LESSON 01
Modelling Rigid Bodies
In mechanics, real objects are often replaced by mathematical models. A rigid body is a model of an object whose shape and dimensions are assumed to remain unchanged when forces act on it.
A body for which the distances between all pairs of points are assumed to remain constant.
Why use the rigid-body model?
Real objects deform slightly. A beam bends, a ladder flexes and a metal rod stretches by a small amount. In many mechanics problems these changes are sufficiently small to ignore.
Shape fixed
The geometry of the body does not change while the forces act.
Forces still act
The body may experience several forces, reactions, tensions and weights.
Rotation matters
Unlike a particle model, a rigid body can have a turning effect about a point.
Equilibrium matters
Both translational and rotational equilibrium must be considered.
When a question says that a rod or beam is rigid, do not introduce bending or deformation into your equations. Treat the geometry as fixed.
Common modelling assumptions
- A light rod has negligible mass, so its weight is ignored.
- A uniform rod has its centre of mass at its midpoint.
- A smooth contact produces a reaction perpendicular to the contacting surface.
- A rough contact can provide a frictional force as well as a normal reaction.
- A particle has size small enough that its rotation does not need to be considered.
LESSON 02
Forces and Lines of Action
A force is completely described by its magnitude, direction and line of action. For moments, the line of action is particularly important.
The straight line extending through the point of application of a force in the direction of the force.
The moment of a force about a point depends on the perpendicular distance between that point and the force's line of action.
where d is the perpendicular distance from the chosen point to the line of action of the force.
Perpendicular distance from the pivot to the line of action.
Distance from the pivot to the point where the force happens to be drawn.
Force through the pivot
If the line of action of a force passes directly through the pivot, its perpendicular distance from the pivot is zero. Therefore its moment is zero.
A force can be very large and still produce zero moment about a point if its line of action passes through that point.
LESSON 03
Moments of Forces
The moment of a force about a point measures its turning effect about that point.
The product of the force and the perpendicular distance from the point to the force's line of action.
Therefore:
Units
Force is measured in newtons and distance in metres, so moment is measured in:
A joule and a newton-metre have the same base dimensions, but they represent different physical quantities. A joule is a unit of energy/work; N m in this context represents a turning effect.
Direction of a moment
A force can produce either a clockwise or anticlockwise turning effect about the chosen point.
Clockwise
The force tends to rotate the body clockwise about the chosen point.
Anticlockwise
The force tends to rotate the body anticlockwise about the chosen point.
Sign convention
In algebraic solutions, one direction can be chosen as positive. For example:
- clockwise = positive
- anticlockwise = negative
or the reverse. Either convention is valid as long as it is used consistently.
LESSON 04
Equilibrium of a Rigid Body
A rigid body is in static equilibrium when it has no translational acceleration and no rotational acceleration.
This requires two independent conditions.
Translational equilibrium
The resultant force is zero.
Rotational equilibrium
The resultant moment is zero.
Horizontal and vertical equilibrium
For a two-dimensional problem, it is often useful to resolve forces horizontally and vertically.
Together with one moment equation, these provide the equilibrium conditions for a planar rigid body.
The resultant force must be zero and the resultant moment about any point must also be zero.
A body can have zero resultant moment about one point while still having a non-zero resultant force. Such a system is not in complete static equilibrium.
LESSON 05
Taking Moments About a Point
Taking moments about a carefully selected point is one of the most powerful techniques in rigid-body statics.
For a body in equilibrium, the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about that point.
Why choose a particular pivot?
Suppose a beam is supported at A and B and the reaction at A is unknown. If moments are taken about A, the reaction at A has zero moment because its line of action passes through A.
This removes that unknown from the moment equation.
Choose a point through which as many unknown forces as possible act. Their moments then become zero.
Taking moments about different points
If a rigid body is in equilibrium, the resultant moment is zero about any point.
The choice of point changes the appearance of the equation, not the physical equilibrium condition.
LESSON 06
Weight and Centre of Mass
The gravitational force acting on a body is called its weight.
where:
- W is weight in newtons
- m is mass in kilograms
- g is acceleration due to gravity
In Edexcel mechanics questions, the value of g should normally be taken from the value specified in the question or from the standard value expected by the paper.
Centre of mass
The weight of a body acts vertically downward through its centre of mass.
For a uniform rod, the centre of mass is at its geometrical midpoint.
A uniform rod of length 6 m has weight 80 N. It is pivoted at one end. Find the moment of its weight about the pivot.
The weight acts at the midpoint:
6 ÷ 2 = 3 m
Therefore:
M = 80 × 3
M = 240 N m
Do not automatically place the weight at the end of a uniform rod. The weight acts through the centre of mass, which is at the midpoint.
LESSON 07
Couples
A couple consists of two equal, opposite and parallel forces whose lines of action are separated.
- The two forces have equal magnitude.
- The forces act in opposite directions.
- The forces are parallel.
- The resultant force is zero.
- The couple still produces a non-zero turning effect.
