Full chapter

Forces and Moments

All 4 topics and the revision summary on one page.

Identify the body and its forces

Choose the body you are analysing. Then identify each interaction that exerts a force on that body, including its source, direction and line of action.

A free-body diagram separates the chosen body from its surroundings and shows the external forces on it. A force the body exerts on something else belongs on that other body's diagram.

Force is a vector measured in newtons, N. Its components are an alternative representation of the same force. The vector methods explain how to resolve a force and combine signed components.

A block held on a smooth slope

A string holds a block stationary on a slope inclined at 30° to the horizontal. The block's weight is 10 N. The slope is smooth, so friction is neglected; the string lies parallel to the slope.

Earth on block: weight W = 10 N vertically down, acting at the centre of gravity.
String on block: tension T = 5.0 N up the slope, along the string.
Slope on block: normal force N = 8.66 N, perpendicular to the contact surface and away from it.

Choose the body, then show the forces on it

1. The physical arrangement

2. Forces on that body

Earth exerts the weight, the string exerts tension and the surface exerts the normal force. The normal force is perpendicular to the slope; it is not equal to the weight.

The physical scene and free-body diagram refer to the same stationary block. The normal is perpendicular to the slope, while the tension is along it. Only forces on the block belong in its force diagram.

Resolving weight gives 10 sin 30° = 5.0 N down the slope and 10 cos 30° = 8.66 N into it. Tension balances the first component; the normal force balances the second. The normal force is not equal to the full weight here.

Use either the weight arrow or its components in a force calculation. Counting the 10 N weight and both its components as three separate forces would count the same interaction twice. Here N is a quantity symbol for the normal force; N after a number is the unit newton.

Optional check A string holds a block stationary on a smooth slope. Which force belongs on the string's free-body diagram, but not on the block's?

Field forces act without contact

Mass in a gravitational field: A mass experiences gravitational force along the field. Near Earth's surface this force is its weight, directed down towards Earth.
Charge in an electric field: A positive charge experiences force along the electric field. A negative charge experiences force in the opposite direction. State the charge sign before choosing the force arrow.
A current-carrying conductor in a magnetic field: The conductor can experience a force perpendicular to both the conventional current and the magnetic field. For current to the right and field into the page, the force is upward. Reversing either direction reverses this force. If current and field are parallel, this magnetic force is zero.

Different fields act on different physical properties

These arrows show directions schematically. Their lengths do not compare field strengths or force magnitudes. Green indicates field direction; purple indicates force.

The arrows distinguish field direction from force on the selected body. The magnetic example uses conventional current to the right and field into the page, giving an upward force on the conductor.

In an end-view symbol, a cross represents a direction into the page and a dot represents a direction out of the page. Read which quantity the symbol labels: it may represent a field or a current, rather than a force.

Contact forces depend on the interaction

Normal force: The support force is perpendicular to the local surface. It is vertical only if that surface is horizontal. Its magnitude follows the force balance and need not equal the weight.
Tension: A taut string pulls along its length, away from the body to which it is attached. A string does not provide a pushing force.
Friction: Friction acts along the contact surface, opposing relative sliding or its tendency there. It can oppose a sliding block's motion, but it can also act forward on a driven wheel whose contact point tends to slip backward on the road.
Viscous resistance or drag: A fluid exerts resistance against an object's motion relative to that fluid. Air resistance is one example. Drag generally changes with speed and depends on the body and fluid; one universal proportionality to speed or speed squared should not be assumed.

Static friction adjusts to what is needed for the stated stationary balance, up to the point where sliding begins. If a horizontal 2 N pull acts on a block that remains stationary, with friction as the only other horizontal force, static friction is 2 N in the opposite direction. It is not automatically one fixed maximum value.

Friction does not always oppose the body's overall direction of travel. Examine the relative sliding, or tendency to slide, at the contact. For drag, examine motion relative to the surrounding fluid.

H2: upthrust from a fluid

Upthrust, or buoyant force, is the net upward force from the surrounding fluid's pressure on an immersed body. In a still fluid, pressure is greater at greater depth, so the pressure forces on lower surfaces can provide a greater upward effect than the downward effect on upper surfaces.

