Equilibrium of forces

2. Key Definitions

• Moment of a Force: The product of the force and the perpendicular distance from the pivot to the line of action of the force. (Unit: $\text{N m}$)
• Principle of Moments: For a body in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.
• Torque of a Couple: The product of one of the forces and the perpendicular distance between the lines of action of the two forces. (Unit: $\text{N m}$)
• Couple: A pair of forces that are equal in magnitude, opposite in direction, and act along different (parallel) lines of action.
• Equilibrium: A state where there is no resultant force (zero translational acceleration) and no resultant torque (zero rotational acceleration) acting on a system.
• Centre of Gravity: The single point through which the entire weight of an object is considered to act.
• Coplanar Forces: A system of forces where all the force vectors lie within the same two-dimensional plane.

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3. Content

3.1 The Principle of Moments

The moment of a force is a measure of its tendency to cause a body to rotate about a specific point (the pivot). It is a vector quantity, though in 2D problems, we simply treat it as clockwise or anticlockwise.

The mathematical definition is: $\text{Moment} = F \times d$

Where:

• $F$ is the magnitude of the applied force ($\text{N}$).
• $d$ is the perpendicular distance from the pivot to the line of action of the force ($\text{m}$).

Forces at an Angle

If a force $F$ is applied at an angle $\theta$ to a beam at a distance $r$ from the pivot, the perpendicular distance $d$ is found using trigonometry: $d = r \sin \theta$ Therefore: $\text{Moment} = F r \sin \theta$

Alternatively, you can resolve the force $F$ into a component perpendicular to the beam ($F \sin \theta$) and a component parallel to the beam ($F \cos \theta$). Only the perpendicular component contributes to the moment.

The Principle in Practice

For any object that is not rotating (or is rotating at a constant angular velocity), the Principle of Moments must hold true: $\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}$

This principle can be applied about any point on or off the object. However, strategic selection of the pivot point (usually at the location of an unknown force) simplifies the calculation by reducing the number of variables in the equation.

3.2 Conditions for Equilibrium

For a rigid body to be in a state of static equilibrium, it must satisfy two independent conditions. If either condition is not met, the object will accelerate.

1. Zero Resultant Force (Translational Equilibrium)

The vector sum of all external forces acting on the body must be zero. $\sum \mathbf{F} = 0$

In a 2D coplanar system, this is usually solved by resolving forces into two perpendicular components (horizontal and vertical):

• $\sum F_x = 0$: (Sum of forces to the right = Sum of forces to the left)
• $\sum F_y = 0$: (Sum of forces upwards = Sum of forces downwards)

2. Zero Resultant Torque (Rotational Equilibrium)

The vector sum of all external torques (moments) about any point must be zero. $\sum \tau = 0$

This ensures the object does not have any angular acceleration. Note that an object can have a resultant force of zero but still have a resultant torque (e.g., a couple), which would cause it to rotate without translating.

3.3 Couples and Torques

A couple is a unique arrangement of forces. Because the two forces are equal and opposite, their resultant force ($\sum F$) is always zero. Consequently, a couple produces pure rotation with no linear translation.

The Torque of a Couple ($\tau$) is calculated as: $\tau = F \times s$

Where:

• $F$ is the magnitude of one of the forces in the pair.
• $s$ is the perpendicular separation (distance) between the two parallel lines of action.

Important Distinction: Unlike the moment of a single force, the torque of a couple is the same about any point in the plane. It does not depend on the position of the pivot.

3.4 Vector Triangles for Forces in Equilibrium

When an object is in equilibrium under the action of exactly three coplanar forces, those forces can be represented as a closed vector triangle.

Characteristics of the Vector Triangle:

1. Tip-to-Tail: The vectors are drawn such that the head of the first force meets the tail of the second, and the head of the second meets the tail of the third.
2. Closed Loop: Because the resultant force is zero ($\sum F = 0$), the "tip" of the final force vector must return exactly to the "tail" of the first force vector.
3. Direction: The arrows must follow a continuous path around the triangle (either all clockwise or all anticlockwise).

Solving the Triangle:

• Right-Angled Triangles: Use basic trigonometry ($\sin, \cos, \tan$) and Pythagoras' Theorem ($a^2 + b^2 = c^2$).
• Non-Right-Angled Triangles:
  • Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
  • Cosine Rule: $a^2 = b^2 + c^2 - 2bc \cos A$

3.5 Worked Example 1: The Uniform Beam

Question

A uniform diving board of length $3.0\text{ m}$ and mass $25\text{ kg}$ is supported by a pillar at end $A$ and a second pillar $1.0\text{ m}$ from end $A$. A diver of mass $60\text{ kg}$ stands at the far end $B$. Calculate the magnitude and direction of the force exerted by the pillar at $A$.

Solution

Step 1: Identify all forces and their positions.

• Weight of board ($W_b$): $mg = 25 \times 9.81 = 245.25\text{ N}$. Acts at the centre ($1.5\text{ m}$ from $A$).
• Weight of diver ($W_d$): $mg = 60 \times 9.81 = 588.6\text{ N}$. Acts at end $B$ ($3.0\text{ m}$ from $A$).
• Force at pillar $C$ ($F_C$): Acts upwards at $1.0\text{ m}$ from $A$.
• Force at pillar $A$ ($F_A$): Acts at $0\text{ m}$ from $A$. (Direction unknown, assume upwards for now).

