Forces movement shape and momentum

Key Units

Effects of Forces

A force is a push or pull that acts on a body and can change its speed, shape, or direction. Forces always arise from an interaction between two bodies.

A force can cause a stationary object to start moving, or a moving object to speed up, slow down, or change direction. A force can also deform an object — that is, change its shape.

These three effects (change in speed, change in direction, change in shape) are the core outcomes that examiners test. A single force may produce more than one effect simultaneously; for example, a tennis racket striking a ball changes both the ball's speed and its shape during contact.

MisconceptionForces do not always cause movement. A force applied to a wall compresses the wall slightly (change in shape) without causing movement. Exam questions sometimes ask students to identify the effect when no motion occurs.

Exam TipAlways consider all three possible effects — speed, direction, and shape — before answering.

Types of Force

Several types of force act between bodies. Each arises from a different physical interaction.

Gravitational force acts at a distance — no contact is needed. Electrostatic and magnetic forces also act at a distance. Friction, normal contact, and tension require physical contact between the bodies.

Examiner InsightEdexcel questions sometimes ask students to identify the type of force in an unfamiliar context. Focus on whether the force requires contact or acts at a distance, and whether it involves masses, charges, or magnets.

Exam TipIf the question says "gravitational," the answer must involve masses; if "electrostatic," it must involve charges.

Scalars and Vectors

Scalar quantities have magnitude (size) only, while vector quantities have both magnitude and direction. This distinction is fundamental because two vectors of equal magnitude can produce different effects if they act in different directions.

In diagrams throughout physics, vector quantities are represented by arrows. The length of the arrow represents the magnitude of the quantity, and the direction of the arrow represents the direction of the quantity. A longer arrow means a larger magnitude.

When a sign convention is used in vector calculations, a positive direction is chosen first. Values in the opposite direction are given a negative sign. For example, if rightward is taken as positive, a leftward force of 5 N is written as −5 N.

MisconceptionSpeed and velocity are not interchangeable. 📌 Speed is the distance travelled per unit time (scalar). 📌 Velocity is speed in a stated direction (vector). A car travelling at 30 m/s north has a velocity; 30 m/s alone is its speed.

Exam TipIf the question asks for velocity, always include a direction in the answer.

Force as a Vector

Force is a vector quantity, which means every force has both a magnitude (measured in newtons) and a direction. When describing or calculating forces, stating the direction is essential. A force of 10 N to the right and a force of 10 N to the left have the same magnitude but produce opposite effects.

Because force is a vector, forces can add together or cancel depending on their directions. This is why the direction of each force must always be stated in exam answers and diagrams.

Resultant Force

The resultant force is the single force that has the same effect as all the individual forces acting on a body combined. When forces act along the same line, the resultant is found by adding forces in one direction and subtracting forces in the opposite direction.

Taking rightward as positive: if a 12 N force acts to the right and a 5 N force acts to the left on the same object, the resultant force is 12 N + (−5 N) = +7 N, meaning 7 N to the right. If equal and opposite forces act on a body, the resultant force is zero and the body is in equilibrium — its motion does not change.

Worked example

A box rests on a surface. A person pushes the box to the right with a force of 80 N. Friction acts to the left with a force of 30 N, and a second person pushes the box to the left with a force of 20 N.

Finding the resultant force

Taking rightward as positive:

Equation used

$${F}_{\text{resultant}}=\text{sum of all forces along the line}$$

Given

$${F}_{\text{right}}=+80\text{ N}$$

$${F}_{\text{left (friction)}}=-30\text{ N}$$

$${F}_{\text{left (person)}}=-20\text{ N}$$

Working

Substitution:

$${F}_{\text{resultant}}=(+80)+(-30)+(-20)$$

$${F}_{\text{resultant}}=+30\text{ N}$$

Answer

$${F}_{\text{resultant}}=30\text{ N to the right}$$

Friction

Friction is a force that opposes motion between surfaces in contact. Friction always acts in the direction opposite to the direction of motion (or attempted motion). Friction arises because surfaces are not perfectly smooth — microscopic roughness on both surfaces interlocks, resisting sliding.

