Moments Atlas
LESSON 01 • FOUNDATIONS
What Is a Moment?
A moment measures the turning effect of a force about a specified point.
The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force.
A force does not only have the ability to make an object move in a straight line. If its line of action does not pass through a suitable point, it can also cause the object to rotate.
Translation
A force can cause an object to accelerate from one position to another.
Rotation
A force can produce a turning effect about a point. This turning effect is measured by a moment.
Particle model vs rigid-body model
In many earlier mechanics questions, an object can be treated as a particle, meaning its size and rotational behaviour are ignored. Moments require us to consider the object's dimensions because the position of a force relative to a pivot matters.
When studying moments, the body must generally be treated as an extended object such as a rod, beam, ladder or plank.
LESSON 02 • CORE FORMULA
Moment = Force × Perpendicular Distance
The magnitude of a moment is therefore controlled by two things: the size of the force and how far its line of action is from the pivot.
Increase F
For a fixed perpendicular distance, increasing the force increases the moment proportionally.
Increase d
For a fixed force, moving the line of action farther from the pivot increases the moment.
Units
Force is measured in newtons (N), while distance is measured in metres (m).
The notation N m should be kept as the unit of moment. Although N m has the same base dimensions as a joule, a moment is not normally expressed as a joule because the physical quantities are different.
Do not use the distance along the rod automatically. The distance required is the perpendicular distance from the pivot to the line of action of the force.
LESSON 03 • GEOMETRY
Line of Action and Perpendicular Distance
The line of action of a force is the straight line passing through the point where the force acts and extending in the direction of the force.
The moment arm is the shortest, perpendicular distance from the pivot to the force's line of action.
Why perpendicular?
Suppose a force acts at the end of a rod. The force may be inclined rather than perpendicular to the rod. The entire length of the rod is not necessarily the moment arm. Only the perpendicular separation between the pivot and the line of action matters.
Three common situations
Force perpendicular to rod
If the force is perpendicular to the rod and acts at distance r from the pivot, then d = r.
Force through pivot
The perpendicular distance is zero, so the moment is zero.
Inclined force
The moment arm is generally smaller than the distance from the pivot to the point where the force acts.
LESSON 04 • DIRECTION
Clockwise and Anticlockwise Moments
A force may tend to rotate a body clockwise or anticlockwise about the chosen pivot.
| Direction | Meaning |
|---|---|
| Clockwise | The force tends to rotate the body in the direction of a clock's hands. |
| Anticlockwise | The force tends to rotate the body opposite to the direction of a clock's hands. |
Sign convention
You may choose either clockwise or anticlockwise as positive when using signed moments, provided you remain consistent.
Signed method
Take anticlockwise moments as positive and clockwise moments as negative, for example.
Magnitude method
Calculate total clockwise moments and total anticlockwise moments separately and equate them in equilibrium.
Imagine the object free to rotate about the pivot. Ask: "If this force acted alone, which way would the object turn?"
LESSON 05 • PRINCIPLE
The Principle of Moments
If a body is in equilibrium, the clockwise and anticlockwise turning effects must balance.
Equivalently, if anticlockwise moments are positive and clockwise moments are negative:
What does equilibrium mean?
For a rigid body to be in complete equilibrium, there must be no resultant force and no resultant turning effect.
Translational equilibrium
The resultant force is zero. The body has no linear acceleration.
Rotational equilibrium
The resultant moment about a point is zero. There is no angular acceleration.
The principle of moments is a rotational-equilibrium condition. In a complete equilibrium problem you may also need to use force equilibrium.
LESSON 06 • EQUILIBRIUM
Equilibrium of a Body
A body is in equilibrium when it has no resultant force and no resultant turning effect.
For a two-dimensional rigid-body problem, this commonly means using both force balance and moment balance.
Horizontal and vertical equilibrium
When forces act in different directions, resolve them into horizontal and vertical components where appropriate.
Then use moments about a strategically chosen point.
In a typical planar equilibrium problem, think: horizontal force balance + vertical force balance + moment balance.
LESSON 07 • STRATEGY
Choosing a Pivot
You can calculate moments about any point. However, some choices make the algebra dramatically easier.
Best pivot strategy
- Choose a point through which one or more unknown forces act.
- Those forces then have zero perpendicular distance from the pivot.
- Their moments disappear from the moment equation.
- Solve for the remaining unknown quantity.
EXAMPLE • STRATEGIC PIVOT
A beam is supported at A and B. Several forces act on the beam. Suppose the reaction at A is unknown and the reaction at B is also unknown.
Taking moments about A makes the reaction at A have zero moment, because its line of action passes directly through A.
The resulting equation can therefore contain the reaction at B without the reaction at A.
Before writing equations, mark the pivot and draw the perpendicular distance for every force that contributes a moment.
LESSON 08 • ZERO MOMENT
Forces Through the Pivot
A force whose line of action passes through the pivot has no turning effect about that pivot.
