A power screw has a lead angle of 6° and a friction angle of 8°. A heavy load is placed on the screw with no torque applied to the screw shaft. What happens?
AThe load back-drives the screw downward because the incline exceeds the friction capacity
BThe load remains stationary — the screw is self-locking because the lead angle (6°) is less than the friction angle (8°)
CThe screw is at the self-locking threshold, so small vibrations will cause it to slip
DThe efficiency is above 50%, so the screw cannot be self-locking
The self-locking condition is λ < φ_s. Here 6° < 8°, so the screw is self-locking. In the inclined-plane analogy, the thread slope is shallower than the friction angle — friction is strong enough to prevent the load from sliding the screw backward. Applying the lowering torque formula M = Wr·tan(λ − φ_s) = Wr·tan(−2°) gives a negative value, meaning you would need to apply torque in the raising direction to make the load descend. Without that torque, the load holds.
Question 2 Multiple Choice
An engineer designing a lifting jack selects a screw with a very small lead angle, reasoning that smaller lead angles improve mechanical efficiency. What error has she made?
AA small lead angle requires a larger friction coefficient to function, which is difficult to achieve
BA small lead angle means λ << φ_s, placing the screw deeply in the self-locking regime where efficiency is well below 50% — if high efficiency is the goal, a larger lead angle is needed (at the cost of requiring a separate brake to prevent back-driving)
CLead angle does not affect efficiency — only the friction coefficient determines how much torque is wasted
DA small lead angle causes problems only during lowering, not during raising
Efficiency η = tan(λ) / tan(λ + φ_s), and for a self-locking screw (λ < φ_s) this is always below 50%. As λ decreases, efficiency drops further — the screw becomes safer against back-driving but wastes more input torque as heat. The efficiency bound of 50% is not coincidental: it is a mathematical consequence of self-locking. So the engineer faces a real tradeoff: safety (self-locking) vs. efficiency. A lifting application like a car jack accepts low efficiency; a precision positioning stage would choose a high-lead-angle ballscrew and a separate brake.
Question 3 True / False
A self-locking power screw always has a mechanical efficiency below 50%.
TTrue
FFalse
Answer: True
This follows directly from the efficiency formula η = tan(λ) / tan(λ + φ_s). For a self-locking screw, λ < φ_s, which means λ + φ_s > 2λ. Using the tangent addition formula and the fact that tan is monotone, η = tan(λ)/tan(λ + φ_s) < tan(λ)/tan(2λ) = 1/2 when λ < φ_s. Intuitively: the same friction force that prevents back-driving (self-locking) also resists the applied torque during raising, dissipating more than half the input energy as heat. This 50% bound is a fundamental law of screw mechanics, not a design deficiency.
Question 4 True / False
A screw with a lead angle of 30° and a friction angle of 15° is self-locking because the large lead angle provides greater mechanical advantage against the load.
TTrue
FFalse
Answer: False
The self-locking condition is λ < φ_s — lead angle must be LESS than friction angle. Here λ = 30° > φ_s = 15°, so the screw is overhauling (back-drivable). The load CAN drive the screw backward without any applied torque. Mechanical advantage from a large lead angle works in the opposite direction: a large lead angle advances the load more per revolution (good for speed) but reduces the friction-to-slope ratio, making back-driving easier. Self-locking requires a shallow thread helix (small λ) so friction dominates the incline geometry.
Question 5 Short Answer
Explain why the self-locking condition requires the lead angle to be less than the friction angle, using the inclined-plane analogy.
Think about your answer, then reveal below.
Model answer: Unwrapping the screw thread produces an inclined plane with slope angle λ (the lead angle), with the load W as a block on the incline. The block will hold without applied force only if friction is strong enough to resist gravity along the slope. The critical condition is when the friction force exactly equals the gravitational component along the plane, which occurs at the friction angle φ_s = arctan(μ_s). For inclinations shallower than φ_s (λ < φ_s), friction wins — the block holds and the screw is self-locking. For inclinations steeper than φ_s (λ > φ_s), the gravitational component wins — the block slides and the screw is overhauling. The lowering torque formula M = Wr·tan(λ − φ_s) encodes this directly: when λ < φ_s the torque is negative, meaning the load cannot descend without external assistance.
The inclined-plane analogy is the conceptual key to all screw analysis. It transforms a seemingly complex 3D helical problem into a familiar 2D statics problem. Once you see the screw as a wrapped inclined plane, the raising formula (load must push up against both gravity and friction), the lowering formula (friction now partially aids descent), and the self-locking condition (friction exceeds the slope component) all follow naturally from force balance on the block.