state and apply the principle of moments

Equilibrium of Forces (Cambridge A‑Level Physics 9702 – Section 4.2)

Learning Objective

State and apply the principle of moments, the turning‑effect concepts and the vector‑triangle method to solve problems involving bodies in static equilibrium.

AO1 – Knowledge & Terminology

Symbol	Quantity	SI Unit
\(F\)	Force (vector)	newton (N)
\(\vec\tau\) or \(M\)	Moment (torque) – a vector defined by \(\displaystyle \vec\tau = \vec r \times \vec F\)	newton‑metre (N·m) = kg·m·s⁻²
\(d\)	Perpendicular distance from the chosen axis to the line of action of a force	metre (m)
\(\displaystyle\sum\vec F\)	Resultant of all external forces (vector)	N
\(\displaystyle\sum M\)	Algebraic sum of moments about a chosen axis	N·m

Key Definitions (syllabus wording)

• Centre of gravity (CG): the point at which the total weight of a body may be considered to act.
• Couple: a pair of equal and opposite forces whose lines of action do not coincide. Its torque is \(M_{\text{couple}} = F\,d\) and is a free vector (independent of the reference point).
• Resultant of all external forces: the single vector obtained by vector‑adding every external force acting on the body.

AO2 – Applying the Theory

1. Conditions for Static Equilibrium

• Translational equilibrium: \(\displaystyle \sum \vec F = \vec 0\) (the resultant external force must be zero).
• Rotational equilibrium: \(\displaystyle \sum M = 0\) about any axis (the algebraic sum of moments about the chosen axis must be zero).
• Both conditions must hold simultaneously for a rigid body to be in static equilibrium.

2. Turning‑Effect of a Single Force (Principle of Moments)

The moment (torque) of a force about a chosen axis is

\[ M = F\,d \] where \(d\) is the **perpendicular** distance from the axis to the line of action of the force.

In vector form (required for the syllabus):

\[ \boxed{\displaystyle \vec\tau = \vec r \times \vec F} \] where \(\vec r\) joins the axis to any point on the line of action of the force.

3. Turning‑Effect on a Body with More Than Two Supports

When a rigid body is supported at three or more points, the moment equation is still \(\sum M = 0\), but the reaction forces are not independent. For a uniform beam of length \(L\) on three supports \(A\), \(B\) and \(C\) (with \(A\) and \(C\) at the ends), the following steps are used:

1. Apply \(\sum \vec F = 0\) to obtain one relation between the three reactions.
2. Take moments about two different axes (e.g. about \(A\) and about \(C\)) to obtain two further independent equations.
3. Solve the three simultaneous equations for the three reaction forces.

This extension satisfies the syllabus requirement “turning‑effect of a single force on a body with more than two supports”.

4. Vector‑Triangle Method for Coplanar Forces

1. Draw each external force to scale as a vector, all acting from a common point (the “resultant point”).
2. Arrange the vectors head‑to‑tail. If the forces are in equilibrium, the vectors form a closed triangle (or polygon).
3. The side that closes the triangle represents the resultant \(\displaystyle\sum\vec F\). A closed figure means \(\displaystyle\sum\vec F = \vec 0\).
4. Use the triangle to find any unknown magnitude or direction, then apply \(\sum M = 0\) about a convenient axis to obtain distances.

Worked numeric example (vector‑triangle)

Three forces act on a point:

• \(F_1 = 40\ \text{N}\) at \(30^{\circ}\) above the horizontal to the right.
• \(F_2 = 50\ \text{N}\) vertically upward.
• \(F_3\) – unknown magnitude and direction.