Submerged equilibrium

A tension reading smaller than the weight

An object of weight 5.0 N is fully submerged in still liquid, held stationary by a string. It does not touch the container. The string tension is 3.8 N.

Upward forces = downward forces:
T + U = W.
Upthrust U = 5.0 - 3.8 = 1.2 N upward.

The object's mass and gravitational weight have not decreased. The string supports part of the weight while the fluid supplies the other upward force.

A submerged object's tension is not its weight

In the force model the long blue upward arrow is tension; the short green upward arrow is upthrust. Both are vertical. The missing 1.2 N is an upward fluid force, not a reduction of gravitational weight.

The submerged body has three forces: 5.0 N weight downward, 3.8 N tension upward and 1.2 N upthrust upward. The string reading is the tension, not the gravitational weight.

For a floating body, upthrust equals weight when they are its only vertical forces and the body is in equilibrium. Include any additional string or contact force before applying that balance.

Optional check An object of weight 5.0 N is stationary, fully submerged in still liquid, supported by a string with tension 3.8 N. It does not touch the container. Which explanation is correct?

Springs and Hooke's law

A spring's extension is its change in length. Within its proportional range, the force needed to stretch it is directly proportional to that extension.

x = loaded length - unloaded length

F = kx

F is the force magnitude in N, x is extension in m and k is the force constant in N/m. A larger k means a greater force is needed for the same extension.

An extended spring exerts a restoring force on its attached load, towards the unstretched position. For a stationary hanging load with no other vertical forces, the upward force from the spring balances the load's downward weight. The load also pulls down on the spring: that force acts on the spring, not on the load. F = kx gives the spring force's magnitude, so its direction must still be shown.

Worked extension

Find the force constant

A spring has unloaded length 16.4 cm and loaded length 20.0 cm under a total applied stretching force of 2.70 N, including any hanger. The spring is within its proportional range.

x = 20.0 - 16.4 = 3.6 cm = 0.036 m.
k = F/x = 2.70/0.036 = 75 N/m.

Using the total length 0.200 m would give the wrong result. The spring responds to extension from its unloaded reference.

Investigate force against extension

Secure a suitable spring vertically beside a fixed ruler. Use a light pointer and record the unloaded length in the same arrangement.
Add known forces in suitable steps, staying within the apparatus's intended range. If using masses, include the hanger in the total mass m and calculate its weight W = mg using the stated gravitational field strength g.
Allow oscillations to settle. Read the pointer with your line of sight perpendicular to the scale, and calculate extension from the unloaded reference.
Record repeated readings where useful, then plot force vertically against extension horizontally. Include units on both axes.
Unload in steps and check whether the spring recovers its original length. Preserve the original readings when evaluating the result.

A hanger already stretches the spring. Do not use a reference measured with the hanger attached while calling subsequent values the total force. Either use a genuinely unloaded reference and total forces, or explicitly analyse changes from a stated pre-load.

Measure extension from a genuinely unloaded spring

Use the same fixed upper reference in both views. The unloaded reference contains no hanger or added load. The loaded force includes the hanger's weight; the pointer is treated as negligible. Extension = 20.0 - 16.4 = 3.6 cm = 0.036 m.

A supplied proportional model

The vertical change divided by the horizontal change gives 3.00 N / 0.040 m = 75 N/m. The graph uses extension, not total spring length.

The ruler compares a genuinely unloaded 16.4 cm reference with a loaded 20.0 cm position. The 2.70 N applied load includes any hanger. The separate graph shows the supplied linear model, rather than experimental readings.

Calculated model values for k = 75 N/m, not measured data
Force F / N	Length / cm	Extension x / m
0	16.4	0
0.75	17.4	0.010
1.50	18.4	0.020
2.25	19.4	0.030
3.00	20.4	0.040

For the straight force-against-extension graph, gradient = ΔF/Δx = (3.00 - 0)/(0.040 - 0) = 75 N/m. Plotting extension vertically against force instead gives gradient 1/k, in m/N.

Actual readings may scatter. Use a suitable best-fit line rather than joining each data point. Do not force a zero intercept unless the reference and evidence justify it: a displaced reference can produce a nonzero intercept without changing the spring's true force constant.

Choose a line by inspection

The crosses below are constructed values for a separate force-extension example. They illustrate scatter and a possible reference offset; they are not observations from the earlier exact model. Which candidate describes the whole pattern more suitably?

