Forces and Newton’s laws — Edexcel A-Level Mathematics
Test yourself on Forces and Newton’s laws with PEARSON EDEXCEL A-Level practice questions.
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Forces and Newton’s laws explained
A force is a push or pull that can change an object’s shape, speed or direction, and it is a vector quantity measured in newtons.
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Forces arise from interactions such as contact, tension, weight and friction, and they combine by vector addition to give a resultant. Newton’s first law states that a body remains at rest or moves with constant velocity unless acted on by a resultant force; equivalently, a body is in equilibrium when the resultant force on it is zero. This law defines an inertial frame and explains why constant velocity, not just rest, indicates zero resultant force. In problems, resolve forces into components, sum them, and set the resultant to zero for equilibrium; a non-zero resultant produces acceleration in the direction of that resultant.
8.2 Understand and use Newton’s second law for motion in a straight line (restricted to forces in two perpendicular directions or simple cases of forces given as 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).
Newton’s second law states that the resultant force on a particle equals mass times acceleration: F = ma. For motion in a straight line, resolve all forces along the direction of motion and perpendicular to it. If forces are given as 2-D vectors, add them component-wise to find the resultant, then apply F = ma to each component. When forces act at angles, resolve each into two perpendicular directions using F cos θ and F sin θ, choosing axes along and perpendicular to the motion. For example, a 5 kg block on a smooth horizontal surface pulled by a 20 N force at 30° above the horizontal: horizontal component = 20 cos 30° ≈ 17.32 N, so a = 17.32 ÷ 5 ≈ 3.46 m s⁻². Perpendicular to motion, forces balance if there is no vertical acceleration. Always state the direction of positive motion and check units: force in N, mass in kg, acceleration in m s⁻².
8.3 Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy. (The inverse square law for gravitation is not required and g may be assumed to be constant, but students should be aware that g is not a universal constant but depends on location).
Weight is the gravitational force on a mass: W = mg, where m is mass in kg and g is gravitational acceleration in m s⁻². Near the Earth’s surface, g ≈ 9.8 m s⁻², often used as 9.81 m s⁻² or 9.8 m s⁻² depending on required accuracy. For motion in a straight line under gravity alone, a particle accelerates downwards at g, so equations of constant acceleration (suvat) apply with a = g. For example, a ball dropped from rest falls 4.9 m in 1 s because s = ½gt² = ½ × 9.8 × 1² = 4.9 m. g is not a universal constant: it varies slightly with location and altitude, being about 9.81 m s⁻² in the UK and about 9.78 m s⁻² at the equator. The inverse square law is not required, but you should recognise that g depends on distance from the Earth’s centre. Always use S.I. units: weight in N, mass in kg, g in m s⁻².
8.4 Understand and use Newton’s third law; equilibrium of forces on a particle and motion in a straight line (restricted to forces in two perpendicular directions or simple cases of forces given as 2-D vectors); application to problems involving smooth pulleys and connected particles; resolving forces in 2 dimensions; equilibrium of a particle under coplanar forces.
Newton’s third law states that if body A exerts a force on body B, then B exerts an equal and opposite force on A; these act on different bodies. For a particle in equilibrium, the resultant force is zero, so forces balance in two perpendicular directions. For motion in a straight line, apply F = ma along the line of motion and set the perpendicular resultant to zero. Resolve forces in 2 dimensions using components. For smooth pulleys and connected particles, treat the system as a whole or separately: tension is the same throughout a light inextensible string over a smooth pulley, and both particles share the same magnitude of acceleration. For example, two masses m₁ and m₂ connected over a smooth pulley: for m₁, T − m₁g = m₁a; for m₂, m₂g − T = m₂a; solving gives a = (m₂ − m₁)g ÷ (m₁ + m₂). Equilibrium under coplanar forces means the vector sum of all forces is zero.
8.5 Understand and use addition of forces; resultant forces; dynamics for motion in a plane.
Forces are vectors, so they add by the triangle or parallelogram rule. The resultant force is the single force that has the same effect as all the original forces combined. For motion in a plane, resolve forces into two perpendicular directions, usually horizontal and vertical or along and perpendicular to the motion. Add components in each direction to find the resultant components, then apply F = ma in each direction. For example, forces of 3 N east and 4 N north give a resultant of 5 N at an angle of arctan(4/3) ≈ 53.1° north of east, since √(3² + 4²) = 5. For a particle of mass 2 kg under this resultant, a = 5 ÷ 2 = 2.5 m s⁻² in the direction of the resultant. When forces balance, the resultant is zero and the particle is in equilibrium or moves with constant velocity. Always state the magnitude and direction of the resultant.