Friction has useful effects: it allows a person to walk without slipping and a car's tyres to grip the road. Friction also has unwanted effects: it causes energy to be transferred to the thermal store of the surfaces, increasing temperature and causing wear. Lubrication (e.g. oil between moving parts) reduces friction by separating the surfaces.

Examiner InsightExaminers often ask for the direction of friction. Friction always opposes relative motion, so if a block slides to the right, friction acts to the left. Never state friction acts "against the object" — state the direction explicitly.

Exam TipAlways draw friction arrows pointing opposite to the direction of motion.

Force, Mass and Acceleration

Acceleration is the change in velocity per unit time. A resultant (unbalanced) force acting on a body causes the body to accelerate in the direction of the resultant force.

Key Equations

Newton's second law:

$$F=m\times a$$

Variables:

• F = resultant force, in newtons (N)
• m = mass, in kilograms (kg)
• a = acceleration, in metres per second squared (m s⁻²)

SI unit of calculated quantity: N

Rearrangements:

Starting from F = ma:

Dividing both sides by m:

$$a=\frac{F}{m}$$

Dividing both sides by a:

$$m=\frac{F}{a}$$

Proportionality

• Force is directly proportional to mass when acceleration is constant. Doubling the mass doubles the force needed for the same acceleration.
• Force is directly proportional to acceleration when mass is constant. Doubling the resultant force doubles the acceleration.
• Acceleration is inversely proportional to mass when force is constant. Doubling the mass halves the acceleration.

Worked example

A cyclist and bicycle have a combined mass of 75 kg. The cyclist accelerates from rest to 54 km/h in 15 s. Calculate the resultant force acting on the cyclist.

Finding the velocity in SI units

$$v=\frac{54}{3.6}$$

$$v=15{\text{ m s}}^{-1}$$

Finding the acceleration

Equation used

$$a=\frac{v-u}{t}$$

Given

$$u=0{\text{ m s}}^{-1}$$

$$v=15{\text{ m s}}^{-1}$$

$$t=15\text{ s}$$

Substitution:

$$a=\frac{15-0}{15}$$

$$a=1.0{\text{ m s}}^{-2}$$

Finding the resultant force

Equation used

$$F=m\times a$$

Given

$$m=75\text{ kg}$$

$$a=1.0{\text{ m s}}^{-2}$$

Substitution:

$$F=75\times 1.0$$

$$F=75\text{ N}$$

Answer

$$F=75\text{ N (in the direction of travel)}$$

MisconceptionStudents often think that a body with no resultant force must be stationary. A body with zero resultant force has zero acceleration — it could be stationary or moving at constant velocity.

Exam TipZero resultant force means no change in velocity, not necessarily zero velocity.

Weight, Mass and Gravitational Field Strength

Weight is the gravitational force on an object. 📌 Mass is a measure of the quantity of matter in an object. These two quantities are frequently confused but are fundamentally different.

Gravitational field strength (g) is the force per unit mass exerted by a gravitational field. On Earth, g ≈ 9.8 N/kg (often rounded to 10 N/kg in calculations). On the Moon, g ≈ 1.6 N/kg, so the same mass weighs less on the Moon.

Key Equations

Weight equation:

$$W=m\times g$$

Variables:

• W = weight, in newtons (N)
• m = mass, in kilograms (kg)
• g = gravitational field strength, in newtons per kilogram (N/kg)

SI unit of calculated quantity: N

Rearrangements:

Dividing both sides by g:

$$m=\frac{W}{g}$$

Dividing both sides by m:

$$g=\frac{W}{m}$$

Proportionality

• Weight is directly proportional to mass when g is constant. Doubling the mass doubles the weight.
• Weight is directly proportional to gravitational field strength. A body on a planet with twice Earth's g would weigh twice as much.

Worked example

An astronaut has a mass of 82 kg. Calculate the astronaut's weight on Mars, where the gravitational field strength is 3.7 N/kg.

