Principle of virtual work

At a Glance

• Frequency: 6 sub-parts across 6 of 13 years (2014, 2015, 2020, 2022, 2023, 2024)
• Priority tier: T2
• Marks (count): 10 (3), 15 (3)
• Average solve time: ~15 min
• Difficulty mix: medium 4, hard 2
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom appears in 6 of the last 13 years with marks ranging from 10 to 15. The questions split almost evenly between two techniques: classical moment balance (taking moments about a hinge or contact point) and the explicit principle of virtual work (giving the system a virtual displacement and requiring zero virtual work). Both techniques appear in Section B, meaning you choose from two optional questions in Q6–Q8 — mastering one or both wins you 10–15 marks reliably. The hard questions (2020, 2024) involve multi-body frameworks and require identifying the single geometric degree of freedom carefully before applying virtual work.

Minimum Theory

Classical equilibrium. For a rigid body or a connected system in static equilibrium, the sum of all forces is zero and the sum of all moments about any point is zero. The key skill is choosing a clever moment axis — typically a hinge, wall contact, or joint — so that unknown reaction forces pass through it and drop out of the moment equation.

Sine rule for moments. If a string of length $l$ is attached to the free end of a rod of length $L$ hinged at $O$, and the string’s far end is at a point $P$ distance $b$ from $O$, then the angle $\phi$ between the rod and the string satisfies $\sin\phi/\sin\theta = b/l$ (sine rule in triangle $OAP$). The moment of the tension $T$ about $O$ is $T \cdot L \cdot \sin\phi$.

Principle of virtual work (PVW). A conservative system in equilibrium satisfies: for any admissible virtual displacement $\delta\mathbf{q}$ of the generalised coordinate $q$, the total virtual work is zero: $\delta W = \sum_i \mathbf{F}_i \cdot \delta\mathbf{r}_i = 0.$ For systems under gravity, this becomes: equilibrium requires $dV/dq = 0$, where $V$ is the total potential energy expressed as a function of the single free coordinate $q$.

Working method for PVW. Identify the single degree of freedom $q$ (e.g., an angle, a length). Express the height of every weight in the system as a function of $q$. Write $V(q) = \sum_i W_i \, y_i(q)$ (taking a fixed datum). For internal forces in a rod or rod under tension $T$, include the work $T \, \delta\ell$ for each such element. Differentiate and set to zero.

When each method excels. Moment balance is fastest for a rod-and-string problem with two or three forces. PVW is cleanest when the system has multiple bodies (framework, chain of rods) because you never need to compute internal joint reactions — they cancel automatically in the virtual work equation.

Question Archetypes

moment-equilibrium (3 question(s); 2015, 2022, 2023)

Recognition Cues

• “A rod is movable in a vertical plane about a hinge at one end … string of length $l$ … find the tension.”
• “A chain of $n$ rods, smoothly jointed, is suspended from one end; a horizontal force $P$ is applied to the other end; find the inclinations.”
• “A solid hemisphere is supported by a string fixed to its rim and to a smooth vertical wall … find the value of $\tan\phi - \tan\theta$.”

Solution Template

1. Draw and label. Identify all forces: weight(s) at their CG locations, tensions/reactions at contact/hinge points. The diagram below shows the standard setup for a hinged-rod-and-string problem.

2. Choose a moment axis. Pick the point through which the most unknown forces pass (hinge, wall-contact). The reaction at that point produces zero moment.
3. Compute each moment. For a force of magnitude $F$ at perpendicular distance $d$ from the axis, moment $= F \cdot d$. If $d$ is not obvious, use $|\vec r \times \vec F|$ with position vector $\vec r$ from the axis.
4. Apply the sine rule to any triangle formed by rod + string + vertical/horizontal to find angles between forces and rod.
5. Set net moment to zero. The desired unknown appears linearly; solve. Check by force balance if needed.

Worked Example(s)

2015 Paper 1, 2015-P1-Q5d (10 marks)

A rod of 8 kg is movable in a vertical plane about a hinge at one end; to the other end is fastened a weight equal to half of the rod; this end is fastened by a string of length $l$ to a point at a height $b$ above the hinge vertically. Obtain the tension in the string.

Let $O$ be the hinge, $A$ the free end, rod length $L$, string length $l$, point $P$ directly above $O$ at height $b$.

Forces on rod: weight $W=8$ kg-wt at midpoint of rod, weight $w=4$ kg-wt at $A$, tension $T$ in string $AP$ at $A$, reaction at $O$.

