Using Moments - Equilibrium Revision Notes for AQA A-Level Mathematics

4.1.2 Using Moments - Equilibrium

Problem: Ladder in Equilibrium

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Given:

A uniform ladder $AB$ is leaning against a smooth vertical wall on rough horizontal ground at an angle of 70° to the horizontal.
The ladder has a length of 8 m.
It is held in equilibrium by a frictional force of magnitude 60 N acting horizontally at $B$ .
The ladder's weight acts at its centre of mass. Find:

The magnitude of the normal reaction of the wall on the ladder at A.
The mass of the ladder.

Solution:

(a): Magnitude of the Normal Reaction at A

Identify the forces:

$R_W$ : Normal reaction force at the wall (horizontal) at $A$ .
$R_C$ : Normal reaction force at the ground (vertical) at $B$ .
Frictional force of 60 N at $B$ .

Horizontal forces:

$R_W = :success[60 N]$ . Result: The normal reaction of the wall on the ladder at $A$ is 60 N.

(b): Mass of the Ladder

Taking moments about point $C$ :

$mg \cos(70^\circ) \times 4 = 60 \times 8 \times \sin(70^\circ)$

Solve for the mass $m$ :

\Rightarrow mg \cos(70^\circ) \times 4 - 60 \times 8 \times \sin(70^\circ)=0\\ \Rightarrow mg \cos(70^\circ) \times 4 = 60 \times 8 \times \sin(70^\circ)

\Rightarrow m = \frac{60 \times 8 \times \sin(70^\circ)}{4 \times \cos(70^\circ)}

Simplifying:

m = \frac{60 \times \sin(70^\circ)}{\cos(70^\circ) \times 4} \approx :success[33.64 kg]

Result: The mass of the ladder is approximately 33.64 kg.

Problem: Tension in the String of a Uniform Rod

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Given:

A uniform rod $AB$ has weight 20 N and length 3 m.
The end $A$ is freely hinged to a point on a vertical wall.
The rod is held horizontally and in equilibrium by a light inextensible string.
One end of the string is attached to the rod at $B$ .
The other end of the string is attached to a point $C$ , which is 1 m directly above $A$ . Find: Calculate the tension in the string.

Solution:

Calculate the angle $\theta$ :

Given that $\tan \theta = \frac{1}{3}$ , use the arctan function:

\theta = \tan^{-1} \left(\frac{1}{3}\right) \approx :highlight[18.435°]

Take moments about point $A$ :

The rod's weight acts at its midpoint, so the distance from $A$ is 1.5 m.
The tension $T$ in the string is at an angle $\theta$ to the horizontal, giving components $T \cos \theta$ and $T \sin \theta$ .

1.5 \times 20 = T \sin(18.435^\circ) \times 3

Solve for $T$ :

T \sin(18.435^\circ) \times 3 = 30

T = \frac{30}{3 \sin(18.435^\circ)} \approx :success[31.62 N]

Result: The tension in the string is approximately 31.62 N.

Tips:

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Identify the pivot point: Choose a point where one or more unknown forces act, as forces through the pivot have no moment. This simplifies the equation by eliminating unknowns.
Apply the principle of moments: In equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any point. Set up an equation for these moments to solve for unknown forces or distances.
Use equilibrium conditions: In addition to moments, apply the conditions for translational equilibrium:

$\sum F_x = 0$ (horizontal forces balance).
$\sum F_y = 0$ (vertical forces balance).

Using Moments - Equilibrium (AQA A-Level Mathematics): Revision Notes

4.1.2 Using Moments - Equilibrium

Problem: Ladder in Equilibrium

Solution:

(a): Magnitude of the Normal Reaction at A

(b): Mass of the Ladder

Problem: Tension in the String of a Uniform Rod

Solution:

Tips:

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