Finding the weight

Equation used

$$W=m\times g$$

Given

$$m=82\text{ kg}$$

$$g=3.7\text{ N/kg}$$

Substitution:

$$W=82\times 3.7$$

$$W=303.4\text{ N}$$

Answer

$$W\approx 303\text{ N (downward)}\text{ (3 s.f.)}$$

MisconceptionStudents sometimes state that mass changes when an object is taken to the Moon. Mass is constant regardless of location because it measures the quantity of matter. Only weight changes because gravitational field strength differs.

Exam TipIf asked "what happens to mass on the Moon?" the answer is always "mass stays the same."

Stopping Distance

The stopping distance of a vehicle is the sum of the thinking distance and the braking distance.

• Thinking distance is the distance the vehicle travels during the driver's reaction time — the time between seeing the hazard and pressing the brake.
• Braking distance is the distance the vehicle travels from the moment the brakes are applied until the vehicle stops.

$$\text{stopping distance}=\text{thinking distance}+\text{braking distance}$$

Factors Affecting Stopping Distance

Several factors affect the stopping distance of a vehicle. These factors influence either the thinking distance, the braking distance, or both.

Examiner InsightExaminers expect students to link speed to kinetic energy explicitly. At double the speed, the kinetic energy is quadrupled (since KE = ½mv²), so the braking distance is approximately four times greater. This squared relationship is frequently tested.

Exam TipAlways mention "kinetic energy is proportional to speed squared" when explaining why braking distance increases with speed.

Forces on Falling Objects and Terminal Velocity

When an object falls, weight acts downward (the gravitational force on the object) and drag (air resistance) acts upward, opposing the motion. At the moment of release, the only force is weight, so the object accelerates downward at g.

As the object's speed increases, the drag force increases because the object collides with more air particles per second and with greater force per collision. The resultant downward force therefore decreases, so the acceleration decreases.

Eventually, the drag force increases until it equals the weight. At this point the resultant force is zero, so the acceleration is zero. The object continues to fall at a constant velocity called terminal velocity.

The full causal chain is:

1. The object is released and accelerates downward due to weight.

2. As speed increases, drag force increases.

3. The resultant force (weight − drag) decreases.

4. Acceleration decreases.

5. Drag continues to increase as speed increases.

6. Drag force equals weight → resultant force = 0 → acceleration = 0 → constant (terminal) velocity.

MisconceptionStudents often write "the forces balance out" when describing terminal velocity. The precise phrasing required is: "the drag force increases until it equals the weight, so the resultant force is zero and acceleration is zero."

Exam TipState "resultant force is zero" and "acceleration is zero" explicitly — both are needed for full marks.

Investigating Extension and Applied Force

PRACTICAL: Investigating How Extension Varies with Applied Force

Aim: To investigate how the extension of helical springs, metal wires, and rubber bands varies with applied force, and to determine whether each material obeys Hooke's law.

Method

1. Clamp a helical spring vertically from a retort stand, with the lower end free to hang. Place a ruler vertically alongside the spring with the zero at the top.

2. Record the natural (unstretched) length of the spring by reading the position of the bottom of the spring on the ruler.

3. Add a known mass (e.g. 100 g = 0.98 N) to the lower end of the spring. Wait for the spring to come to rest, then read the new position of the bottom of the spring.

4. Calculate the extension: extension = new length − original length.

5. Add further equal masses one at a time, recording the total force and the total extension each time. Use at least six different loads.

6. Remove the masses and check whether the spring returns to its original length (testing for elastic behaviour).

7. Repeat the entire procedure for a metal wire and a rubber band, using the same method and range of forces.

8. Plot a graph of extension (y-axis) against force (x-axis) for each material.

Variables

Independent variable (IV): Applied force (N) — changed by adding masses in equal increments

Dependent variable (DV): Extension (m or mm) — measured using a ruler, calculated as stretched length minus original length

Control variables (CV): Same spring/wire/rubber band used throughout; same type and size of masses; temperature kept constant (thermal expansion could affect wire length)

Expected results

For the helical spring, extension is directly proportional to force up to a point (a straight line through the origin on the graph). This linear region represents Hooke's law. Beyond a certain force, the graph curves — the spring extends more per unit force. A metal wire shows a similar linear region but with much smaller extensions. A rubber band does not show a clear linear region — its force–extension graph is curved, and the loading and unloading curves differ (hysteresis).