Crosses: constructed points. Candidate A is the solid line; candidate B is dashed.

Place a ruler through the central trend over the whole plotted range. Rotate and shift it until the vertical deviations on either side are reasonably balanced, with no systematic run from below to above the line. Do not require equal point counts on each side or force the line through the first and last readings.
A is a suitable by-eye line. B is too flat: the low-extension points are mostly below it and the high-extension point is above it. A by-eye line is an estimate; it need not match one unique calculator regression exactly.
Choose two widely separated points on the chosen line to calculate its gradient. They need not be original data points. For A, (0, 0.15 N) and (0.040 m, 3.15 N) give (3.15 - 0.15)/0.040 = 75 N/m.

For a straight line, y = mx + c: m is its gradient and c its vertical intercept. Here y is F, the horizontal variable is extension x, and m is a graph coefficient, not mass. Candidate A therefore represents:

F = (75 N/m)x + 0.15 N

The 0.15 N intercept is the line's predicted reading at x = 0. Check the force zero and unloaded reference before interpreting that offset. A nonzero intercept is not by itself evidence that the spring fails Hooke's law. Keep every original point unless a recorded reason justifies excluding one.

Find where the two lines agree

Candidate B is F = (50 N/m)x + 0.90 N. At their intersection, both predict the same force at the same extension:

(75 N/m)x + 0.15 N
= (50 N/m)x + 0.90 N
(25 N/m)x = 0.75 N
x = 0.030 m; F = 2.40 N

This crossing lies within the plotted range. Agreement at one point does not make B a good description of the full data pattern, or make the crossing a measured reading. Predict with the chosen line only over a range where the spring model remains supported.

Proportionality and elasticity are different tests

The limit of proportionality marks where force and extension cease to be directly proportional. The elastic limit concerns recovery of the original length after the load is removed. Returning to the original length alone does not prove that the loading graph was a straight line.

Pointer viewed at an angle: Parallax changes the reading. Bring the eye level with the pointer and look perpendicular to the ruler scale.
Pointer still oscillating: The length is changing during the reading. Allow it to settle before recording a stationary value.
Ruler or reference has shifted: Check that the ruler remains fixed and recheck the unloaded reference. Repeating a shifted reading does not repair the reference.
Different unloading behaviour: Record it and investigate whether the original length is recovered. Do not assume the linear model continues beyond the tested range.

Estimate and choose a range

Check the scale before measuring

For an assumed 0.3 kg total load and supplied g ≈ 10 N/kg, expect a force around 3 N. If the spring extends roughly 4 cm, estimate k ≈ 3/0.04 ≈ 75 N/m.

A 0 to 1 N force scale would be too small for that load. A suitable 0 to 5 N scale could cover it, provided its resolution and uncertainty meet the task and the spring and stand are rated for the load. These are planning estimates. Use measured values to determine the force constant with a justified precision.

Optional check A spring has force constant 75 N/m and unloaded length 16.4 cm. Its loaded length is 21.2 cm, within its proportional range. What is the magnitude of the stretching force?

Moments, couples and centre of gravity

A force's turning effect depends on its line of action as well as its magnitude. State the point about which the moment is being calculated.

Moment of a force = Fd

Here d is the perpendicular distance from the chosen point to the force's line of action. It is the shortest distance to that line, which may be extended beyond the body. Moment has unit N m.

For a planar diagram, choose a sign convention such as anticlockwise positive in the stated view. A force whose line of action passes through the pivot has zero moment about it, even when the force itself is not zero.

Worked angled force

Two ways to find the same moment

A horizontal arm extends 0.30 m to the right of a pivot. A 20 N force at its end acts 30° above the arm.

Use the full force: perpendicular distance d = 0.30 sin 30° = 0.15 m. Moment = 20 × 0.15 = 3.0 N m anticlockwise.
Alternatively, resolve the force: its upward component is 20 sin 30° = 10 N, acting 0.30 m from the pivot. Its moment is 10 × 0.30 = 3.0 N m anticlockwise. The horizontal component acts along a line through the pivot, so its moment is zero.

Use the perpendicular distance to the line of action

The green distance is perpendicular to the dashed line of action. Moment = 20 N × 0.15 m = 3.0 N m anticlockwise. Resolving the force is an alternative way to obtain the same result.

