8.4 Three-Dimensional Equilibrium, Cables, and Pulley Systems

8.4 Three-Dimensional Equilibrium, Cables, and Pulley Systems

NCEES lists "Equilibrium of rigid bodies (e.g., support reactions)" and "Internal forces in rigid bodies (e.g., trusses, frames, machines)" without restricting them to two dimensions, and three-dimensional items appear regularly — usually a mast or boom held by two or three guy cables. The mathematics is not harder than 2-D, but the bookkeeping is, and unit vectors are the tool that keeps it manageable.

The Six Equations of 3-D Equilibrium

$\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum F_z = 0$ $\sum M_x = 0 \qquad \sum M_y = 0 \qquad \sum M_z = 0$

Six equations means a 3-D problem is statically determinate with at most six unknown reaction components. More than six is indeterminate; fewer generally means the body is unstable in some direction.

Cable Forces via Unit Vectors

This single procedure handles every 3-D cable problem, and it is worth executing in exactly this order:

Step 1. Write the position vector from the attachment point on the body toward the anchor: $\vec{r} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$

Step 2. Compute its magnitude — the cable's length: $|\vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$

Step 3. Form the unit vector: $\hat{u} = \frac{\vec{r}}{|\vec{r}|} = \cos\theta_x\,\hat{i} + \cos\theta_y\,\hat{j} + \cos\theta_z\,\hat{k}$

The components of $\hat{u}$ are the direction cosines, and they satisfy $\cos^2\theta_x + \cos^2\theta_y + \cos^2\theta_z = 1$ — a free check on your arithmetic.

Step 4. The cable force is the scalar tension times the unit vector: $\vec{T} = T\,\hat{u}$

Sign discipline: a cable pulls. The unit vector must point from the body toward the anchor. Reversing it is the most common 3-D error and flips the sign of every component.

Worked Example: Mast Held by Two Guy Cables

A vertical mast $OA$ stands 12 m tall, with a ball-and-socket at the base $O = (0,0,0)$ and top at $A = (0,0,12)$. Two cables run from $A$ to ground anchors $B = (8,-6,0)$ and $C = (-8,-6,0)$. A horizontal force $\vec{P} = 5{,}000,\hat{j}$ N is applied at $A$. Find the cable tensions.

Cable AB: $\vec{r}_{AB} = (8-0)\hat{i} + (-6-0)\hat{j} + (0-12)\hat{k} = 8\hat{i} - 6\hat{j} - 12\hat{k}$ $|\vec{r}_{AB}| = \sqrt{64+36+144} = \sqrt{244} = 15.62\ \text{m}$ $\hat{u}_{AB} = 0.5122\,\hat{i} - 0.3842\,\hat{j} - 0.7683\,\hat{k}$

Check: $0.5122^2 + 0.3842^2 + 0.7683^2 = 0.2624 + 0.1476 + 0.5903 = 1.000$ ✓

Cable AC is the mirror image in $x$: $\hat{u}_{AC} = -0.5122\,\hat{i} - 0.3842\,\hat{j} - 0.7683\,\hat{k}$

Equilibrium at $A$ in the $y$ direction. By symmetry $T_{AB} = T_{AC} = T$, and the $x$ components cancel automatically:

$\sum F_y = 0: \quad 5{,}000 - 0.3842\,T - 0.3842\,T = 0$ $0.7683\,T = 5{,}000 \;\Rightarrow\; T = \boxed{6{,}508\ \text{N in each cable}}$

Note the amplification. Each cable carries 6{,}508 N to resist a 5{,}000 N load — 30% more than the applied force — because only 38% of each cable's tension acts in the useful direction. The steeper the cable, the worse the ratio. The compressive load driven into the mast follows from the $z$ components:

$F_{\text{mast}} = 2(0.7683)(6{,}508) = 10{,}000\ \text{N compression}$

Twice the applied horizontal load, which is why guyed masts are checked for buckling.