Precaution

Masses hanging below the spring could fall and cause injury; place a cushion or sand tray beneath the setup to absorb impact. Parallax error when reading the ruler is a significant source of inaccuracy — read the ruler at eye level, perpendicular to the scale.

Measurement technique

Use a millimetre ruler (resolution 1 mm) positioned vertically beside the spring. A pointer or set square attached to the bottom of the spring helps align the reading with the ruler scale, reducing parallax. Take each reading at eye level.

SafetyMechanical hazard: falling masses. Place a sand tray or cushion below the hanging masses to absorb impact if the spring breaks or a mass slips off.

Exam cue: Annotate the original length and the extended length, with extension clearly marked as the difference.

Hooke's Law

The initial linear region of a force–extension graph is associated with Hooke's law. In this region, extension is directly proportional to the applied force — the graph is a straight line passing through the origin.

Hooke's law can be written as:

$$F=k\times e$$

where F is the applied force (N), k is the spring constant (N/m), and e is the extension (m). The spring constant k describes the stiffness of the spring — a larger k means the spring is stiffer and requires more force per unit extension.

The limit of proportionality is the point on the graph beyond which the extension is no longer proportional to the force. Beyond this point, the graph curves and Hooke's law no longer applies.

Examiner InsightEdexcel asks students to identify the Hooke's law region on a graph. The linear region starts at the origin and ends where the line begins to curve. Students must state "extension is directly proportional to force" — not just "the graph is a straight line."

Exam Tip"Directly proportional" requires a straight line through the origin — a straight line that does not pass through the origin shows a linear relationship, not proportionality.

Elastic Behaviour

Elastic behaviour is the ability of a material to recover its original shape after the forces causing deformation have been removed. A spring that returns to its original length after the load is taken off displays elastic behaviour. If the spring does not return to its original length, it has been permanently deformed — it has exceeded its elastic limit and the deformation is described as plastic.

Rubber bands show elastic behaviour over a wide range of extensions — they return to their original length even after large stretches. However, their force–extension relationship is not linear, so they do not obey Hooke's law even in the elastic region.

Momentum

Momentum is the product of mass and velocity. Momentum is a vector quantity — it has both magnitude and direction.

Key Equations

Momentum:

$$p=m\times v$$

Variables:

• p = momentum, in kilogram metres per second (kg m s⁻¹)
• m = mass, in kilograms (kg)
• v = velocity, in metres per second (m s⁻¹)

SI unit of calculated quantity: kg m s⁻¹

Rearrangements:

$$m=\frac{p}{v}$$

$$v=\frac{p}{m}$$

Proportionality

• Momentum is directly proportional to mass when velocity is constant. Doubling the mass doubles the momentum.
• Momentum is directly proportional to velocity when mass is constant. Doubling the velocity doubles the momentum.

Because momentum is a vector, direction must be included. A 2 kg ball moving at 5 m s⁻¹ to the right has momentum +10 kg m s⁻¹ (taking right as positive). The same ball moving at 5 m s⁻¹ to the left has momentum −10 kg m s⁻¹.

Worked example

A trolley of mass 850 g moves to the right at 2.4 m s⁻¹. Calculate its momentum.

Converting mass to SI units:

$$m=\frac{850}{1000}$$

$$m=0.850\text{ kg}$$

Finding the momentum

Equation used

$$p=m\times v$$

Given

$$m=0.850\text{ kg}$$

$$v=2.4{\text{ m s}}^{-1}\text{ (to the right)}$$

Substitution:

$$p=0.850\times 2.4$$

$$p=2.04{\text{ kg m s}}^{-1}$$

Answer

$$p\approx 2.0{\text{ kg m s}}^{-1}\text{ to the right (2 s.f.)}$$

Momentum and Safety Features

The idea of momentum explains how safety features in vehicles reduce the force on passengers during a collision. In a collision, the momentum of a passenger decreases to zero (the passenger stops). The change in momentum is fixed for a given crash speed. The force on the passenger depends on how quickly this change in momentum occurs.