Step 2: Apply the Principle of Moments about pillar $C$. Choosing $C$ as the pivot eliminates $F_C$ from the equation.

• Clockwise moments about $C$: $W_b \times (1.5 - 1.0) + W_d \times (3.0 - 1.0)$
• Anticlockwise moments about $C$: $F_A \times 1.0$

$\sum \text{Clockwise} = \sum \text{Anticlockwise}$ $(245.25 \times 0.5) + (588.6 \times 2.0) = F_A \times 1.0$ $122.625 + 1177.2 = F_A$ $F_A = 1299.825\text{ N}$

Step 3: Determine direction. In our moment equation, we placed $F_A$ on the anticlockwise side. For $F_A$ (at the left end) to provide an anticlockwise moment about $C$ (to its right), $F_A$ must act downwards.

Final Answer: $F_A = 1300\text{ N}$ acting downwards (to 2 s.f.).

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3.6 Worked Example 2: Vector Triangle (Tension in Cables)

Question

A sign of weight $150\text{ N}$ is supported by two cables. Cable 1 is at an angle of $35^\circ$ to the horizontal. Cable 2 is at an angle of $55^\circ$ to the horizontal. Calculate the tension $T_1$ and $T_2$ in the cables.

Solution

Step 1: Analyze the geometry. The angles to the horizontal are $35^\circ$ and $55^\circ$. The angle between the two cables is $180^\circ - (35^\circ + 55^\circ) = 90^\circ$. Since the cables meet at a right angle, we can use a right-angled vector triangle.

Step 2: Construct the triangle.

• The Weight ($W = 150\text{ N}$) acts vertically down.
• $T_1$ acts at $35^\circ$ to the horizontal, which is $90 - 35 = 55^\circ$ to the vertical (the weight).
• $T_2$ acts at $55^\circ$ to the horizontal, which is $90 - 55 = 35^\circ$ to the vertical.

Step 3: Use trigonometry. In the right-angled triangle where $W$ is the hypotenuse: $T_1 = W \cos(55^\circ) = 150 \times \cos(55^\circ) = 86.03\text{ N}$ $T_2 = W \cos(35^\circ) = 150 \times \cos(35^\circ) = 122.87\text{ N}$

Final Answer: $T_1 = 86\text{ N}$, $T_2 = 120\text{ N}$ (to 2 s.f.).

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4. Key Equations

Equation	Description	Status
$\text{Moment} = F \times d$	$d$ is perpendicular distance from pivot.	Memorise
$\tau = F \times s$	Torque of a couple; $s$ is separation.	Memorise
$\sum \text{Moments} = 0$	Rotational equilibrium condition.	Memorise
$\sum F_x = 0, \sum F_y = 0$	Translational equilibrium conditions.	Memorise
$W = mg$	Weight calculation ($g = 9.81\text{ m s}^{-2}$).	Data Sheet
$\frac{a}{\sin A} = \frac{b}{\sin B}$	Sine rule for non-right-angled triangles.	Memorise

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5. Common Mistakes to Avoid

• ❌ Using the wrong distance: Using the length of a beam instead of the perpendicular distance from the pivot to the line of action of the force.
  • ✅ Right: Always draw the line of action of the force. The distance $d$ must be the shortest distance from the pivot to that line.
• ❌ Ignoring the weight of the object: Forgetting to include the weight of a beam or rod in the moment equation.
  • ✅ Right: If the word "uniform" is used, add a weight vector acting at the exact geometric centre. If "non-uniform," the question will specify the centre of gravity.
• ❌ Incorrect Torque of a Couple: Multiplying the total force ($2F$) by the distance, or using the distance from a central pivot ($s/2$).
  • ✅ Right: The torque is $F \times s$. It is the magnitude of one force multiplied by the entire distance between them.
• ❌ Sign errors in resolution: Mixing up sine and cosine when resolving forces.
  • ✅ Right: The component adjacent to the angle $\theta$ is always $F \cos \theta$. The component opposite the angle is $F \sin \theta$.
• ❌ Misinterpreting "Equilibrium": Assuming that because an object is moving, it isn't in equilibrium.
  • ✅ Right: Equilibrium means zero acceleration. An object moving at a constant velocity in a straight line is in equilibrium.

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6. Exam Tips

1. The "Magic" Pivot: If you have two unknown forces, place your pivot directly on top of one of them. This makes its distance zero, its moment zero, and allows you to solve for the other unknown immediately.
2. Free-Body Diagrams (FBD): Never attempt an equilibrium problem without a diagram. Draw all forces as arrows starting from the point they act. Use different colours for distances and forces to avoid confusion.
3. Check your Units: Moments and Torques are in $\text{N m}$. Ensure all distances are converted to metres ($\text{m}$) and all masses to weights ($\text{N}$) using $g = 9.81\text{ m s}^{-2}$.
4. Vector Triangle Direction: When drawing a vector triangle, ensure the arrows follow each other around the perimeter. If two arrows meet head-to-head, you have drawn a resultant force, not an equilibrium state.
5. Read the wording carefully:
   • "Light" means the mass of the object is negligible (ignore its weight).
   • "Uniform" means the weight acts at the centre.
   • "Smooth" means there is no friction.
   • "Rough" means you must include a friction force parallel to the surface.
6. Three-Force Rule: If an object is in equilibrium under three non-parallel forces, the lines of action of all three forces must pass through the same point. This is a very useful check for ladder or wall-hinge problems.