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What forces are, the common types of force, Newton's first law, finding the resultant of forces given as vectors, and using the fact that a particle in equilibrium has zero resultant force. This is the foundation for every forces question on Paper 2.
Key facts
Newton's first law: zero resultant force ⇔ at rest or constant velocity
Equilibrium: resolve in two perpendicular directions, each sum = 0
Tension pulls along a string; thrust pushes along a rod
Notes
Forces and Newton's first law
diagram
A force is a push or a pull. It is a vector, measured in newtons (N).
Newton's first law: a particle stays at rest, or moves with constant velocity, unless a resultant force acts on it.
So a particle in equilibrium (zero resultant force) is either at rest or moving with constant velocity.
Forces you need: weight\(mg\), vertically downwards; normal reaction, perpendicular to a surface; tension, pulling along a string or rod; thrust (compression), pushing along a rod; friction, along a rough surface; resistance to motion, such as air resistance; the driving force of an engine.
Draw every force on the object, each as an arrow in its own direction
Resultant forces
The resultant is the single force that has the same effect as all the forces together. Add the forces as vectors.
In \(\mathbf{i}\), \(\mathbf{j}\) form, add the components: \((3\mathbf{i} + 2\mathbf{j}) + (\mathbf{i} - 5\mathbf{j}) = 4\mathbf{i} - 3\mathbf{j}\)
The magnitude of \(x\mathbf{i} + y\mathbf{j}\) is \(\sqrt{x^2 + y^2}\); its angle with \(\mathbf{i}\) comes from \(\tan\theta = \dfrac{y}{x}\). Sketch it to get the right quadrant.
A resultant 'parallel to \(a\mathbf{i} + b\mathbf{j}\)' has its components in the ratio \(a : b\)
Equilibrium of a particle
In equilibrium the resultant force is zero: resolve in two perpendicular directions and set each sum to zero.
For forces given as vectors, the sum of the forces is \(0\mathbf{i} + 0\mathbf{j}\)
Choose directions that make the working easy, e.g. horizontal and vertical, or perpendicular to a force you don't know.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Resultant of forces in vector form
4 to 7 marksGrade D
Add the \(\mathbf{i}\) components and the \(\mathbf{j}\) components.
Find the magnitude using Pythagoras.
Find the angle using \(\tan\), and say which direction it is measured from.
For equilibrium, the extra force needed is minus the resultant.
Example. Three forces, \((3\mathbf{i} - 2\mathbf{j})\) N, \((-5\mathbf{i} + 6\mathbf{j})\) N and \((4\mathbf{i} + 3\mathbf{j})\) N, act on a particle. (a) Find the magnitude of the resultant force. (b) Find the angle between the resultant force and the vector \(\mathbf{i}\). (c) A fourth force \(\mathbf{F}\) is added so that the particle is in equilibrium. Find \(\mathbf{F}\)
(c) \(\mathbf{F} = (-2\mathbf{i} - 7\mathbf{j})\) N
(B1)
Resultant parallel to a given vector
3 to 5 marksGrade C
Write the resultant in terms of the unknown.
Parallel to \(a\mathbf{i} + b\mathbf{j}\) means (\(\mathbf{j}\) component) \(= \dfrac{b}{a}\) × (\(\mathbf{i}\) component).
Solve, then find anything else asked for, such as the magnitude.
Example. Two forces, \((-\mathbf{i} + 8\mathbf{j})\) N and \((p\mathbf{i} + p\mathbf{j})\) N, where \(p\) is a constant, act on a particle. The resultant of the two forces is parallel to \((\mathbf{i} + 2\mathbf{j})\). (a) Find the value of \(p\). (b) Find the magnitude of the resultant.
Leaving a force off the diagram, especially the weight or the normal reaction.
Giving an angle without saying what it is measured from.
Assuming the normal reaction always equals \(mg\)
Thinking equilibrium means 'at rest' only: moving with constant velocity is equilibrium too.
More tips
Memory tricks
Resolve perpendicular to a force you don't know, to remove it from the equation.
Sense checks: a tension cannot be negative, and a resultant can never be larger than the sum of the magnitudes.
Exam technique
Draw and label a force diagram before resolving, even when one is given.
Show the direction you resolve in, e.g. 'R(→)' or 'Resolving vertically'.
Work with exact values such as \(3g\) and \(\sqrt{3}\), rounding only at the end; give answers using \(g = 9.8\) to 2 or 3 s.f.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Draw a clear, labelled force diagramShow the weight, normal reaction, tension or thrust, friction and any applied forces, each with an arrow.
Grade E
Add forces given in vector formAdd the \(\mathbf{i}\) components and the \(\mathbf{j}\) components separately to get the resultant.
Grade D
Find the magnitude and direction of a forceUse Pythagoras for the magnitude and \(\tan\theta\) for the angle with \(\mathbf{i}\)
Grade C
Use equilibrium to find unknown forcesIn equilibrium the resultant force is zero, so the components in each direction add to zero.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
[In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Three forces \(\mathbf{F}_1 = (-5\mathbf{i} - 4\mathbf{j})\text{ N}\), \(\mathbf{F}_2 = (6\mathbf{i} + 3\mathbf{j})\text{ N}\) and \(\mathbf{F}_3\) act on a particle \(P\). The particle is in equilibrium.
(a) Find \(\mathbf{F}_3\).[2]
(b) Find the magnitude of \(\mathbf{F}_3\). Give your answer to 3 significant figures.[2]
M1 for \(\sqrt{(-1)^2 + 1^2}\) using their \(\mathbf{F}_3\)
A1 for \(\sqrt{2}\) or awrt 1.41
Worked solution: \(|\mathbf{F}_3| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} = 1.41\) N
Question 2Medium8 marks
[In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.] Three forces \(\mathbf{F}_1 = (p\mathbf{i} + q\mathbf{j})\text{ N}\), \(\mathbf{F}_2 = (-2q\mathbf{i} + p\mathbf{j})\text{ N}\) and \(\mathbf{F}_3 = (8\mathbf{i} + 2\mathbf{j})\text{ N}\), where \(p\) and \(q\) are constants, act on a particle. The particle is in equilibrium.
(a) Find the value of \(p\) and the value of \(q\).[4]
(b) The force \(\mathbf{F}_2\) is now removed. Find the magnitude of the resultant of the forces that remain. Give your answer to 3 significant figures.[2]
(c) Find the angle between the direction of this resultant and the vector \(\mathbf{j}\), giving your answer in degrees to one decimal place.[2]
Show the answer and mark scheme
(a)Answer: \(p = -4, \ q = 2\)
M1 for equating the \(\mathbf{i}\) components to zero: \(p - 2q + 8 = 0\)
M1 for equating the \(\mathbf{j}\) components to zero: \(p + q + 2 = 0\)
dM1 for solving the simultaneous equations (dependent on both M marks)