From the equation F = Δp / t, if the time over which the momentum changes is increased, the force on the passenger is reduced. Safety features work by increasing the time of deceleration:

In every case, the mechanism is the same: the same change in momentum occurs over a longer time, so the force is smaller.

Examiner InsightA common 6-mark question asks students to explain how a named safety feature reduces injury. The full chain required is: the passenger has momentum → in a collision, momentum decreases to zero → the change in momentum is fixed → the safety feature increases the time taken for this change → from F = Δp/t, a longer time means a smaller force → smaller force reduces injury.

Exam TipAlways quote F = Δp/t and state explicitly that "same change in momentum over longer time reduces force."

Conservation of Momentum

The conservation of momentum states that in a closed system (where no external forces act), the total momentum before an event equals the total momentum after the event. This principle applies to collisions and explosions.

For a collision between two objects:

$${m}_{1}{v}_{1}+{m}_{2}{v}_{2}={m}_{1}{v}_{1}^{′}+{m}_{2}{v}_{2}^{′}$$

where the primed (') quantities are the velocities after the collision. In an explosion from rest, the total momentum before is zero, so the total momentum after must also be zero — the momenta of the fragments are equal in magnitude and opposite in direction.

A sign convention must be established before any calculation. Taking rightward as positive: objects moving to the right have positive velocity, and objects moving to the left have negative velocity.

Worked example

A 3.0 kg trolley moving at 4.0 m s⁻¹ to the right collides with a stationary 2.0 kg trolley. After the collision, the 3.0 kg trolley moves at 1.0 m s⁻¹ to the right. Calculate the velocity of the 2.0 kg trolley after the collision.

Taking rightward as positive.

Finding total momentum before

$${p}_{\text{before}}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$$

$${p}_{\text{before}}=(3.0\times 4.0)+(2.0\times 0)$$

$${p}_{\text{before}}=12.0{\text{ kg m s}}^{-1}$$

Using conservation of momentum

$${p}_{\text{after}}={p}_{\text{before}}=12.0{\text{ kg m s}}^{-1}$$

$${m}_{1}{v}_{1}^{′}+{m}_{2}{v}_{2}^{′}=12.0$$

$$(3.0\times 1.0)+(2.0\times {v}_{2}^{′})=12.0$$

$$3.0+2.0{v}_{2}^{′}=12.0$$

Rearranging for v₂':

$$2.0{v}_{2}^{′}=12.0-3.0$$

$${v}_{2}^{′}=\frac{9.0}{2.0}$$

$${v}_{2}^{′}=4.5{\text{ m s}}^{-1}$$

Answer

$${v}_{2}^{′}=4.5{\text{ m s}}^{-1}\text{ to the right}$$

The positive value confirms the 2.0 kg trolley moves to the right after the collision.

Force, Change in Momentum and Time

A resultant force causes a change in an object's momentum. The larger the force or the longer it acts, the greater the change in momentum.

Key Equations

Force and momentum change:

$$F=\frac{mv-mu}{t}$$

Variables:

• F = resultant force, in newtons (N)
• m = mass, in kilograms (kg)
• v = final velocity, in metres per second (m s⁻¹)
• u = initial velocity, in metres per second (m s⁻¹)
• t = time taken, in seconds (s)
• (mv − mu) = change in momentum, in kg m s⁻¹

SI unit of calculated quantity: N

Rearrangements:

Multiplying both sides by t:

$$mv-mu=F\times t$$

Dividing both sides gives:

$$t=\frac{mv-mu}{F}$$

Proportionality

• Force is directly proportional to the change in momentum when time is constant. A greater change in momentum requires a greater force.
• Force is inversely proportional to time when the change in momentum is constant. Doubling the time halves the force — this is the basis of vehicle safety features.

Worked example

A 150 g cricket ball travelling at 36 m s⁻¹ is hit by a bat. The ball leaves the bat at 24 m s⁻¹ in the opposite direction. The ball is in contact with the bat for 0.0020 s. Calculate the average force exerted by the bat on the ball.

