Equilibrium of Forces in Three Dimensions

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2023)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: A | Dominant type: application

Why This Chapter Matters

This atom appeared once (2023) in Section A. UPSC uses it to test whether you can set up and solve the six scalar equilibrium equations for a body acted on by a three-dimensional force system, typically finding unknown reactions or tensions. The key skill is systematic resolution along three axes and taking moments about three axes.

Minimum Theory

Equilibrium conditions for a rigid body

A rigid body is in equilibrium if and only if the resultant force and resultant moment about any point are both zero:

$\sum \mathbf{F} = \mathbf{0} \qquad \text{and} \qquad \sum \mathbf{M}_O = \mathbf{0}$

In scalar component form (resolving along $x$, $y$, $z$ axes):

$\sum F_x = 0,\quad \sum F_y = 0,\quad \sum F_z = 0$ $\sum M_x = 0,\quad \sum M_y = 0,\quad \sum M_z = 0$

These six equations are independent for a general 3D force system.

Moment of a force about a point

The moment (torque) of force $\mathbf{F}$ applied at position $\mathbf{r}$ about origin $O$:

$\mathbf{M}_O = \mathbf{r} \times \mathbf{F}$

In components, if $\mathbf{r} = (x,y,z)$ and $\mathbf{F} = (F_x, F_y, F_z)$:

$\mathbf{M}_O = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ F_x & F_y & F_z\end{vmatrix} = (yF_z - zF_y)\hat{i} - (xF_z - zF_x)\hat{j} + (xF_y - yF_x)\hat{k}$

Moment of a force about an axis

The moment of $\mathbf{F}$ about a line through $O$ with unit direction $\hat{n}$:

$M_{\hat{n}} = \hat{n}\cdot(\mathbf{r}\times\mathbf{F})$

Common reactions in 3D problems

• Smooth surface / knife-edge: reaction normal to surface only.
• Smooth hinge (pin joint): three unknown force components; zero moment.
• Fixed support (encastré): three force components + three moment components.
• Ball-and-socket joint: three force components, no moment.
• String / rod: tension along the string/rod axis only.

Strategy for 3D equilibrium

1. Draw a free-body diagram; label all forces with their lines of action and magnitudes.
2. Express each force in component form $\mathbf{F}_i = F_{ix}\hat{i}+F_{iy}\hat{j}+F_{iz}\hat{k}$.
3. Write the three force equations $\sum F_x = 0$, $\sum F_y = 0$, $\sum F_z = 0$.
4. Choose a convenient moment point (typically where several unknowns’ lines of action meet) to reduce simultaneous equations.
5. Write the three moment equations; solve the system.

Question Archetypes

3d-force-reaction (1 question; 2023)

Recognition Cues

• A rigid body (rod, plate, frame) is supported at multiple points in 3D space.
• Some forces or reactions are unknown; equilibrium is given or assumed.
• You must resolve along three coordinate axes and/or take moments about three axes.
• The geometry involves coordinates given explicitly in 3D.

Solution Template

1. Set up a coordinate system; write position vectors of all key points.
2. Express all applied loads and unknown reactions as vectors.
3. Apply $\sum \mathbf{F} = \mathbf{0}$: write three scalar equations.
4. Choose a moment point; compute $\mathbf{r}_i \times \mathbf{F}_i$ for each force.
5. Apply $\sum \mathbf{M}_O = \mathbf{0}$: write three scalar equations.
6. Solve the (up to 6) equations for the unknowns; check for consistency.

Worked Example

2023 Paper 1, 2023-P1-Q3c (10 marks)

A uniform rod $AB$ of weight $W$ and length $2a$ is smoothly hinged at $A$ to a fixed point. The rod is held in equilibrium in a horizontal position by a vertical string attached at $B$ and by a horizontal string at the midpoint $M$ of the rod, the horizontal string being perpendicular to $AB$ and lying in the horizontal plane through $AB$. Find the tension in each string and the reaction at the hinge.

Step 1. Set up coordinates.

