Applications of vectors - forces and equilibrium
Apply vectors to forces and equilibrium in Year 11 Specialist Mathematics for Queensland (QCAA). Because a force has both size and direction it is a vector, so combining forces and balancing a block on a slope come down to adding and resolving vectors.
You will learn to find the resultant of concurrent forces, the equilibrant that holds a particle at rest, and how to model a particle on a smooth inclined plane — core mechanics skills for the Specialist course.
Theory
Forces are vectors, so the vector algebra of Year 11 Specialist Mathematics (QCAA, Queensland) models them directly. This page shows how to find the resultant of concurrent forces, what equilibrium means (the forces sum to the zero vector), how to find the equilibrant, and how to resolve forces into components — including a particle on a smooth inclined plane.
A force has both a size (measured in newtons, N) and a direction, so it is a vector. Forces acting at the same point are called concurrent, and can be added like any other vectors — in component form \(\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}\), or by the triangle rule.
The resultant \(\mathbf{R}\) is the single force equal to the vector sum of all the forces acting. Its magnitude is \(|\mathbf{R}|=\sqrt{F_x^{\,2}+F_y^{\,2}}\) and its direction is found from \(\tan\theta=\dfrac{F_y}{F_x}\).
A particle is in equilibrium when it stays at rest, which happens exactly when the resultant force is the zero vector: \(\sum\mathbf{F}=\mathbf{0}\). The single extra force that would produce equilibrium is the equilibrant — it is equal in magnitude but opposite in direction to the resultant, \(\mathbf{E}=-\mathbf{R}\).
To analyse a slope we resolve each force into two perpendicular directions. For a particle on a smooth (frictionless) plane inclined at \(\theta\), the weight \(W\) splits into a component \(W\sin\theta\) down the line of greatest slope and \(W\cos\theta\) into the surface, balanced by the normal reaction \(N\).
For concurrent forces \(\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}\), the magnitude and direction of the resultant are:
A force of magnitude \(F\) at angle \(\theta\) to the horizontal resolves into perpendicular components:
A particle is in equilibrium when the forces sum to the zero vector; the equilibrant is the opposite of the resultant:
How to solve a forces problem
- Draw a clear diagram showing every force acting at the point (or on the particle), with its direction.
- Resolve each force into components — either \(\mathbf{i}\) and \(\mathbf{j}\), or along and perpendicular to the slope for an incline.
- Add or set to zero: for a resultant, add the components; for equilibrium, set the totals in each direction equal to zero.
- Solve for the unknown force or angle, then state the magnitude \(|\mathbf{R}|=\sqrt{F_x^{\,2}+F_y^{\,2}}\) and direction where asked.
Perpendicular forces combine by Pythagoras — the resultant is the diagonal:
| \(|\mathbf{R}|\) | \(=\) | \(\sqrt{9^2+12^2}\) |
| \(=\) | \(\sqrt{81+144}\) | |
| \(=\) | \(\sqrt{225}\) | |
| \(=\) | \(15\) |
\(|\mathbf{R}| = 15\) N.
Horizontal component is \(F\cos\theta\):
| \(F_x\) | \(=\) | \(20\cos 60^\circ\) |
| \(=\) | \(20\times\dfrac{1}{2}\) | |
| \(=\) | \(10\) |
Vertical component is \(F\sin\theta\):
| \(F_y\) | \(=\) | \(20\sin 60^\circ\) |
| \(=\) | \(20\times\dfrac{\sqrt{3}}{2}\) | |
| \(=\) | \(10\sqrt{3}\) |
\(F_x = 10\) N and \(F_y = 10\sqrt{3}\) N.
First add the two known forces, component by component:
| \(\mathbf{F}_1+\mathbf{F}_2\) | \(=\) | \((4-2)\mathbf{i}+(-1+5)\mathbf{j}\) |
| \(=\) | \(2\mathbf{i}+4\mathbf{j}\) |
For equilibrium the total must be \(\mathbf{0}\), so \(\mathbf{F}_3\) is the equilibrant (equal and opposite):
| \(\mathbf{F}_3\) | \(=\) | \(-(2\mathbf{i}+4\mathbf{j})\) |
| \(=\) | \(-2\mathbf{i}-4\mathbf{j}\) |
\(\mathbf{F}_3 = -2\mathbf{i}-4\mathbf{j}\) N.
Along the slope, equilibrium needs the force to balance \(W\sin\theta\):
| \(T\) | \(=\) | \(W\sin\theta\) |
| \(=\) | \(50\times\dfrac{7}{25}\) | |
| \(=\) | \(\dfrac{350}{25}\) | |
| \(=\) | \(14\) |
Perpendicular to the slope, the normal reaction balances \(W\cos\theta\):
| \(N\) | \(=\) | \(W\cos\theta\) |
| \(=\) | \(50\times\dfrac{24}{25}\) | |
| \(=\) | \(\dfrac{1200}{25}\) | |
| \(=\) | \(48\) |
\(T = 14\) N up the slope and \(N = 48\) N.
Common pitfalls
Frequently asked questions
How do you find the resultant of two forces at right angles?
Add them as vectors. The magnitude is \(\sqrt{F_x^{\,2}+F_y^{\,2}}\) by Pythagoras, and the direction comes from \(\tan\theta=\dfrac{F_y}{F_x}\).
What does equilibrium mean for forces?
A particle is in equilibrium when it stays at rest, which happens exactly when the resultant force is the zero vector, so all the forces sum to \(\mathbf{0}\).
What is the equilibrant of a set of forces?
It is the single extra force that produces equilibrium. It is equal in magnitude but opposite in direction to the resultant, \(\mathbf{E}=-\mathbf{R}\).
What is the force needed to hold a block on a smooth slope?
Along the line of greatest slope the block would slide with a force \(W\sin\theta\), so a force of \(W\sin\theta\) up the slope holds it in equilibrium.
What is the normal reaction on an inclined plane?
The normal reaction is the push of the surface perpendicular to the plane. On a smooth incline it balances the perpendicular part of the weight, so \(N=W\cos\theta\).
Why do we resolve forces into components?
Splitting each force into two perpendicular directions turns a two-dimensional problem into two simple one-dimensional equations that can be solved separately.