Converting mass to SI units:

$$m=\frac{150}{1000}=0.150\text{ kg}$$

Establishing the sign convention: taking the initial direction of the ball (toward the bat) as positive, the ball arrives at +36 m s⁻¹ and leaves at −24 m s⁻¹.

Finding the force

Equation used

$$F=\frac{mv-mu}{t}$$

Given

$$m=0.150\text{ kg}$$

$$u=+36{\text{ m s}}^{-1}$$

$$v=-24{\text{ m s}}^{-1}$$

$$t=0.0020\text{ s}$$

Substitution:

$$F=\frac{(0.150\times (-24))-(0.150\times 36)}{0.0020}$$

$$F=\frac{-3.6-5.4}{0.0020}$$

$$F=\frac{-9.0}{0.0020}$$

$$F=-4500\text{ N}$$

Answer

$$F=4500\text{ N (in the direction the ball leaves the bat)}$$

The negative sign indicates the force acts in the direction the ball travels after being hit — opposite to its original direction.

Newton's Third Law

Newton's third law states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A. These two forces are called a Newton's third law pair. They are always:

• Equal in magnitude
• Opposite in direction
• The same type of force (e.g. both gravitational, both electrostatic, both contact)
• Acting on two different bodies (never on the same body)

For example, when a person stands on the ground, the person pushes down on the ground with a contact force equal to their weight. The ground pushes up on the person with a normal contact force of equal magnitude. These forces act on different bodies — one on the ground, one on the person.

A common error is to confuse a Newton's third law pair with balanced forces. Balanced forces act on the same body and can result in equilibrium. A Newton's third law pair always acts on two different bodies.

MisconceptionThe weight of a book resting on a table and the normal contact force from the table on the book are NOT a Newton's third law pair. They are balanced forces acting on the same body (the book). The third law pair for the book's weight is the gravitational pull of the book on the Earth.

Exam TipA third law pair must act on two different objects and be the same type of force.

Moments

A moment is the product of a force and the perpendicular distance from the pivot. The moment of a force measures its turning effect about a point.

Key Equations

Moment of a force:

$$M=F\times d$$

Variables:

• M = moment, in newton-metres (N m)
• F = force, in newtons (N)
• d = perpendicular distance from the pivot to the line of action of the force, in metres (m)

SI unit of calculated quantity: N m

Rearrangements:

$$F=\frac{M}{d}$$

$$d=\frac{M}{F}$$

Proportionality

• Moment is directly proportional to force when distance is constant. Doubling the force doubles the moment.
• Moment is directly proportional to perpendicular distance when force is constant. Doubling the distance from the pivot doubles the moment.

The word "perpendicular" is critical. The distance used must be measured at right angles from the pivot to the line along which the force acts. Using the wrong distance (e.g. the length of the beam rather than the perpendicular distance) is a common error.

Worked example

A spanner is used to turn a bolt. A force of 35 N is applied at the end of the spanner, 28 cm from the bolt. Calculate the moment about the bolt.

Converting distance to SI units:

$$d=\frac{28}{100}=0.28\text{ m}$$

Finding the moment

Equation used

$$M=F\times d$$

Given

$$F=35\text{ N}$$

$$d=0.28\text{ m}$$

Substitution:

$$M=35\times 0.28$$

$$M=9.8\text{ N m}$$

Answer

$$M=9.8\text{ N m (clockwise)}$$

Examiner InsightExaminers penalise students who forget to state the direction of the moment (clockwise or anticlockwise) and who use a distance that is not perpendicular to the force.

Exam TipAlways state whether the moment is clockwise or anticlockwise, and always check the distance is perpendicular.

Centre of Gravity

The weight of a body acts through its centre of gravity. The centre of gravity is the single point at which the entire weight of the body can be considered to act. For a uniform, regular-shaped object, the centre of gravity is at its geometric centre. For example, the centre of gravity of a uniform metre rule is at the 50 cm mark.

If an object is suspended freely from a point, it rotates until its centre of gravity is directly below the point of suspension. This principle allows the centre of gravity to be located experimentally by suspending the object from two or more different points and marking the vertical line below each point — the intersection of the lines is the centre of gravity.