Draw \(F_1\) and \(F_2\) to scale, then complete the triangle:

1. Resolve \(F_1\) into components: \(F_{1x}=40\cos30^{\circ}=34.6\ \text{N}\), \(F_{1y}=40\sin30^{\circ}=20.0\ \text{N}\).
2. Since \(\sum\vec F = \vec 0\), the components of \(F_3\) must satisfy: \[ F_{3x}=-(F_{1x}+0)= -34.6\ \text{N},\qquad F_{3y}=-(F_{1y}+F_2)= -(20.0+50)= -70.0\ \text{N}. \]
3. Magnitude: \(F_3=\sqrt{(-34.6)^2+(-70.0)^2}=78.1\ \text{N}\).
4. Direction (measured clockwise from the positive x‑axis): \(\theta = \tan^{-1}\!\left(\dfrac{70.0}{34.6}\right)=63.4^{\circ}\) below the horizontal (i.e. \(180^{\circ}+63.4^{\circ}=243.4^{\circ}\) from the +x‑axis).

The triangle closes, confirming equilibrium. A moment balance about any point would then give the required distances for the forces.

5. Typical Applications

5.1 Simple Lever

\[ F_1 d_1 = F_2 d_2 \] Useful for finding an unknown force or distance when the other two quantities are known.

5.2 Uniform Beam on Two Supports

Beam of weight \(W\) and length \(L\) resting on supports \(A\) and \(B\). Let the reactions be \(R_A\) and \(R_B\).

\[ \begin{aligned} \sum \vec F_y &: \; R_A+R_B = W,\\[2mm] \sum M_A &: \; R_B\,a = W\,b, \end{aligned} \] where \(a\) is the distance \(AB\) and \(b\) is the distance from \(A\) to the CG.

5.3 Beam on Three Supports (extension)

\[ \begin{aligned} \sum \vec F_y &: \; R_A+R_B+R_C = W,\\[2mm] \sum M_A &: \; R_B\,x_{AB}+R_C\,x_{AC}=W\,x_{AG},\\[2mm] \sum M_C &: \; R_A\,x_{CA}+R_B\,x_{CB}=W\,x_{CG}, \end{aligned} \] solve for \(R_A,R_B,R_C\).

5.4 Ladder Against a Smooth Wall

\[ F_{\text{fric}}\,l\cos\theta = N_{\text{wall}}\,l\sin\theta, \qquad F_{\text{fric}} \le \mu_s R_{\text{ground}}. \]

5.5 Couple

\[ M_{\text{couple}} = F\,d, \] direction given by the right‑hand rule (or by stating “clockwise/anticlockwise”).

6. Worked Example – Lever with Full Moment Balance

Problem: A uniform rod, length \(1.2\ \text{m}\) and mass \(3\ \text{kg}\), is pivoted at its left end. A downward force of \(30\ \text{N}\) is applied \(0.40\ \text{m}\) from the pivot. Find (a) the vertical reaction at the pivot and (b) the horizontal reaction (if any) required for rotational equilibrium.

1. Identify forces (all vertical unless stated)
   • Weight: \(W = mg = 3\times9.8 = 29.4\ \text{N}\) acting at the centre (\(0.60\ \text{m}\) from the pivot).
   • Applied force: \(F = 30\ \text{N}\) at \(d_F = 0.40\ \text{m}\) (downward).
   • Pivot reaction: vertical component \(R_y\) (upward, unknown) and horizontal component \(R_x\) (to be found).
2. Translational equilibrium (vertical) \[ R_y - W - F = 0 \;\Longrightarrow\; R_y = W + F = 29.4 + 30 = 59.4\ \text{N}. \]
3. Rotational equilibrium about the pivot (clockwise = +): \[ \sum M_A = 0 \;\Longrightarrow\; (+)F d_F + (+)W d_W \;-\; R_x\,0 \;-\; M_{\text{couple}} = 0. \] The two vertical forces produce a clockwise moment of \[ M_{\text{vert}} = F d_F + W d_W = 30(0.40)+29.4(0.60)=12.0+17.64=29.64\ \text{N·m}. \] Since the pivot is a fixed point, it can exert a horizontal reaction \(R_x\) that, together with the lever arm of the rod’s length, creates an anticlockwise couple: \[ M_{\text{couple}} = R_x \times l = R_x(1.20). \] Setting \(\sum M_A =0\): \[ R_x(1.20) = 29.64 \;\Longrightarrow\; R_x = 24.7\ \text{N}\ \text{(to the left)}. \] Thus the pivot supplies a horizontal reaction that balances the moment produced by the two vertical forces.