If the force magnitude is F and the perpendicular distance between the two lines of action is d:
Why is a couple special?
A single force may produce both translation and rotation. A pure couple produces rotation without a resultant force.
Two equal and opposite forces of 50 N act with perpendicular separation 0.8 m. Find the moment of the couple.
M = Fd
M = 50 × 0.8
M = 40 N m
The moment of a couple is independent of the point about which moments are taken.
LESSON 08
Rods, Beams and Rigid Members
Many statics questions involve a rod or beam acted on by several forces. The most important first step is to identify every force acting on the body.
Light rods
A light rod has negligible mass, so its own weight is ignored. Only the external forces shown or described in the question are included.
Uniform rods
A uniform rod has its centre of mass at its midpoint, so its weight acts vertically downward at the midpoint.
General force balance
For a horizontal beam:
For example, if upward reactions are R₁ and R₂ and downward loads are W₁ and W₂:
A second equation is normally obtained by taking moments about one support.
A light horizontal beam is 8 m long. It carries a 120 N load 4 m from support A. The supports are at A and B. Find the reactions at A and B.
Take moments about A:
RB × 8 = 120 × 4
8RB = 480
RB = 60 N
Now use vertical equilibrium:
RA + 60 = 120
RA = 60 N
Unequal loads
If a load is not centred between two supports, the reactions are generally unequal. The support closer to the larger moment-producing load may carry the larger reaction.
LESSON 09
Support Reactions
Supports exert forces on bodies. The direction and type of reaction depend on the model used in the question.
Pin / hinge
A hinged support can provide horizontal and vertical reaction components but does not allow the body to translate through the support.
Smooth contact
The reaction acts perpendicular to the surface at the point of contact.
String / cable
A string provides tension along the string and pulls rather than pushes.
Support reaction
The reaction direction should be determined from the geometry and contact conditions given.
Two-support strategy
- Draw all known forces and unknown reactions.
- Use the moment equation to find one reaction.
- Use force equilibrium to find the remaining reaction.
- Check that the final values satisfy both equations.
Taking moments about one support usually removes the reaction at that support from the equation. This is often the fastest route to the other reaction.
LESSON 10
Ladders and Contact Problems
A ladder is modelled as a rigid body. The forces acting on it depend on the contact conditions at its ends.
For a ladder resting against a wall, identify:
- the weight of the ladder
- the reaction at the ground
- the reaction at the wall
- any frictional forces if the surfaces are rough
- any applied forces or loads
Smooth surfaces
At a smooth surface there is no frictional force. The reaction is normal to the surface.
For example, at a smooth vertical wall the reaction is horizontal. At a smooth horizontal floor the normal reaction is vertical.
Rough surfaces
A rough surface may provide both a normal reaction and friction. The direction of friction opposes the tendency of relative motion.
Do not automatically assume that friction always acts towards the wall or away from the wall. Determine the direction by considering the tendency of motion.
Ladder method
- Draw the ladder as a rigid body.
- Mark its weight through its centre of mass.
- Draw every contact reaction in the correct direction.
- Resolve forces horizontally and vertically if necessary.
- Take moments about the most convenient point.
- Solve the simultaneous equations.
LESSON 11
Forces at an Angle
Not every force acts perpendicular to the rod. When a force is inclined, the moment can be found using either the perpendicular distance to the line of action or a resolved component.
Using the perpendicular component
If a force F acts at angle θ to the position vector from the pivot to the point of application:
because the perpendicular component of the force is F sin θ.
Using perpendicular distance
Alternatively, determine the perpendicular distance d from the pivot to the line of action:
Both methods give the same physical moment when the geometry is handled correctly.
A 40 N force acts at an angle of 30° to a rod. The point of application is 3 m from the pivot. Find the moment.
M = Fr sin θ
M = 40 × 3 × sin 30°
M = 120 × 0.5
M = 60 N m
Make sure θ is the angle between the force and the line from the pivot to the point of application. If the diagram gives a different angle, convert it before using F r sin θ.
LESSON 12
The Complete Exam Method
A strong statics solution is systematic. Use the following process rather than immediately writing an equation.
- Identify the body. Decide exactly which object you are analysing.
- Draw all forces. Include weight, reactions, tension, applied forces and friction where appropriate.
- Locate the centre of mass. For a uniform rod, this is the midpoint.
- Choose a moment point. Look for a point that removes one or more unknown reactions.
- Write a moment equation. Use perpendicular distances or resolve angled forces.
- Use force equilibrium. Resolve horizontally and vertically where required.
- Solve algebraically. Keep exact values until the final stage where practical.
- Check signs and units. Re-read the geometry and make sure every moment direction is correct.
Before calculating, ask: “What point should I take moments about?” This single decision often determines how simple the entire solution becomes.