8.6 Understand and use the F ≤ μR model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.
Friction opposes relative motion or tendency to move. The model F ≤ μR relates friction F to normal reaction R and coefficient μ. When a body is in limiting equilibrium, F = μR; otherwise F < μR. For a body on a rough surface, resolve perpendicular to the surface to find R, then apply F = μR at limiting equilibrium or F ≤ μR when stationary. In statics, friction balances the component of weight or applied force along the surface. For motion, use Newton’s second law along the surface with F = μR if sliding. Example: a block of mass 5 kg on a rough horizontal plane with μ = 0.3 has R = 5g N, so limiting friction is 0.3 × 5g = 1.5g N. If the applied horizontal force is 10 N and g = 9.8 m s⁻², then 1.5g = 14.7 N, so the block remains stationary and friction is 10 N.
Your focus
- Define force and describe its vector nature and effects on a body.
- State and apply Newton’s first law to explain rest, constant velocity and equilibrium.
- Resolve forces and determine whether a body is in equilibrium from its free-body diagram.
Show all 18 objectives
- Apply F = ma to find acceleration, mass or resultant force for motion in a straight line.
- Resolve forces into two perpendicular directions and use the components in Newton’s second law.
- Handle simple cases where forces are given as 2-D vectors by adding components and applying F = ma.
- Calculate weight from mass using W = mg with an appropriate value of g.
- Solve problems involving vertical motion under gravity using constant acceleration equations.
- Explain that g depends on location and select an appropriate value for a given context.
- Apply Newton’s third law to identify force pairs acting on different bodies.
- Solve equilibrium problems for a particle under coplanar forces by resolving in two perpendicular directions.
- Analyse connected particles over a smooth pulley using equations of motion and uniform tension.
- Add two or more forces as vectors to find the resultant magnitude and direction.
- Resolve forces in two perpendicular directions and apply F = ma to motion in a plane.
- Interpret a zero resultant as equilibrium or motion with constant velocity.
- State the friction model F ≤ μR and explain the meaning of limiting friction and the coefficient of friction.
- Resolve forces perpendicular to a rough surface to determine the normal reaction R in static and dynamic problems.
- Apply F = μR or F ≤ μR correctly to solve problems involving statics and motion of a body on a rough surface.
Forces and Newton’s laws exam tips
Marking Points
- Defines force as a vector quantity that can change an object’s motion or shape, measured in newtons.
- States Newton’s first law: a body remains at rest or in uniform motion in a straight line unless acted on by a resultant force.
- Explains equilibrium as the condition of zero resultant force, covering both rest and constant velocity.
- Resolves forces into perpendicular components and sums them to find a resultant or to impose equilibrium.
- Distinguishes mass as a scalar measure of inertia from weight as the force mg acting downwards.
- Applies the law to identify forces acting on a body and to predict or explain its motion.
- State Newton’s second law as F = ma, where F is the resultant force in newtons, m is mass in kilograms and a is acceleration in m s⁻².
- Resolve forces into two perpendicular directions, typically along the line of motion and perpendicular to it, using trigonometry.
- For 2-D vector forces, add components in each direction separately to obtain the resultant force vector.
- Apply F = ma independently in each perpendicular direction, setting the perpendicular resultant to zero when there is no acceleration in that direction.
- Use consistent sign conventions for forces acting in opposite directions along the line of motion.
- Calculate weight using W = mg, ensuring mass is in kg and g is in m s⁻².
- Use g = 9.8 m s⁻² or 9.81 m s⁻² as appropriate to the accuracy required, and state the value used.
- Apply the constant acceleration equations with a = g for vertical motion under gravity, taking downwards as positive or negative consistently.
- Recognise that g varies with location and is not a universal constant, though it may be treated as constant near the Earth’s surface.
- Convert between units where necessary, for example grams to kilograms, and keep S.I. units throughout.
- State Newton’s third law correctly, noting that the paired forces act on different bodies and are equal in magnitude and opposite in direction.
- For equilibrium, set the resultant force to zero in two perpendicular directions and solve the resulting equations.
- Resolve forces in 2 dimensions using trigonometry and apply F = ma along the direction of motion.
- For smooth pulleys and connected particles, use the same tension throughout the string and the same acceleration magnitude for both particles.
- Set up separate equations of motion for each particle and solve them simultaneously.
- Add forces as vectors using components or the triangle/parallelogram rule to find the resultant.
- Resolve forces into two perpendicular directions and sum components to obtain the resultant components.
- Calculate the magnitude of the resultant using Pythagoras and its direction using trigonometry.
- Apply F = ma in each perpendicular direction for motion in a plane.
- Recognise that a zero resultant means equilibrium or constant velocity.