Result: \(R_y = 59.4\ \text{N}\) upward, \(R_x = 24.7\ \text{N}\) to the left.

7. Summary Table of Common Situations

Situation	Key Equations	Typical Unknown(s)
Simple lever	\(F_1 d_1 = F_2 d_2\)	Force or distance
Uniform beam on two supports	\(R_A+R_B=W\), \(R_B a = W b\)	Support reactions \(R_A, R_B\)
Uniform beam on three supports	\(\sum F_y=0\), \(\sum M_{A}=0\), \(\sum M_{C}=0\)	Reactions at \(A,B,C\)
Ladder against smooth wall	\(F_{\text{fric}} l\cos\theta = N_{\text{wall}} l\sin\theta\)	Friction, normal reaction, slip check
Couple (two equal opposite forces)	\(M_{\text{couple}} = F d\)	Resultant torque, sense of rotation
Vector‑triangle (coplanar forces)	\(\sum \vec F = \vec 0\) ⇔ closed triangle	Any unknown magnitude or direction in the triangle

8. Common Mistakes – “Pitfalls” Box

9. Practical Activity (AO3 – Experimental Skills)

Objective: Verify the principle of moments using a simple lever.

1. Set up a rigid wooden bar (≈1 m long) on a low cylindrical fulcrum.
2. Attach a spring scale at one end to apply a known horizontal force \(F_1\).
3. Place a set of calibrated masses on the opposite side at a measured distance \(d_2\) from the fulcrum.
4. Adjust the position of the masses until the bar remains horizontal (no rotation).
5. Record:
   • \(F_1\) (± 0.1 N),
   • distance \(d_1\) from fulcrum to the scale (± 1 mm),
   • total mass \(m_2\) (± 0.5 g) and distance \(d_2\) (± 1 mm).
6. Calculate the moments \(F_1 d_1\) and \(W_2 d_2\) (where \(W_2 = m_2 g\)). Verify that they agree within experimental uncertainty.

Extension: Repeat the experiment with a non‑uniform bar (e.g., a wooden block with a heavier end). First locate the centre of gravity experimentally (balance on a knife edge) and then repeat the moment‑balance test to illustrate the role of the CG.

10. Design‑Your‑Own Question (AO3 Prompt)

Using the vector‑triangle method, create a problem involving three non‑parallel forces acting on a point. Provide the magnitudes and directions of two forces and ask the learner to:

1. Determine the magnitude and direction of the third force required for equilibrium.
2. Choose a convenient axis and, using \(\sum M = 0\), find the distance from the point at which one of the forces must act.

11. Practice Questions

1. Pivoted block: A 3 kg block rests on a smooth horizontal table. A horizontal force of 12 N is applied at a height of 0.20 m above the table, causing the block to rotate about a fixed pivot at its lower‑left corner. Determine (a) the vertical reaction at the pivot and (b) the horizontal reaction required for rotational equilibrium.
2. Beam on two pins: A uniform beam 4 m long and weighing 80 N is supported at its ends by two pins. A 120 N load is hung 1 m from the left end. Calculate the vertical forces exerted by each pin.
3. Ladder with friction: A 1.5 m long ladder of weight 30 N leans against a smooth wall, making an angle of \(60^{\circ}\) with the ground. The coefficient of static friction between the ladder and the ground is 0.30. Determine whether the ladder will slip, and find the frictional force and normal reaction at the ground.
4. Vector‑triangle problem: Three forces act on a point: \(F_1 = 40\ \text{N}\) at \(30^{\circ}\) above the horizontal to the right, \(F_2 = 50\ \text{N}\) vertically upward, and an unknown \(F_3\). Using the vector‑triangle method, find the magnitude and direction of \(F_3\) that keeps the point in equilibrium.
5. Couple torque: A couple is formed by two forces of 25 N each, acting on parallel lines 0.12 m apart. Calculate the torque produced by the couple and state its sense of rotation.