Large/small contact surfaces and screws

A car jack lifts a two-tonne vehicle with one hand turning a small screw, and stays put the instant you let go — no motor, no brake, no ratchet. The secret is a shape trick: a thin wedge or a helical thread quietly turns a small applied force into a much larger normal force, while ordinary Coulomb friction at the inclined contact does double duty as both the resistance you must push through and, if the incline is shallow enough, the lock that holds the load afterward. Wedges and square-threaded screws look like separate machine elements, but they solve the same triangle of forces. A wedge is a small block with an inclined face; a screw thread is the same inclined face wrapped helically around a cylinder. Once you "unroll" the thread into an equivalent incline, raising or lowering a load with a screw becomes exactly the block-on-incline problem you already know, just with the incline angle replaced by the helix angle and the pushing force replaced by a torque. This article extends the single-surface Coulomb friction result to these two-surface and helical geometries: how to set up wedge equilibrium, how to derive the screw-jack torque formulas, why self-locking happens, and how to compute jack efficiency.

The core insight

A wedge or a screw thread converts Coulomb friction on an inclined contact into a large mechanical advantage, and the same friction angle $\phi=\tan^{-1}\mu_s$ that resists raising the load can also make the incline self-locking against lowering.

The core idea

Wedges as force multipliers. A wedge has two or more inclined faces. Pushed horizontally, it slides beneath (or between) other parts and raises them, converting a small horizontal force into a large normal/vertical force. Because motion is impending at every sliding contact simultaneously, each contact carries friction at its limiting value, $F=\mu_s N$ , acting to oppose the relative sliding at that surface. Solving a wedge problem means drawing a separate free body diagram for the wedge and for the body it lifts, writing $F=\mu_s N$ at each contact, and solving the resulting simultaneous equilibrium equations — there is no single universal formula because the number of frictional surfaces (one, two, or three) depends on the geometry.

Screws as a wrapped incline. A square-threaded screw of mean thread radius $r$ with lead $L$ (axial advance per revolution) is geometrically an inclined plane wrapped around a cylinder. Unrolling one turn of the thread gives a right triangle of base $2\pi r$ (the circumference) and height $L$ , so the helix angle is

\theta = \tan^{-1}\left(\frac{L}{2\pi r}\right)

where: $\theta$ is the helix (lead) angle, $L$ is the lead, and $r$ is the mean thread radius. Raising or lowering the load $W$ by turning the screw is then equivalent to pushing a block of weight $W$ up or down this unrolled incline, with the equivalent horizontal push producing a torque $M = Fr$ about the screw axis, and friction acting at angle $\phi=\tan^{-1}\mu_s$ exactly as in ordinary Coulomb friction.

Figure

A screw thread unrolled into an inclined plane triangle with the lead as height and the thread circumference as base, showing the helix angle theta

Raising and lowering torques. Because raising the load is impending motion up the equivalent incline, and lowering (while the screw remains self-locking) is impending motion down it, the standard incline-with-friction results carry over directly, replacing force with torque and the pushing force with $Wr$ :

M_{raise} = Wr\tan(\theta+\phi), \qquad M_{lower} = Wr\tan(\phi-\theta)

where: $M_{raise}$ and $M_{lower}$ are the torques needed to just start raising or lowering the load, $W$ is the load, $r$ the mean thread radius, $\theta$ the helix angle, and $\phi=\tan^{-1}\mu_s$ the friction angle.

Self-locking and efficiency

A screw is self-locking — it holds the load without unwinding when you release the handle — exactly when friction alone can supply enough resistance to balance the load on the equivalent incline, i.e. when $\phi\ge\theta$ , or equivalently $\mu_s\ge\tan\theta = L/2\pi r$ . Physically, self-locking needs a shallow helix (small lead relative to radius) or higher friction; a steep, fast-advancing thread ( $\theta$ large) tends not to be self-locking and would need a holding torque to stop it running back under load.

Derivation AScrew-jack torque to raise a load, from the unrolled-incline model

Derive the torque $M_{raise}$ needed to just start raising a load $W$ on a square-threaded screw of mean radius $r$ and helix angle $\theta$ against friction angle $\phi=\tan^{-1}\mu_s$ , by treating the screw as a block sliding up an equivalent inclined plane under an equivalent horizontal force.

Step 1. Unroll the thread contact into a plane inclined at angle $\theta$ to the horizontal, carrying a block of weight $W$ (the axial load). Turning the screw one revolution corresponds to sliding this block once up the full length of the unrolled incline while the load $W$ stays vertical, exactly as in ordinary wedge-and-incline problems.

Step 2. At impending motion up the incline, three forces act on the block: the weight $W$ vertically down, the normal reaction $N$ perpendicular to the incline, and the friction force $F=\mu_s N$ acting down the incline (opposing the impending upward slip), together with the equivalent applied force $P$ acting horizontally (this represents the effect of the torque, since $P$ acts at radius $r$ from the screw axis).

Step 3. Resolve along axes parallel and perpendicular to the incline. Perpendicular: $N = W\cos\theta + P\sin\theta$ . Parallel (up positive): $P\cos\theta = W\sin\theta + F = W\sin\theta+\mu_s N$ . Substituting $N$ and using $\mu_s=\tan\phi$ throughout gives, after combining terms and dividing by $\cos\theta\cos\phi$ , the compact form $P = W\tan(\theta+\phi)$ — friction simply adds the friction angle $\phi$ onto the geometric incline angle $\theta$ .