- State that friction acts parallel to the contact surface and opposes relative motion or the tendency to move.
- Define the coefficient of friction μ as the constant in the relationship F = μR at limiting equilibrium, and note that μ is dimensionless.
- Resolve forces perpendicular to the rough surface to determine the normal reaction R, including cases where an applied force has a component perpendicular to the surface.
- Apply F = μR when the body is in limiting equilibrium or sliding, and apply F ≤ μR when the body is stationary but not at the point of sliding.
- For statics on a rough inclined plane, resolve parallel and perpendicular to the plane and use F = μR at the point of impending motion.
- For motion on a rough surface, use Newton’s second law along the direction of motion with friction F = μR, and include acceleration or deceleration as appropriate.
Examiner Tips
- 💡Draw a free-body diagram showing every force acting on the object before writing equations.
- 💡State the direction of motion and the direction of the resultant force explicitly when explaining motion.
- 💡For equilibrium problems, resolve parallel and perpendicular to a convenient direction and set each sum to zero.
- 💡Draw a clear force diagram with all forces labelled and axes chosen before writing equations.
- 💡Show the resolution step explicitly, writing component expressions such as F cos θ and F sin θ with the angle identified.
- 💡Check that the final acceleration has the correct magnitude and direction, and include units in your answer.
- 💡Write down the value of g you are using at the start of your solution.
- 💡List the suvat variables you know and the one you need before substituting into an equation.
- 💡Check that your final answer has the correct unit and a sensible magnitude for the context.
- 💡Draw separate force diagrams for each particle in a connected system.
- 💡Label the direction of motion and take it as positive for both particles.
- 💡Solve the simultaneous equations carefully and check the sign of the acceleration.
- 💡Draw a vector triangle or parallelogram to visualise the addition of forces.
- 💡Show the component sums explicitly before calculating the resultant magnitude and direction.
- 💡State the direction of the resultant clearly, for example as a bearing or angle from a named axis.
- 💡Draw a clear force diagram showing weight, normal reaction, friction and any applied forces, with friction opposing the direction of motion or impending motion.
- 💡State whether the body is in limiting equilibrium, stationary but not limiting, or moving before choosing F = μR or F ≤ μR.
- 💡Show the resolution perpendicular to the surface to find R, then substitute into F = μR if appropriate; keep g as a symbol until the final numerical evaluation unless a numerical value is required.
- 💡Check the direction of friction on an inclined plane by considering which way the body would slide if friction were absent.
Common Mistakes
- Believing that a moving object must have a resultant force acting on it; the correction is that constant velocity means zero resultant force.
- Treating force as a scalar and adding magnitudes directly; the correction is to add forces as vectors, using components where necessary.
- Confusing mass with weight; the correction is that mass is measured in kilograms and weight is the force mg measured in newtons.
- Adding force magnitudes directly without resolving when forces act at angles; correct by resolving each force into perpendicular components before summing.
- Forgetting to include weight or normal reaction in the perpendicular direction; correct by drawing a complete force diagram showing all forces.
- Using the total force instead of the resultant force in F = ma; correct by subtracting opposing forces to find the resultant first.
- Using g = 10 m s⁻² when the question expects 9.8 m s⁻² or 9.81 m s⁻²; correct by reading the required accuracy and using the stated value.
- Confusing mass and weight; correct by remembering mass is in kg and weight is a force in N calculated as mg.
- Mixing sign conventions in suvat equations; correct by choosing a positive direction and applying it consistently to displacement, velocity and acceleration.
- Treating Newton’s third law pairs as acting on the same body; correct by identifying the two different bodies involved.
- Assuming tension differs on each side of a smooth pulley; correct by recognising tension is uniform for a light inextensible string over a smooth pulley.
- Forgetting that connected particles have the same acceleration magnitude; correct by linking the accelerations in the equations.
- Adding force magnitudes arithmetically without considering direction; correct by resolving into components or using vector addition.
- Forgetting to give the direction of the resultant; correct by calculating the angle relative to a stated axis.
- Applying F = ma to individual forces instead of the resultant; correct by finding the resultant first.
- Assuming friction always equals μR. Correction: use F = μR only at limiting equilibrium or when sliding; otherwise friction is whatever value is needed to maintain equilibrium, up to μR.
- Using R = mg without resolving perpendicular to the surface. Correction: resolve perpendicular to the surface; on an inclined plane R = mg cos θ for a body with no other perpendicular forces.
- Treating μ as having units. Correction: μ is dimensionless because it is the ratio of two forces.
- Forgetting that friction can act up or down an inclined plane depending on the direction of impending motion. Correction: determine the direction of relative motion or tendency to move before assigning the direction of friction.