Take moments about $O$ (reaction drops out). Let $\theta$ be the angle $\angle POA$.

Gravity moment (clockwise): $W\cdot(L\sin\theta/2) + w\cdot L\sin\theta = (4L + 4L)\sin\theta = 8L\sin\theta$.

Tension moment (anticlockwise): by the sine rule in triangle $OAP$, $\sin\phi/b = \sin\theta/l$, where $\phi = \angle OAP$. Perpendicular from $A$ to the rod has magnitude $l\sin\phi = lb\sin\theta/l\cdot (1)$… more precisely $\sin\phi = b\sin\theta / l$. The moment of $T$ about $O$ is $T\cdot L\cdot\sin\phi = TLb\sin\theta/l$.

Setting moments equal: $TLb\sin\theta/l = 8L\sin\theta$.

Both $L$ and $\sin\theta$ cancel: $\boxed{T = \frac{8l}{b} \text{ kg-wt.}}$

Common traps: The rod length $L$ must cancel — if it doesn’t, check the perpendicular-distance calculation. The total clockwise moment is $(W/2 + W)\cdot L\sin\theta = 8L\sin\theta$ because $W=8$ acts at $L/2$ arm and $w=4$ acts at $L$ arm, giving $4L + 4L$.

2022 Paper 1, 2022-P1-Q8b (15 marks)

A chain of $n$ equal uniform rods is smoothly jointed and suspended from one end $A_1$. A horizontal force $P$ is applied to the other end $A_{n+1}$. Find the inclinations of the rods to the downward vertical in equilibrium.

Sub-chain method. Consider rod $k$ in isolation. The tension at joint $A_{k+1}$ (pulled by the $n-k$ rods below) has:

• Horizontal component $= P$ (the applied force propagates unchanged up the chain).
• Vertical component $= (n-k)w$ downward (weight of the $n-k$ rods below $A_{k+1}$).

Take moments about $A_k$ for rod $k$ alone (weight $w$ at midpoint, arm $l\sin\theta_k/2$; tension at $A_{k+1}$ with components $(P, -(n-k)w)$, position $(l\sin\theta_k, -l\cos\theta_k)$):

$-\frac{w\sin\theta_k}{2} - (n-k)w\sin\theta_k + P\cos\theta_k = 0.$

$\tan\theta_k = \frac{P}{w\bigl(\tfrac{1}{2} + n - k\bigr)} = \frac{2P}{w(2n - 2k + 1)}.$

The denominator $(2n-2k+1)$ decreases as $k$ increases: the bottom rod ($k=n$) is most inclined with $\tan\theta_n = 2P/w$; the top rod is least inclined.

2023 Paper 1, 2023-P1-Q7b (15 marks)

A solid hemisphere supported by a string fixed to a point on its rim and to a point on a smooth vertical wall with which the curved surface is in contact. If $\theta$ is the inclination of the string with the vertical and $\phi$ is the inclination of the base to the vertical, find $\tan\phi - \tan\theta$.

Place the wall-contact at the origin, so the centre $O$ of the hemisphere (radius $a$) is at $(a, 0)$. The axis of the hemisphere (perpendicular to the base, pointing toward the curved side) makes angle $\phi$ with the horizontal, so it points in direction $(-\cos\phi, -\sin\phi)$.

CG of a solid hemisphere is $3a/8$ from the flat base along the axis: $G = O + \tfrac{3a}{8}(-\cos\phi, -\sin\phi) = \bigl(a - \tfrac{3a\cos\phi}{8},\; -\tfrac{3a\sin\phi}{8}\bigr).$

Rim attachment $R$ (upper rim point in cross-section): $R = O + a(-\sin\phi, \cos\phi) = (a - a\sin\phi,\; a\cos\phi).$

Force balance. Tension $T$ makes angle $\theta$ with the vertical toward the wall, direction $(-\sin\theta, \cos\theta)$. Vertical: $T\cos\theta = W$. Horizontal: $R_w = T\sin\theta = W\tan\theta$.

Moments about the wall-contact $(0,0)$. Moment from $W$ (at $G$): $-Wx_G = -Wa(8-3\cos\phi)/8$. Moment from $T$ at $R$: $a[(1-\sin\phi)T\cos\theta + \cos\phi\cdot T\sin\theta]$. Setting total to zero and substituting $T = W/\cos\theta$:

$(1 - \sin\phi) + \cos\phi\tan\theta = 1 - \tfrac{3\cos\phi}{8}.$

$-\sin\phi + \cos\phi\tan\theta = -\tfrac{3\cos\phi}{8}.$

Divide by $\cos\phi$: $\tan\theta - \tan\phi = -3/8$, so $\boxed{\tan\phi - \tan\theta = 3/8}$.