The perpendicular distance to the 20 N force's line of action is 0.15 m. The 0.30 m arm can instead be paired with the 10 N perpendicular component. The two calculations describe the same turning effect.

Use one method or the other. Adding the full-force moment to its component moment would count the same force twice. Multiplying 20 N by 0.30 m would wrongly treat the force as perpendicular to the arm.

A couple has zero resultant force but a turning effect

A couple consists of two equal, opposite, parallel forces on the same body, acting along different lines. The forces cancel as vectors, but their moments give a tendency to rotate. The torque of the couple is:

τ = Fd

F is the magnitude of one force, and d is the perpendicular separation of the two lines of action. Torque may be written τ or T; in another context T can mean tension, so state the quantity and unit.

Worked couple

Equal forces can turn in the same sense

A 12 N force acts upward on the right and a 12 N force downward on the left. Their parallel lines are 0.18 m apart horizontally.

Resultant force = 0.
Couple torque = 12 × 0.18 = 2.16 N m anticlockwise.

About the midpoint, each force contributes 12 × 0.09 = 1.08 N m anticlockwise. The two moments add. About the left force's line, that force has zero moment and the right force contributes all 2.16 N m. The total couple torque is the same about any chosen point.

A couple has zero resultant force and a nonzero torque

Torque = 12 N × 0.18 m = 2.16 N m anticlockwise. Both forces act on this one body; a third-law pair acts on different bodies.

Both forces act on the same body. Their moments have the same anticlockwise sense, although their resultant force is zero. The 0.18 m distance is the full separation between their lines of action.

A third-law pair acts on different bodies, so it is not a couple on one body's free-body diagram. Also, two equal opposite forces acting on the same line have no couple torque: their line separation is zero.

Moment and torque have the same dimensions as energy, but they describe a different physical quantity. Report torque in N m, rather than joules.

Optional check A body experiences a 12 N upward force on the right and a 12 N downward force on the left, on parallel lines 0.18 m apart. With no other forces or moments, is it in equilibrium?

Replace distributed weight by one equivalent force

Every part of an extended body has weight. For force and moment calculations, their combined effect can be represented by a single weight acting at the body's centre of gravity.

A uniform straight beam in a uniform gravitational field has its centre of gravity at its midpoint. A nonuniform or loaded body need not. In some shapes the centre of gravity lies outside the material, so do not automatically place the weight at a convenient contact point.

For scale, a few newtons acting through a perpendicular arm of a few tenths of a metre give a moment of order 1 N m. For example, 5 N × 0.2 m = 1 N m. State the assumptions and convert centimetres to metres before interpreting a calculated torque.

Combine force and moment balance

Equilibrium requires both zero resultant force and zero resultant torque. A body can satisfy one condition while failing the other.

ΣF_x = 0 and ΣF_y = 0

Σmoments = 0

For forces in a plane, resolve along two perpendicular axes and add signed components. For rotation, choose a point and one positive turning sense. The principle of moments states that in rotational equilibrium, total clockwise moment equals total anticlockwise moment about the chosen point.

A couple shows why zero resultant force is insufficient. Likewise, balanced moments alone do not establish zero resultant force. Equilibrium does not require forces to be absent, and force balance alone does not show that the body is at rest. The worked arrangements below are explicitly stationary.

A beam with supports away from its ends

A uniform rigid beam is 1.20 m long and weighs 10 N. Its two resting supports are at x = 0.20 m and x = 1.00 m, measured from the left end. A 20 N downward load acts at x = 1.10 m. Both support forces act vertically upward.

Balance the beam's forces and moments

All force arrows share one scale, so the small left reaction is deliberately short. Taking moments about the left support, use arms 0.40 m for the beam's weight, 0.80 m for the right reaction and 0.90 m for the added load. These differ from the labelled positions.

The beam's own weight acts at x = 0.60 m. Positions measured from the left end are distinct from moment arms measured from a chosen support. Both reactions and both downward forces belong in the beam model.