Exam TipThe weight arrow must originate from the centre of gravity, not from the edge of the object.

Principle of Moments

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.

$$\text{sum of clockwise moments}=\text{sum of anticlockwise moments}$$

This principle applies to a system of parallel forces acting in one plane. To solve a moments problem:

1. Choose (or use the given) pivot point.

2. Calculate each moment: M = F × d (using the perpendicular distance from the pivot).

3. Classify each moment as clockwise or anticlockwise.

4. Set the total clockwise moments equal to the total anticlockwise moments.

5. Solve for the unknown.

Worked example

A uniform beam of length 2.0 m and weight 40 N is supported at its centre. A 60 N weight is placed 0.30 m from the left end. Where must a 20 N weight be placed to balance the beam?

The pivot is at the centre (1.0 m from each end). The 60 N weight is 0.30 m from the left end, so its perpendicular distance from the pivot is 1.0 − 0.30 = 0.70 m. The weight of the beam acts at the pivot, so its moment about the pivot is zero.

Let d be the distance from the pivot to the 20 N weight on the opposite side.

Setting clockwise moments equal to anticlockwise moments:

The 60 N weight creates an anticlockwise moment. The 20 N weight must be on the right side to create a clockwise moment.

$$\text{anticlockwise moment}=60\times 0.70=42\text{ N m}$$

$$\text{clockwise moment}=20\times d$$

$$20\times d=42$$

Rearranging for d:

$$d=\frac{42}{20}$$

$$d=2.1\text{ m from the pivot}$$

This distance (2.1 m) exceeds the beam's half-length (1.0 m), so the 20 N weight cannot balance the beam when placed on it. This result makes physical sense — a small force at a realistic distance cannot balance a much larger force close to the pivot on a short beam.

MisconceptionStudents sometimes forget that the weight of the beam itself produces a moment unless the pivot is at the centre of gravity. If the beam is uniform and the pivot is at its centre, the beam's weight has zero moment. If the pivot is not at the centre, the beam's weight must be included as a force acting at the centre of gravity.

Exam TipAlways check whether the beam's own weight produces a moment — it does unless the pivot is at the beam's centre of gravity.

Exam TipEnsure distances are measured from the pivot, not from the end of the beam.

Support Forces on a Beam

When a heavy object is placed on a light beam supported at both ends, the two support forces (upward reactions at each end) vary depending on the position of the object. The total upward force from both supports always equals the total downward force (the weight of the object plus the weight of the beam, if not negligible).

If the object is placed at the centre of the beam, each support provides half of the total downward force. As the object moves toward one end, the support force at that end increases and the support force at the other end decreases. When the object is directly above one support, that support carries the entire weight of the object.

This behaviour follows from the principle of moments. Taking moments about one support: the moment from the object's weight must be balanced by the moment from the upward force at the other support. As the object moves closer to one support, the perpendicular distance from that support decreases, so the moment about that support decreases — which means the other support needs to provide a smaller upward force to maintain balance. Conversely, taking moments about the other support shows the nearer support must provide a greater force.

Worked example

A light beam of length 4.0 m is supported at both ends A and B. A 200 N weight is placed 1.0 m from end A. Calculate the support forces at A and B.

Taking moments about A to find the reaction at B:

The 200 N weight creates a clockwise moment about A (distance from A = 1.0 m).

The upward reaction at B (call it $R_{\mathrm{B}}$) creates an anticlockwise moment about A (distance from A = 4.0 m).

Applying the principle of moments about A:

$${R}_{B}\times 4.0=200\times 1.0$$

$${R}_{B}=\frac{200}{4.0}$$

$${R}_{B}=50\text{ N}$$

Finding the reaction at A:

The total upward force equals the total downward force:

$${R}_{A}+{R}_{B}=200$$

$${R}_{A}=200-50$$

$${R}_{A}=150\text{ N}$$

Answer

$${R}_{A}=150\text{ N (upward)}, {R}_{B}=50\text{ N (upward)}$$

The support closer to the weight carries the greater share of the load.

Exam TipThe reaction arrow closer to the weight must be drawn longer than the one further away.