Equilibrium (AQA A-Level Further Maths): Revision Notes

Worked Example: Ladder Against a Wall

Problem: A uniform ladder $AB$ weighing 120 N rests on a smooth floor with its upper end $B$ on a smooth wall. End $A$ is held by a string $AC$ fixed to the base of the wall at $C$. A weight of 80 N is attached three-quarters of the way up the ladder at $D$. If $AC = 4$ m and $BC = 6$ m, find the tension $T$ in the string and the reaction $S$ at the ground.

Solution:

Since both surfaces are smooth, the reactions $R$ and $S$ are normal to the wall and ground respectively.

Step 1: Resolve vertically

For vertical equilibrium: $S - 120 - 80 = 0$

or equivalently (upward force balances downward force): $S = 120 + 80$

Therefore: $S = 200$ N

Step 2: Take moments about point $A$

Taking moments about $A$ eliminates both $S$ and $T$ from appearing in certain forms, simplifying the calculation.

The moment equation gives us information about the reaction $R$ at the wall.

Step 3: Resolve horizontally

For horizontal equilibrium: $T\cos\theta = R\sin 2\theta$

Using the geometry of the problem and solving gives: $T = \frac{0.15W \times \sin 73.8°}{\cos 36.9°} = 0.18W$

Key insight: Choosing to take moments about point $A$ was strategic because it eliminated forces $S$ and $T$ from that equation, making the algebra simpler.

Worked Example: Cylindrical Roller with Plank

Problem: A cylindrical roller with centre $O$ and radius 63 cm is fixed to smooth, horizontal ground. A uniform plank rests against it at point $C$. One end, $A$, lies on the ground and the other end, $B$, projects beyond the roller. The plank is perpendicular to the horizontal axis of the roller. A light string $AO$ of length 1.05 m keeps the plank from slipping. If the plank has weight $W$ and length 90 cm, find the tension $T$ in the string in terms of $W$.

Solution:

Step 1: Draw a diagram showing all forces

$G$ is the midpoint of the uniform plank $AB$ (where the weight acts). There are smooth contacts at $C$ and $A$, so reactions $R$ and $S$ are perpendicular to $AB$ and $AD$ respectively.

Step 2: Use geometry to find angles

Let $\angle GAO = \angle DAO = \theta$ using the symmetry of the circle.

In $\triangle AOC$: $\sin\theta = \frac{63}{105} \Rightarrow \theta = 36.9°$

Using Pythagoras' theorem: $AC = \sqrt{AO^2 - CO^2} = \sqrt{105^2 - 63^2} = 84 \text{ cm}$

Step 3: Take moments about $A$

This eliminates $S$ and $T$ from the equation: $R \times 84 = W \times \frac{1}{2} \times 90 \times \cos 2\theta$

$\Rightarrow R = \frac{45\cos 73.8°}{84}W = 0.15W$

Step 4: Resolve horizontally

Avoid using $S$ in this equation: $T\cos\theta = R\sin 2\theta$

Therefore: $T = \frac{0.15W \times \sin 73.8°}{\cos 36.9°} = 0.18W$

Worked Example: Hinged Rod

Problem: A uniform rod $AB$ of weight 10 N and length $2a$ is hinged at $A$. Its end $B$ is held by a string so that the rod and string are both inclined at 30° to the horizontal. Find the tension in the string and the total reaction at the hinge.

Solution:

Step 1: Resolve horizontally

Let the tension in the string be $T$ and components of reaction at hinge be $X$ and $Y$.

$X = T\cos 30°$ $\Rightarrow 2X = \sqrt{3}T \quad [1]$

Step 2: Resolve vertically

$Y + T\sin 30° = 10$ $\Rightarrow 2Y + T = 20 \quad [2]$

Step 3: Take moments about $B$

Taking moments about $B$ means $T$ does not appear in this equation:

$X \times 2a\sin 30° + 10 \times a\cos 30° = Y \times 2a\cos 30°$

$\Rightarrow 2X + 10\sqrt{3} = 2Y\sqrt{3} \quad [3]$

Step 4: Solve simultaneously

Substitute equations [1] and [2] into [3]: $\sqrt{3}T + 10\sqrt{3} = \sqrt{3}(20 - T)$

$\Rightarrow T = 5 \text{ N}$

From equations [1] and [2]: $X = \frac{5}{2}\sqrt{3} \text{ N} \text{ and } Y = \frac{15}{2} \text{ N}$

Step 5: Find total reaction at hinge

$\text{Reaction at hinge} = \sqrt{X^2 + Y^2} = \frac{5}{2}\sqrt{3 + 9} = 5\sqrt{3} \text{ N}$

Alternative method: The algebra can be simplified by taking moments about $A$ instead, with $T$ resolved into two components. This means $X$ and $Y$ do not appear in the equation:

$10 \times a\cos 30° = T\sin 30° \times 2a\cos 30° + T\cos 30° \times 2a\sin 30°$

$\Rightarrow T = 5 \text{ N}$

Worked Example: Couple Using Vector Product

Problem: Two forces $\mathbf{F}_1 = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k}$ N and $\mathbf{F}_2 = 3\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}$ N act at points $P(2, -2, 3)$ and $Q(0, 4, -3)$ respectively. Their resultant forms a couple with a third force $\mathbf{F}_3$ which acts at the origin $O$.

Find: a) The force $\mathbf{F}_3$ and its magnitude b) The moment M of the couple and its magnitude

Solution:

Part a: Find $\mathbf{F}_3$

$\mathbf{F}_1 + \mathbf{F}_2 = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) + (3\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}) = 5\mathbf{i} + \mathbf{j} - 3\mathbf{k} \text{ N}$

Since the overall resultant is a couple (not a single force): $\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -5\mathbf{i} - \mathbf{j} + 3\mathbf{k} \text{ N}$

Magnitude: $|\mathbf{F}_3| = \sqrt{25 + 1 + 9} = \sqrt{35} \text{ N}$

Part b: Find moment M of couple

The moment M of the couple equals the total moment of all three forces about any point. Taking moments about $O$:

$\mathbf{M} = (\mathbf{p} \times \mathbf{F}_1) + (\mathbf{q} \times \mathbf{F}_2) + (\mathbf{0} \times \mathbf{F}_3)$

where $\mathbf{p} = 2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{q} = 0\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$

Using the determinant method for cross products:

$\mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 3 \\ 2 & 4 & -1 \end{vmatrix} + \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 4 & -3 \\ 3 & -3 & -2 \end{vmatrix} + \mathbf{0}$

$= -27\mathbf{i} + 17\mathbf{j} + 24\mathbf{k} \text{ Nm}$

Magnitude of the moment: $|\mathbf{M}| = \sqrt{(-27)^2 + 17^2 + 24^2} = 39.9 \text{ Nm}$