Worked support forces

Choose a point that removes one unknown moment

Place the beam's weight: its centre of gravity is at the midpoint, x = 0.60 m.
Take moments about the left support: the left reaction has zero moment there. The right reaction's arm is 1.00 - 0.20 = 0.80 m; the weight's arm is 0.40 m; the load's arm is 0.90 m.
Balance the moments: R_right(0.80) = 10(0.40) + 20(0.90). Thus R_right = 27.5 N.
Balance vertical forces: R_left + 27.5 = 10 + 20. Thus R_left = 2.5 N.

Check about the right support: 10(0.40) = 2.5(0.80) + 20(0.10). Both sides are 4.0 N m.

The right reaction can exceed the added 20 N load because it also helps support the beam's own weight and balance the turning effect. The reactions need not be equal just because there are two supports.

A negative answer for an assumed upward reaction would mean that a downward force is needed under the proposed conditions. A simple resting support cannot pull the beam down; reconsider whether contact can be maintained, rather than accepting the assumed arrangement unchanged.

Check the model against a measurement

Mark the beam's centre of gravity and measure load and support positions from one fixed reference. Keep applied loads vertical, record the support readings with their units, and compare their sum with the total weight. Use the same points to calculate perpendicular moment arms. Finite support widths or a shifted load position affect the assumed lines of action; address that alignment rather than only repeating the force readings.

Optional check A uniform 1.20 m beam weighs 10 N and rests on supports at x = 0.20 m and 1.00 m. Its 20 N load is moved to x = 0.90 m. What are the left and right support forces in equilibrium?

A stationary knot and a closed force triangle

A light knot supports a 50 N load through a third string. The knot's own weight is neglected. The third string pulls the knot down with 50 N; do not also add the load's gravitational force to the knot's diagram.

The left string points four parts left for three parts up. The right string points three parts right for four parts up. Each direction uses a 3-4-5 triangle: horizontal and vertical components are the corresponding fractions of the tension.

The spatial arrangement and the force triangle answer different questions

1. The strings set the force directions

2. Place the force arrows head to tail

The three vectors are (-24, +18) N, (+24, +32) N and (0, -50) N. Closing the triangle establishes force balance. For an extended body, also check the moments and lines of action.

The spatial diagram gives string directions; the separate force triangle gives force magnitudes and directions. Its arrows are (-24, +18) N, (+24, +32) N and (0, -50) N, placed head to tail with equal scales on both axes.

Worked tensions

Balance components before comparing magnitudes

Take right and up as positive. Write the tensions as T_left and T_right.

Horizontal balance:
0.8T_left = 0.6T_right.

Vertical balance:
0.6T_left + 0.8T_right = 50.

The first equation gives T_left = 0.75T_right. Substituting gives 1.25T_right = 50, so T_right = 40 N and T_left = 30 N.

The components are (-24, +18) N and (+24, +32) N. Horizontally, -24 + 24 = 0. Vertically, 18 + 32 - 50 = 0.

Placed head to tail, the three force vectors form a closed triangle: the final tip returns to the initial tail. The tensions are unequal, but their horizontal components are equal and opposite.

Here all forces act at the same point, so the light-knot model has no remaining couple. For an extended body, a closed force triangle establishes zero resultant force but does not by itself establish zero resultant moment. Check the actual positions and lines of action too.

Optional check At a light stationary knot, the left tension has components (-24, +18) N and the right tension (+24, +32) N. A third string pulls down with 50 N. What do these values show?

Use sine and cosine rules for a non-right triangle

A force triangle need not be right-angled. For any triangle with side lengths a, b and c opposite interior angles α, β and γ:

a/sin α = b/sin β = c/sin γ

c² = a² + b² - 2ab cos γ

Match each side to its opposite angle. In the cosine rule, γ is the interior angle between the sides a and b. In a force triangle the side lengths represent force magnitudes, not physical string lengths.

Worked non-right triangle

Two 20 N tensions support a load

A different stationary light knot has two equal 20 N tensions, each directed 60° above the horizontal, one to the left and one to the right. A third string pulls down with the load force.

The tensions are 60° apart when drawn from the same origin. When drawn head to tail, their closed force triangle has a 120° interior angle between the two 20 N sides and two other interior angles of 30°.

Calling the downward load magnitude W, the cosine rule gives:

W² = 20² + 20² - 2(20)(20) cos 120°
= 1200 N².

Therefore W = 20√3 N ≈ 34.6 N.

