Questions: Friction Applications: Wedges, Screws, and Belts

Question 1

A rope is wrapped around a post with contact angle β = π radians (half a turn), with μ = 0.3. The slack-side tension is T_slack = 50 N. A student doubles the wrap to β = 2π (one full turn). How does the tight-side tension T_tight change?

• A. It doubles — the tension ratio is proportional to the wrap angle
• B. It quadruples — doubling the wrap angle squares the achievable tension ratio
• C. It increases by e^(0.3π) — adding one more half-turn multiplies the ratio by the same factor
• D. It stays the same — T_tight depends on the applied force, not the wrap angle

Answer: (C)

The belt friction equation is T_tight/T_slack = e^(μβ). At β = π: ratio = e^(0.3π) ≈ 2.57, so T_tight ≈ 128 N. At β = 2π: ratio = e^(0.6π) ≈ 6.59, so T_tight ≈ 330 N. Doubling β from π to 2π squares the ratio (e^(0.6π) = (e^(0.3π))²), not doubles it. This exponential relationship is why a few turns of rope around a capstan can hold enormous loads — adding half a turn doesn't add a fixed amount, it multiplies the existing ratio by e^(μπ).

Question 2

A square-threaded screw has lead angle λ = 8° and the friction angle φ_s = arctan(μ_s) = 12°. What happens when the driving torque is removed while the screw is loaded?

• A. The screw back-drives — the load pushes it backward because the lead angle is less than 45°
• B. The screw self-locks — friction is strong enough to prevent back-driving because λ < φ_s
• C. The screw back-drives — the load always overcomes static friction unless the thread is locked mechanically
• D. The screw self-locks only if the load is applied axially; radial loads always cause back-driving

Answer: (B)

The self-locking condition for a screw is λ < φ_s. Here λ = 8° < φ_s = 12°, so the screw self-locks. The lead angle measures how steeply the thread helix rises; the friction angle measures the angle at which friction force can balance the load's tendency to push the nut backward down the thread. When the lead angle is shallower than the friction angle, friction wins and the screw stays put. Standard fasteners are designed this way — if λ > φ_s, the screw would unscrew under vibration.

Question 3

In a belt friction problem, the tight side always carries higher tension than the slack side, and identifying which side is tight requires knowing the direction of impending motion.

• T. True
• F. False

Answer: True

The belt friction equation T_tight/T_slack = e^(μβ) always gives a ratio ≥ 1 (since e^(μβ) > 1 for μ, β > 0), so the tight side always has higher tension. Which side is tight depends on the physical setup: the tight side is the side the load or motion tends to pull the belt toward. Identifying tight vs. slack from the direction of impending motion must happen before applying the formula — applying it backward gives an inverted ratio that is less than 1, predicting a physically impossible configuration.

Question 4

Doubling the contact angle β in a belt friction problem doubles the achievable tension ratio T_tight/T_slack.

• T. True
• F. False

Answer: False

The relationship is exponential, not linear: T_tight/T_slack = e^(μβ). Doubling β replaces e^(μβ) with e^(2μβ) = (e^(μβ))², which squares the original ratio. For example, if the original ratio is 3, doubling the wrap angle gives ratio 9, not 6. This exponential behavior is the engineering power of capstans and bollards — a small number of additional turns produces a dramatically larger holding force, not a proportional increase.

Question 5

Explain why the self-locking condition for a screw (λ < φ_s) means the screw will not back-drive under axial load. What physical mechanism keeps the screw from unscrewing when the driving torque is removed?

Model answer: A screw thread is geometrically a wedge wrapped around a cylinder. When an axial load tries to push the screw backward (unscrew it), the load's component along the thread helix must overcome the friction force at the thread surface. The lead angle λ controls how much of the axial load resolves into a back-driving force along the thread; the friction angle φ_s controls the maximum friction available to resist it. When λ < φ_s, the friction force is larger than the component of the load trying to slide the thread backward, so the thread stays in place — friction self-locks the system.

This is exactly the wedge self-locking analysis applied to a helical geometry. Standard bolts and machine screws use fine threads (small lead angle) specifically to ensure self-locking. Power screws used for lifting or pressing (like a car jack) may deliberately have larger lead angles to allow back-driving, trading self-locking for mechanical advantage. The condition λ = φ_s is the tipping point: below it the screw holds, above it the screw unwinds under load.