Common Traps

• The horizontal propagation of force $P$ up a jointed chain: the horizontal component of tension at every joint equals the applied horizontal force $P$ — a fact many students miss.
• The CG of a solid hemisphere is at $3a/8$ from the flat face (not $a/2$, which is the shell result).
• The denominator in $\tan\theta_k$ is $(2n-2k+1)$, an arithmetic sequence $2n-1, 2n-3, \dots, 3, 1$ from top to bottom — the bottom rod has the largest tilt.

virtual-work (3 question(s); 2014, 2020, 2024)

Recognition Cues

• “Using the principle of virtual work, find …”
• “Two rods on a smooth vertical circle … find the relation between $l$, $r$, $\theta$.”
• “A square framework hung from a corner … find the thrust in the horizontal diagonal rod.”
• “A regular tetrahedron of six light rods … ring resting on slant sides … find the stress.”

Solution Template

The diagram below shows the square-framework problem — the canonical PVW setup. The single DOF is angle $\theta$; all weights are expressed in terms of $\theta$; differentiating $V(\theta)$ and applying virtual work gives the thrust directly.

1. Identify the single degree of freedom $q$ (angle, base-edge length, vertical span). Every other geometric quantity must be expressible as a function of $q$.
2. Express all heights as $y_i(q)$. For each weight $W_i$ in the system, write its vertical position (positive upward) as $y_i(q)$.
3. Write the total potential energy $V(q) = \sum_i W_i y_i(q)$.
4. Include internal-rod work. If an internal rod of force $T$ (tension positive, compression negative) has length $\ell(q)$, contribute $-T\,d\ell/dq$ to the virtual work (tension opposes stretching). Equivalently, write $T = -dV/d\ell \cdot d\ell/dq / (dV/dq)$… simpler: use $T \delta\ell + \delta V_{\text{gravity}} = 0$.
5. Set $dV/dq = 0$ (for gravity-only problems) or $\delta W_{\text{total}} = 0$ (when internal forces are involved): $T \frac{d\ell}{dq} + \frac{dV}{dq} = 0 \;\Rightarrow\; T = -\frac{dV/dq}{d\ell/dq}.$
6. Evaluate at the given configuration (specific angle, specific $q$).

Worked Example(s)

2014 Paper 1, 2014-P1-Q5d (10 marks)

Two equal uniform rods $AB$ and $AC$, each of length $l$, are freely jointed at $A$ and rest on a smooth fixed vertical circle of radius $r$. If $2\theta$ is the angle between the rods, find the relation between $l$, $r$ and $\theta$ via virtual work.

DOF: angle $\theta$ (half the angle between the rods). By symmetry, $A$ sits on the vertical through the circle’s centre $O$. The tangency condition (each rod tangent to the circle) gives $h = r/\sin\theta$ for the height of $A$ above $O$.

CG of each rod is at its midpoint, at height $y_\text{CG} = r/\sin\theta - (l/2)\cos\theta$ above $O$.

Total PE: $V(\theta) = 2W\,y_\text{CG} = \frac{2Wr}{\sin\theta} - Wl\cos\theta.$

Equilibrium $dV/d\theta = 0$: $\frac{dV}{d\theta} = -\frac{2Wr\cos\theta}{\sin^2\theta} + Wl\sin\theta = 0.$

$\boxed{l\sin^3\theta = 2r\cos\theta.}$

2020 Paper 1, 2020-P1-Q7c-i (10 marks)

A square framework of four uniform rods each of weight $W$, jointed together, is hung from corner $A$. A weight $W$ is suspended from each of the three lower corners. The shape is preserved by a light rod along the horizontal diagonal $BD$. Find the thrust in the light rod.

DOF: angle $\theta$ that each rod makes with the downward vertical. With rod length $2a$:

Heights above $A$ (negative = below $A$):

• Midpoints of upper rods $AB$, $AD$: $y = -a\cos\theta$ each.
• Midpoints of lower rods $BC$, $DC$: $y = -3a\cos\theta$ each.
• Corners $B$, $D$ (hung weights): $y = -2a\cos\theta$.
• Corner $C$ (hung weight): $y = -4a\cos\theta$.