This remains true regardless of the size of the force. A very large force can still have zero moment if its line of action passes through the pivot.
Do not automatically multiply the force by the distance to the point where it is applied. Check whether the force's line of action actually passes through the pivot.
LESSON 09 • INCLINED FORCES
Moments of Inclined Forces
If a force acts at an angle, the perpendicular distance to its line of action can be used directly.
Alternatively, the force can be resolved into components and the component perpendicular to the radius can be used.
Understanding the sine factor
Only the component of the force perpendicular to the line joining the pivot to the point of application produces a moment.
Therefore:
θ = 90°
sin 90° = 1, so M = Fr. The moment is maximum.
θ = 0°
sin 0° = 0, so M = 0. The force acts directly along the line joining the pivot to the point of application.
LESSON 10 • RESOLUTION
Resolving Forces Before Taking Moments
For an inclined force, resolving the force into horizontal and vertical components can make a complicated moment problem easier.
Which component contributes a moment depends on the geometry. A component whose line of action passes through the pivot has zero moment.
EXAMPLE • COMPONENT METHOD
A 20 N force acts at the end of a 3 m horizontal beam at an angle of 30° above the beam.
Only the vertical component produces a moment about the left pivot.
Vertical component:
20 sin 30° = 10 N
Moment:
M = 10 × 3
M = 30 N m
LESSON 11 • RODS
Uniform Rods
A uniform rod has its mass distributed evenly along its length. Its centre of mass is therefore at its midpoint.
If the rod has length L, its weight acts at a distance L/2 from either end.
EXAMPLE • UNIFORM ROD
A uniform horizontal rod is 6 m long and has weight 80 N. Find the moment of its weight about one end.
The weight acts through the midpoint:
d = 6/2 = 3 m
Therefore:
M = 80 × 3
M = 240 N m
For a uniform rod, its weight acts through its centre of mass, which is the midpoint.
LESSON 12 • CENTRE OF MASS
Centre of Mass and Weight
The weight of a body acts vertically downward through its centre of mass.
For a uniform rod, the centre of mass is at the geometric centre of the rod.
This means that when taking moments, the weight can be represented by a single downward force acting through the centre of mass.
Examples
- Uniform rod → centre of mass at the midpoint.
- Uniform rectangular plate → centre of mass at its geometric centre.
- For a non-uniform body → the centre of mass need not be at the geometric centre.
If the question explicitly states that a rod is uniform, immediately mark its weight at the midpoint.
LESSON 13 • BEAMS
Beams and Supports
A beam can be supported at one or more points. The supports provide reaction forces that prevent the beam from accelerating.
For a horizontal beam, vertical reactions commonly act upward, while weights act downward.
Typical beam workflow
- Draw the beam and mark every force.
- Mark the weight at the centre of mass.
- Mark all support reactions.
- Choose a convenient pivot.
- Calculate clockwise and anticlockwise moments.
- Solve the moment equation.
- Use force equilibrium for remaining unknown reactions.
EXAMPLE • SIMPLY SUPPORTED BEAM
A 4 m uniform beam weighs 100 N. It is supported at both ends. A 200 N load is placed 1 m from the left end. Find the reaction at the right end.
Take moments about the left end.
The beam's own weight acts at 2 m.
Let the right reaction be R.
Anticlockwise moment:
4R
Clockwise moments:
200(1) + 100(2)
Therefore:
4R = 200 + 200
4R = 400
R = 100 N
LESSON 14 • HINGES
Hinges and Reaction Forces
A hinge can exert a force on a body while allowing rotation about the hinge point.
The reaction at a hinge may need to be resolved into horizontal and vertical components.
If moments are taken about the hinge itself, both hinge components have zero moment because their lines of action pass through the pivot.
Taking moments about the hinge often eliminates the unknown hinge reaction completely.
LESSON 15 • TENSION
Cables and Tensions
A cable or string under tension pulls on the body along the direction of the cable.
When a cable is inclined, its tension may contribute both horizontal and vertical components.
Whether the tension contributes a moment depends on the perpendicular distance from the pivot to its line of action.
The tension force always acts along the cable. If the cable is inclined, the force is inclined too.
LESSON 16 • MULTIPLE FORCES
Multiple-Force Moment Problems
Realistic Mechanics questions often contain several forces acting simultaneously.
The safest approach is to build a complete force diagram before doing any algebra.
Systematic method
- Identify the body being analysed.
- Mark the pivot.
- Draw every external force.
- Identify each line of action.
- Calculate each perpendicular distance.
- Classify each moment as clockwise or anticlockwise.
- Write the moment equation.
- Use force equilibrium if necessary.
The resultant moment is the algebraic sum of all individual moments about the chosen point.
LESSON 17 • EXAM METHOD
How to Solve Moments Questions
A large percentage of mistakes in moments questions come from diagram interpretation rather than difficult algebra.
- Draw the force diagram. Do not start calculating immediately.
- Choose a pivot. Prefer a point that removes an unknown reaction from the moment equation.