Place $A$ at the origin. Let $AB$ lie along the positive $x$-axis, so:

$A = (0,0,0),\quad M = (a,0,0),\quad B = (2a,0,0)$

The weight $W$ acts downward ($-z$ direction) at the midpoint $M = (a, 0, 0)$.

Step 2. Identify forces and unknowns.

• Vertical string at $B$: tension $T_1$ in the $+z$ direction, i.e., $\mathbf{T}_1 = T_1\hat{k}$.
• Horizontal string at $M$ perpendicular to $AB$: tension $T_2$ in the $+y$ (or $-y$) direction, i.e., $\mathbf{T}_2 = T_2\hat{j}$.
• Weight at $M$: $\mathbf{W} = -W\hat{k}$.
• Hinge reaction at $A$: $\mathbf{R} = R_x\hat{i} + R_y\hat{j} + R_z\hat{k}$ (three unknown components).

Step 3. Force equations.

$\sum F_x = 0: \quad R_x = 0$ $\sum F_y = 0: \quad R_y + T_2 = 0$ $\sum F_z = 0: \quad R_z + T_1 - W = 0$

Step 4. Moment equations about $A$.

Compute $\mathbf{r}\times\mathbf{F}$ for each force (excluding $\mathbf{R}$ at $A$ since $\mathbf{r} = \mathbf{0}$):

For weight $-W\hat{k}$ at $(a,0,0)$: $\mathbf{r}_M\times\mathbf{W} = a\hat{i}\times(-W\hat{k}) = -aW(\hat{i}\times\hat{k}) = -aW(-\hat{j}) = aW\hat{j}$

For $T_1\hat{k}$ at $(2a,0,0)$: $\mathbf{r}_B\times\mathbf{T}_1 = 2a\hat{i}\times T_1\hat{k} = 2aT_1(\hat{i}\times\hat{k}) = 2aT_1(-\hat{j}) = -2aT_1\hat{j}$

For $T_2\hat{j}$ at $(a,0,0)$: $\mathbf{r}_M\times\mathbf{T}_2 = a\hat{i}\times T_2\hat{j} = aT_2(\hat{i}\times\hat{j}) = aT_2\hat{k}$

Step 5. Apply $\sum \mathbf{M}_A = \mathbf{0}$.

$aW\hat{j} - 2aT_1\hat{j} + aT_2\hat{k} = \mathbf{0}$

• $\hat{j}$ component: $aW - 2aT_1 = 0 \implies T_1 = \dfrac{W}{2}$
• $\hat{k}$ component: $aT_2 = 0 \implies T_2 = 0$
• $\hat{i}$ component: automatically $0$ (no $\hat{i}$ moments present)

Step 6. Back-substitute for hinge reaction.

From $\sum F_y$: $R_y = -T_2 = 0$. From $\sum F_z$: $R_z = W - T_1 = W - W/2 = W/2$.

$\boxed{T_1 = \frac{W}{2},\quad T_2 = 0,\quad \mathbf{R} = \frac{W}{2}\hat{k}}$

The hinge exerts a vertical reaction of $W/2$ upward; the horizontal string has zero tension (the geometry is self-supporting in the horizontal direction).

Common Traps

• Taking moments about a point that is not on the line of action of $\mathbf{R}$, leaving the hinge reaction in the moment equation and complicating the algebra.
• Errors in the cross-product $\hat{i}\times\hat{k} = -\hat{j}$ (not $+\hat{j}$); always verify the right-hand rule.
• Missing the $\hat{i}$ component moment equation — writing only two moment equations and overlooking the third.
• Treating the hinge as having only one or two force components instead of three in a general 3D setting.

Marks-Aware Writing

This is a 10-mark application question. Examiners look for:

1. Free-body diagram described with all forces named and directions stated (2 marks).
2. Three force equations written and used (2 marks).
3. Three moment equations set up with correct cross products (4 marks).
4. Correct final values for all unknowns (2 marks).

Label every equation; keep algebra tidy by computing cross products one at a time.

Practice Set

Only one historical question on this atom (shown above).