
Vertically: RL + 82.5 = 120 + 60 → RL = 97.5 N
4. Three forces of 5.0 N, 5.0 N and 5.0 N act on a point at 0°, 120° and 240°. Is the point in equilibrium?
Method: Check that the vector sum is zero. By symmetry, equal forces at 120° intervals cancel out.
Yes, the forces form an equilateral triangle and the point is in equilibrium.
Final checklist for equilibrium problems
- Draw a clear free-body diagram showing all forces on the chosen object.
- Label angles and identify which forces are known and unknown.
- Choose perpendicular axes and resolve angled forces.
- Write ΣFx = 0 and ΣFy = 0.
- Choose a pivot that removes unknown forces from the moment equation.
- Write Σclockwise moments = Σanticlockwise moments.
- Check that your answers have sensible magnitudes and units.