The sine rule gives the same value:
20/sin 30° = W/sin 120°, so W = 20 sin 120°/sin 30° ≈ 34.6 N.

Use the interior angles of the closed force triangle

The downward side is W = 20√3 N, approximately 34.6 N. Its opposite interior angle is 120°. Each 20 N side is opposite a 30° angle. The 60° between the common-origin forces is not the interior angle to insert into the subtraction form of the cosine rule.

The 60-degree angle between tensions in the spatial arrangement becomes a 120-degree interior angle in the head-to-tail force triangle. The side opposite 120 degrees is the 20 square-root-3 N load force.

The subtraction form of the cosine rule uses the triangle's interior angle. Substituting the 60° tail-to-tail angle into that form would give the wrong load.

Finally, keep the system boundary consistent. If a body and an attached load are analysed together, forces between them are internal to that combined system. If analysing one alone, those interaction forces cross its boundary and must be included.

Revision summary

Choose the body, draw its external forces, then decide whether you need force balance, moment balance or both.

Force models

Name the source and recipient. Include forces on the chosen body, not forces it exerts on other bodies.
Weight acts along the gravitational field. Electric force is along the field for a positive charge and opposite it for a negative charge.
Magnetic force on a current-carrying conductor is perpendicular to current and field. In the given view, current right and field into the page give force up. Parallel current and field give zero magnetic force.
Normal force is perpendicular to the contact surface. Tension pulls along a taut string.
Friction opposes relative sliding or its tendency at the contact. Drag opposes motion relative to the fluid. Do not infer a universal drag-speed law.
Components represent the original force; they are not extra forces to count alongside it.

H2 upthrust: the surrounding fluid exerts a resultant buoyant force. In the submerged example, T + U = W. A smaller tension reading does not mean a smaller gravitational weight.

Springs

Extension: x = loaded length - unloaded length. Convert to metres when k is in N/m.
Hooke's law: F = kx within the limit of proportionality. Show the restoring direction separately.
Force-extension graph: Force vertically and extension horizontally: gradient k. Reversed axes: gradient 1/k.
Practical reference: Include the hanger in total force and use a genuinely unloaded reference, or state a consistent incremental model. Read a settled pointer perpendicular to a fixed ruler.

Proportionality tests the linear force-extension relationship; elasticity tests recovery after unloading. Repeats, calibration, alignment and loading/unloading checks address different limitations.

For a scattered force-extension graph, choose a line that follows the whole trend. In y = mx + c, the gradient gives force per extension and the intercept is the reading at zero extension. At the intersection of two candidate lines, equate their force expressions; one crossing does not establish a good fit.

Moments and couples

Moment of a force: Fd about a stated point, with d perpendicular to the force's line of action. A line through the pivot gives zero moment.
Couple: Equal opposite parallel forces on the same body, with different lines of action. Resultant force zero; torque τ = Fd, where d is their full perpendicular separation.
Centre of gravity: Replace distributed weight by one equivalent force at the centre of gravity. A uniform straight beam in a uniform field has it at the midpoint.

Use a consistent clockwise/anticlockwise sign. A third-law pair acts on different bodies and is not a couple on one body.

Quantity and unit reference
Quantity	Common symbol	Unit
Force; weight	F; W	N
Tension	T	N
Extension	x	m
Perpendicular arm or separation	d	m
Force constant	k	N/m
Moment or torque	T or τ	N m

Distinguish tension T from torque T by context and units. Although torque has the same dimensions as energy, report it in N m, not J.

Equilibrium

Draw the free-body diagram, including the body's own weight where relevant.
Use ΣF_x = 0 and ΣF_y = 0 for force balance.
Use equal clockwise and anticlockwise moments. Choosing a point on an unknown force's line removes its moment from that equation.
Check another balance and interpret any unexpected sign against the actual support or string model.

A closed force triangle means zero resultant force. For an extended body, also check moments using actual lines of action. In a non-right triangle, use opposite side-angle pairs for the sine rule and the interior included angle for the cosine rule:

a/sin α = b/sin β = c/sin γ

c² = a² + b² - 2ab cos γ

The angle between two vectors drawn from a common origin need not be their head-to-tail triangle's interior angle.

Back to identifying the body and its forces