Step 4. Convert the equivalent force to a torque about the screw axis: since $P$ acts effectively at the mean thread radius $r$ , the torque is $M_{raise}=Pr=Wr\tan(\theta+\phi)$ , which is the required result. Running Step 3 for impending motion down the incline instead (friction now reverses to act up the incline, resisting the descent) gives $P=W\tan(\phi-\theta)$ and so $M_{lower}=Wr\tan(\phi-\theta)$ , valid whenever $\phi>\theta$ (self-locking); if $\phi<\theta$ the load descends on its own and a torque is instead needed to hold it back.

The mechanical efficiency compares the ideal, frictionless torque ( $\theta$ only, $\phi=0$ ) to the actual torque required:

\eta = \frac{Wr\tan\theta}{Wr\tan(\theta+\phi)} = \frac{\tan\theta}{\tan(\theta+\phi)}

Self-locking screws ( $\phi\ge\theta$ ) are always comparatively inefficient, because the same friction that stops the screw unwinding also eats most of the input work as heat while raising the load — you cannot have "easy to turn" and "holds itself" from friction alone at the same time.

Worked example

A square-threaded screw jack has mean thread radius $r=20$ mm and lead $L=8$ mm, with $\mu_s=0.15$ between screw and nut, supporting a load $W=6000$ N. Find the helix angle, check self-locking, find the torque to raise the load, and find the efficiency — then find the torque to lower the load.

Step 1 — Helix angle. $\theta=\tan^{-1}(L/2\pi r)=\tan^{-1}(8/125.7)=\tan^{-1}(0.0637)=3.64°$ .

Step 2 — Self-locking check. $\phi=\tan^{-1}(0.15)=8.53°$ . Since $\phi(8.53°)>\theta(3.64°)$ , the condition $\mu_s\ge\tan\theta$ holds and the screw is self-locking: it holds the 6000 N load without a holding torque.

Step 3 — Torque to raise. $M_{raise}=Wr\tan(\theta+\phi)=6000(0.020)\tan(12.17°)=120(0.2157)=25.9\ \text{N}\,\cdot\,\text{m}$ .

Step 4 — Efficiency. $\eta=\tan\theta/\tan(\theta+\phi)=0.0637/0.2157=0.295\approx29.5\%$ , typically low because this screw is comfortably self-locking.

Step 5 — Torque to lower. Since $\phi>\theta$ , use $M_{lower}=Wr\tan(\phi-\theta)=120\tan(8.53°-3.64°)=120\tan(4.89°)=120(0.0855)=10.3\ \text{N}\,\cdot\,\text{m}$ , a positive torque still needed with the load (in the direction of lowering) because friction continues to resist motion even though it is not large enough to prevent it entirely once that torque is applied.

The torque required to raise the 6000 N load is 25.9 N·m, the jack is self-locking, its efficiency is about 29.5%, and it takes 10.3 N·m to lower the load.

Competitive stretch

If the same screw instead had a double-start thread with lead $L=16$ mm (thread advances twice as fast per turn, same $r$ and $\mu_s$ ), recompute: $\theta=\tan^{-1}(16/125.7)=\tan^{-1}(0.1273)=7.25°$ . Now $\phi(8.53°)$ still exceeds $\theta(7.25°)$ , so it remains self-locking, but only just — $M_{raise}=120\tan(15.78°)=120(0.2826)=33.9\ \text{N}\,\cdot\,\text{m}$ and efficiency rises to $\eta=\tan7.25°/\tan15.78°=0.1273/0.2826=0.451\approx45.1\%$ . Doubling the lead nearly doubled the torque needed but also nearly doubled the efficiency — illustrating the fundamental trade-off between self-locking margin and efficiency.

Common traps

Mixing up raising vs. lowering formulas. Raising uses $\tan(\theta+\phi)$ ; lowering (while still self-locking) uses $\tan(\phi-\theta)$ — swapping them gives a negative or nonsensical torque.
Using the outer diameter instead of the mean thread radius. The screw formulas need the mean (pitch) radius $r$ , not the major or minor diameter — always check the diagram or problem statement.
Comparing $\mu_s$ to $\theta$ directly. The self-locking test is $\mu_s\ge\tan\theta$ , not $\mu_s\ge\theta$ ; also remember a larger lead angle makes a screw less likely to self-lock, not more.
Assuming every wedge has the same number of friction surfaces. Some wedge problems have friction at one surface, others at two or three (as in the guided-block example) — always draw both FBDs before reaching for a memorized formula.

Ask the tutor: If the incline-unrolling picture for a screw thread or the two-FBD setup for a wedge still feels shaky, ask the AI tutor to walk through the geometry step by step or to compare a self-locking and a non-self-locking case side by side.

Why do we bother unrolling the screw thread into an incline instead of analyzing it as a 3D helix directly

Because the thread's local contact behaves exactly like a block sliding on a straight incline once you go around one turn: the same weight, the same friction law, and the same "add phi to theta" result. Unrolling turns an intimidating 3D helix problem into a 2D incline problem you already know how to solve, and it is exact for a square thread with all the load carried on the flat, non-inclined-in-the-other-direction thread face.

Why can a screw jack hold a huge load with no motor or brake, when a smooth ramp cannot

A smooth (frictionless) ramp always lets the load slide back down, because nothing resists motion along it. A screw thread has friction at the contact, and if the helix angle theta is shallow enough that the friction angle phi exceeds it, that friction alone supplies more resisting force than the load's tendency to slide back — so the screw is self-locking. A steep thread (large lead) removes that margin and the screw can unwind on its own, the same way a steep frictional ramp lets a block slide even with friction present.

Is a wedge always more efficient than a self-locking screw since it has fewer friction angles added

Not necessarily — efficiency depends on the specific geometry and friction coefficients in each case, not on being a wedge versus a screw. What is true in general is the same trade-off shown in the worked example: any self-locking device, wedge or screw, sacrifices efficiency for the safety of holding its load without slipping, because the friction doing the holding is the same friction eating the input work while raising.