- Mark the centre of mass. For a uniform rod, this is the midpoint.
- Find perpendicular distances. Do not automatically use the rod length.
- Determine directions. Label each moment clockwise or anticlockwise.
- Write the equation. Use either signed moments or clockwise = anticlockwise.
- Solve algebraically. Keep units visible.
- Check the result. Ask whether the direction and magnitude make physical sense.
- Using the wrong distance.
- Forgetting the weight of a uniform rod.
- Putting the weight at the wrong position.
- Incorrectly identifying clockwise direction.
- Including a force through the pivot in the moment equation.
- Using force equilibrium but forgetting moment equilibrium.
- Using degrees/radians incorrectly in trigonometry.
LESSON 18 • WORKED EXAMPLES
Exam-Style Worked Examples
EXAMPLE 01 • BASIC MOMENT
A 15 N force acts perpendicular to a beam at a distance of 2.4 m from a pivot.
M = Fd
M = 15 × 2.4
M = 36 N m
EXAMPLE 02 • FINDING A FORCE
A force acts perpendicular to a pivot at a distance of 5 m. Its moment is 75 N m. Find the force.
75 = 5F
F = 75/5
F = 15 N
EXAMPLE 03 • TWO OPPOSING MOMENTS
A 20 N force acts 3 m to the left of a pivot and a 12 N force acts 4 m to the right. Determine whether the moments balance.
Left moment: 20 × 3 = 60 N m
Right moment: 12 × 4 = 48 N m
The moments are not equal.
Resultant moment = 60 − 48 = 12 N m toward the direction of the larger moment.
EXAMPLE 04 • UNIFORM ROD
A uniform 8 m rod weighs 120 N. Find the moment of its weight about an end.
Centre of mass = 4 m from the end.
M = 120 × 4
M = 480 N m
EXAMPLE 05 • UNKNOWN SUPPORT REACTION
A 5 m uniform beam weighs 150 N and is supported at both ends. A 250 N load acts 2 m from the left end. Find the reaction at the right support.
Take moments about the left end.
The beam weight acts at 2.5 m.
Let the right reaction be R.
5R = 250(2) + 150(2.5)
5R = 500 + 375
5R = 875
R = 175 N
EXAMPLE 06 • INCLINED FORCE
A 50 N force acts at the end of a 2 m rod. The angle between the force and the rod is 30°. Find the moment about the end.
M = Fr sin θ
M = 50 × 2 × sin 30°
M = 100 × 0.5
M = 50 N m
EXAMPLE 07 • FORCE THROUGH PIVOT
A 200 N force acts along a straight line passing through the pivot. Find its moment.
The perpendicular distance is zero.
M = 200 × 0
M = 0 N m
EXAMPLE 08 • PRINCIPLE OF MOMENTS
A beam is in rotational equilibrium. The clockwise moments total 260 N m. Find the total anticlockwise moment.
For equilibrium:
clockwise = anticlockwise
Anticlockwise moment = 260 N m
EXAMPLE 09 • FINDING AN UNKNOWN LOAD
A 40 N load acts 1.5 m from a pivot. An unknown load W acts 3 m from the pivot in the opposite direction. The system is in rotational equilibrium. Find W.
40(1.5) = 3W
60 = 3W
W = 20 N
EXAMPLE 10 • COMPLETE EQUILIBRIUM
A uniform 6 m beam weighs 120 N. It is supported at both ends. A 180 N load acts 2 m from the left support. Find the two support reactions.
Let left reaction = A and right reaction = B.
Taking moments about the left support:
6B = 120(3) + 180(2)
6B = 360 + 360
6B = 720
B = 120 N
Now use vertical equilibrium:
A + B = 120 + 180
A + 120 = 300
A = 180 N
LESSON 19 • EXAM TRAPS
Things You Must Not Get Wrong
Trap 01
Using the distance to the point where the force acts rather than the perpendicular distance to its line of action.
Trap 02
Forgetting that a uniform rod's weight acts at its midpoint.
Trap 03
Assuming every force contributes a moment about every pivot.
Trap 04
Calling a force "clockwise" instead of identifying the turning effect it produces about the chosen pivot.
Trap 05
Forgetting that complete equilibrium also requires force balance.
Trap 06
Using the full inclined force instead of its perpendicular component when the geometry requires resolution.
For every force, ask: Where is its line of action, and what is the shortest distance from the pivot to that line?
FORMULA ATLAS
Moments Formula Sheet
CHECKPOINT
Quick Practice
Show answer
M = 12 × 5 = 60 N m.
Show answer
The perpendicular distance is zero, so the moment is 0 N m.
Show answer
At the midpoint, 5 m from either end.
Show answer
75 N m.
Show answer
The hinge reaction acts through the hinge, so its perpendicular distance is zero and its moment about the hinge is zero.
FINAL ASSESSMENT
100-Question Moments Exam
A comprehensive Mechanics 1 assessment covering definitions, calculations, equilibrium, beams, inclined forces and exam traps.