Cables and Flexible Members

A cable is a tension-only member. Three consequences follow, and the exam relies on all three:

1. It cannot push, so if analysis returns a negative tension the cable has gone slack and carries zero — the assumed configuration is wrong.
2. It cannot resist a moment, so it transmits force only along its own length.
3. Its direction changes as the load moves, so the geometry must be recomputed for each configuration.

Concentrated Loads on a Cable

For a cable with concentrated loads, apply equilibrium at each load point. The horizontal component of tension is constant along the whole cable when all applied loads are vertical — because no horizontal force is applied anywhere between the supports:

$T\cos\theta = H = \text{constant} \qquad\Longrightarrow\qquad T = \frac{H}{\cos\theta}$

Tension is therefore maximum where the cable is steepest (largest $\theta$, smallest $\cos\theta$), which is at the supports, and minimum at the lowest point, where the cable is horizontal and $T = H$.

Parabolic Cable (Uniform Load per Horizontal Distance)

For a cable carrying a uniformly distributed load $w$ per unit horizontal length with span $L$ and midspan sag $h$:

$H = \frac{wL^2}{8h}, \qquad T_{\max} = \sqrt{H^2 + \left(\frac{wL}{2}\right)^2}$

The sag-tension trade-off: $H$ is inversely proportional to sag. Halving the sag doubles the horizontal tension and therefore roughly doubles the force on the supports. A "tight" cable is a highly loaded cable — the reason transmission lines and suspension bridges are deliberately allowed to sag.

Pulleys

For an ideal pulley — frictionless and massless — the governing rule is short:

The pulley changes the cable's direction but not its tension. The tension is the same on both sides.

A pulley therefore contributes no unknowns of its own beyond its support reaction, and the reaction is the vector sum of the two cable pulls. For a pulley redirecting a cable through angle $\theta$ between the two segments, the bearing force is:

$R = 2T\cos\left(\frac{\theta}{2}\right)$

so a pulley turning a cable through 180° (both segments parallel) carries $2T$, while one turning it 90° carries $T\sqrt{2}$.

Block and Tackle: Mechanical Advantage

Count the rope segments that actually support the movable block. With $n$ supporting segments and one continuous rope at uniform tension $T$:

$T = \frac{W}{n}, \qquad IMA = n$

Energy conservation fixes the trade-off — the free end must move $n$ times as far as the load:

$d_{\text{effort}} = n\,d_{\text{load}}, \qquad W \cdot d_{\text{load}} = T \cdot d_{\text{effort}}$

Real systems have friction, so:

$\eta = \frac{W/n}{T_{\text{actual}}} = \frac{IMA}{AMA}^{-1} \quad\text{, that is}\quad \eta = \frac{W}{n\,T_{\text{actual}}}$

Worked Example: Block and Tackle with Efficiency

A tackle with 4 rope segments supporting the movable block lifts a 6{,}000 N load. The measured pull is 1{,}750 N. Find the ideal pull, the efficiency, and the rope travel for a 2.0 m lift.

Ideal pull: $T_{\text{ideal}} = \frac{6{,}000}{4} = 1{,}500\ \text{N}$

Efficiency: $\eta = \frac{1{,}500}{1{,}750} = 0.857 = 85.7\%$

Rope travel: $d_{\text{effort}} = 4(2.0) = 8.0\ \text{m}$

Work check: output $= 6{,}000(2.0) = 12{,}000$ J; input $= 1{,}750(8.0) = 14{,}000$ J. Ratio $= 12{,}000/14{,}000 = 0.857$ ✓ — the same efficiency, confirming that mechanical advantage buys force at the exact cost of distance, and friction is the only true loss.

Trap: count the segments supporting the movable block, not every rope you can see and not the number of pulleys. A two-pulley arrangement can give $n = 2, 3,$ or $4$ depending on which pulleys move with the load and where the rope is anchored.