LESSON 13
Common Mistakes and Exam Traps
| Mistake | Correct approach |
|---|---|
| Using distance to the force rather than perpendicular distance | Use the shortest perpendicular distance to the line of action. |
| Forgetting a force through the pivot has zero moment | Its perpendicular distance is zero. |
| Putting a uniform rod's weight at the end | Place the weight at the midpoint. |
| Mixing clockwise and anticlockwise signs | Choose one sign convention and maintain it. |
| Using moments without force equilibrium | For full static equilibrium, also require ΣF = 0. |
| Using the wrong angle in Fr sin θ | θ must be the angle between the force and the position vector. |
| Calling a couple a single force | A couple consists of two equal, opposite, parallel forces. |
| Forgetting support reactions | Draw a complete force diagram before calculating. |
| Wrong units | Moment is measured in N m. |
| Rounding too early | Keep sufficient accuracy throughout the calculation. |
LESSON 14
Worked Examples — Foundation to Exam Level
A 20 N force acts perpendicular to a rod at a distance of 3 m from a pivot. Find its moment.
M = Fd
M = 20 × 3
M = 60 N m
A 100 N force acts along a rod through its pivot. Find its moment.
The line of action passes through the pivot.
Therefore d = 0.
M = 100 × 0
M = 0 N m
A 30 N force acts 2 m from a pivot. An unknown force F acts 3 m from the pivot on the opposite side. The rod is in equilibrium. Find F.
Clockwise moment = anticlockwise moment.
30 × 2 = F × 3
60 = 3F
F = 20 N
A uniform rod of length 5 m and weight 100 N is hinged at one end. Find the moment of its weight about the hinge.
The weight acts at the midpoint.
Distance = 5/2 = 2.5 m
M = 100 × 2.5
M = 250 N m
A 10 m beam carries a 200 N load at a point 4 m from the left support A. The right support is B. Find the reaction at B.
Take moments about A.
RB × 10 = 200 × 4
10RB = 800
RB = 80 N
Using the previous example, find the reaction at A.
Vertical equilibrium:
RA + RB = 200
RA + 80 = 200
RA = 120 N
A 60 N force acts at 40° to a rod. Its point of application is 2 m from the pivot. Find its moment.
M = Fr sin θ
M = 60 × 2 × sin 40°
M ≈ 120 × 0.6428
M ≈ 77.1 N m
Two equal and opposite forces of 80 N form a couple whose perpendicular separation is 0.5 m. Find the couple moment.
M = Fd
M = 80 × 0.5
M = 40 N m
A light horizontal beam has a pivot at its left end. A 20 N force acts downward at 2 m and a 35 N force acts upward at 5 m. Find the resultant moment about the pivot.
Take anticlockwise as positive.
35 × 5 − 20 × 2
= 175 − 40
Resultant moment = 135 N m anticlockwise
A beam is supported at A and B. Both support reactions are unknown. Explain why taking moments about A is useful.
The reaction at A acts through A.
Therefore its perpendicular distance from A is zero.
Its moment about A is therefore zero.
This leaves an equation involving the reaction at B and the known loads, allowing B to be found directly.
A horizontal beam is 6 m long. It is supported at both ends. A 120 N downward load acts 2 m from the left support. The beam itself has weight 60 N acting at its midpoint. Find both support reactions.
Let the reactions be RA and RB.
Take moments about A:
RB(6) = 120(2) + 60(3)
6RB = 240 + 180
6RB = 420
RB = 70 N
Vertical equilibrium:
RA + 70 = 120 + 60
RA = 110 N
A 50 N load is placed at a distance x from a pivot. A 20 N load acts 4 m from the pivot on the opposite side. The system is balanced. Find x.
50x = 20 × 4
50x = 80
x = 1.6 m
About a point O, a 25 N force produces a clockwise moment of 100 N m. A second force produces an anticlockwise moment of 60 N m. What additional anticlockwise moment is required for equilibrium?
Total clockwise moment = 100 N m.
Existing anticlockwise moment = 60 N m.
Required additional moment:
100 − 60 = 40
40 N m anticlockwise
A 100 N force acts at 60° to a 4 m rod from its pivot. Find its moment.
Perpendicular component = 100 sin 60°.
= 86.60 N approximately.
Moment = 86.60 × 4
≈ 346.4 N m
Two forces have equal magnitude and act at the same perpendicular distance from a pivot. One is clockwise and one is anticlockwise. What is the resultant moment?
The two moments have equal magnitude and opposite direction.
Therefore they cancel.
Resultant moment = 0
LESSON 15
Statics Formula Atlas
For a uniform rod, the weight acts through the midpoint. For a force through the chosen pivot, the moment about that pivot is zero.
CHECKPOINT
Quick Statics Check
Show answer
M = 15 × 3 = 45 N m.
Show answer
Its line of action passes through the pivot, so the perpendicular distance is zero.
Show answer
The resultant force is zero and the resultant moment is zero.
Show answer
At the midpoint, which is the centre of mass of the uniform rod.
Show answer
M = 40 × 0.75 = 30 N m.
FINAL ASSESSMENT
Statics of Rigid Bodies — 100 Question Exam
A complete interactive assessment covering moments, equilibrium, support reactions, centre of mass, couples, rods, beams, ladders, angled forces, modelling and exam traps.