$V(\theta) = W(-2a\cos\theta) + W(-6a\cos\theta) + 2W(-2a\cos\theta) + W(-4a\cos\theta) = -16Wa\cos\theta.$

Length of diagonal $BD = 4a\sin\theta$, so $d\ell_{BD}/d\theta = 4a\cos\theta$.

Virtual work equation ($T$ = thrust, which does positive work when $BD$ lengthens): $T\frac{d\ell_{BD}}{d\theta} + \frac{dV}{d\theta} = 0 \;\Rightarrow\; T(4a\cos\theta) + 16Wa\sin\theta = 0.$

$T = -\frac{16Wa\sin\theta}{4a\cos\theta} = -4W\tan\theta.$

The negative sign means the rod is in compression (thrust). At $\theta = 45°$ (square configuration): $\boxed{\text{Thrust} = 4W.}$

Pitfall: account for all weights — the coefficient $16$ comes from $2 + 6 + 4 + 4$ (two upper CGs, two lower CGs, $B$ and $D$ hung weights, $C$ hung weight).

2024 Paper 1, 2024-P1-Q6a (15 marks)

A regular tetrahedron of six light rods, each of length $l$, rests on a smooth horizontal plane. A ring of weight $W$ and radius $r$ is supported by the slant sides. Find the stress in any horizontal side by virtual work.

DOF: base-edge length $s$ (initially $s = l$; slant rods have fixed length $l$). Apex height $h(s) = \sqrt{l^2 - s^2/3}$. The ring contacts each slant rod at horizontal distance $r$ from the axis, giving height above base: $z(s) = h(s)\!\left(1 - \frac{r\sqrt{3}}{s}\right).$

At $s = l$: $h = l\sqrt{6}/3$, $dh/ds = -1/\sqrt{6}$, and $\frac{dz}{ds}\Big|_{s=l} = -\frac{1}{\sqrt{6}} + \frac{3r\sqrt{2}}{2l}.$

Virtual work (three base rods, each with tension $T$; ring weight $W$ at height $z$): $-3T\,\delta s - W\frac{dz}{ds}\delta s = 0 \;\Rightarrow\; T = -\frac{W}{3}\cdot\frac{dz}{ds}.$

$\boxed{T = \frac{W}{3\sqrt{6}} - \frac{Wr\sqrt{2}}{2l} = \frac{W(l\sqrt{6} - 9r\sqrt{2})}{18l}.}$

Sign: $T > 0$ (tension) when $r < l\sqrt{3}/9$; compression otherwise. The three base rods hold the base against the spreading force of the ring.

Common Traps

• DOF identification: the single degree of freedom must capture the full configuration. For the circle problem it is the half-angle $\theta$; for the square it is the rod angle $\theta$ from the vertical; for the tetrahedron it is the base-edge length $s$.
• All weights contribute to $V$: in the square framework, seven separate weights (four rod weights at midpoints plus three hung weights) must all be included.
• Sign of internal rod work: a rod under tension $T > 0$ opposes lengthening, so it contributes $-T\,\delta\ell$ to $\delta W$. Setting total $\delta W = 0$ gives $T = -dV/d\ell \cdot (dq/d\ell)^{-1}$.
• Evaluate at the specific configuration: the virtual-work equation gives a general result; substitute the given geometry ($\theta = 45°$ for the square, $s = l$ for the tetrahedron) only at the end.

Marks-Aware Writing

10-mark question: A complete answer needs a labelled diagram (2 min), the equilibrium equation (force or moment balance / PVW set-up), and the final result derived cleanly in 3–4 steps. Stating the principle used (e.g., “Taking moments about the hinge…”) earns method marks even if subsequent arithmetic goes wrong.

15-mark question: Expect a harder geometry step (locating the CG of the hemisphere, deriving heights in the framework, or computing $dz/ds$ for the tetrahedron). Show each step explicitly; the examiner awards marks at each junction. For PVW problems, write the expression for $V(q)$ in full before differentiating.

Practice Set

• 2023-P1-Q5c (10 m) — — Moment balance with a string-and-weight system; identify the effective arm using the sine rule.
• 2020-P1-Q6c (15 m) — — Virtual work on a mechanical system; start by writing total PE as a function of the single angle.
• 2016-P1-Q7b (15 m) — — Chain of rods or hinged system; use the sub-chain tension approach to find inclinations.