Question

A self-locking screw is one which

• A. Has locking arrangement
• B. Has a hole drilled through for inserting locking pin
• C. Has coefficient of friction equal to or greater than the tangent of the load angle
• D. Has fine pitch screw threads

Hint:

Efficiency of a self-locking screw is always less than 50%. If efficiency exceeds 50%, the screw is "overhauling" and will unwind under its own load.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This problem pertains to the mechanics of power screws, which are used to convert rotary motion into linear motion. A "self-locking" screw is a critical engineering component that stays in place under a load without needing an external brake. If a screw is not self-locking, the weight of the load might cause the screw to rotate backwards and descend on its own (a condition called "overhauling"). The question asks for the mathematical criteria involving friction and geometry that ensures this stable, self-locking behavior.
Step 2: Key Formulas and approach:
The behavior of a screw can be modeled as a block on an inclined plane, where the incline angle is the helix angle ($\alpha$). The key variables are:
1. $\alpha$: The helix angle or lead angle of the thread.
2. $\phi$: The friction angle, defined such that $\tan(\phi) = \mu$ (the coefficient of friction).
For a screw to be self-locking, the torque required to lower the load must be positive. This occurs mathematically when the friction angle is greater than or equal to the helix angle: $\phi \geq \alpha$.
Step 3: Detailed Explanation:

Consider the forces acting on the thread: the axial load, the normal force, and the frictional force.

The tendency of the load to slide down the thread is proportional to $\sin(\alpha)$.

The frictional force resisting this motion is proportional to $\mu \cos(\alpha)$.

For the screw to remain stationary (self-lock), the resisting frictional force must be greater than or equal to the downward component of the load.

This leads to the condition: $\mu \geq \tan(\alpha)$.

Since $\mu$ is the coefficient of friction and $\tan(\alpha)$ is the tangent of the load (helix) angle, the condition is that the coefficient of friction must be at least as large as the tangent of the helix angle.

Furthermore, it can be proven that for a screw to be self-locking, its mechanical efficiency must be less than 50%. If efficiency is higher, the screw will overhaul.

Step 4: Final Answer:
The mechanical condition for self-locking is that the coefficient of friction is equal to or greater than the tangent of the load angle, which is option (C).

Answer 2

We can derive the self-locking condition directly from a force balance on the thread, treated as a block resting on an inclined plane whose incline angle equals the helix angle \(\alpha\) of the screw.

1. Setting up the forces: The axial load \(W\) presses the block onto the incline. Its component trying to slide the block back down the slope is \(W\sin\alpha\), and the normal force pressing the block into the incline is \(W\cos\alpha\).
2. Maximum available friction: The maximum friction force resisting that sliding is \(\mu\times(\text{normal force})=\mu W\cos\alpha\), where \(\mu\) is the coefficient of friction between the threads.
3. The self-locking condition: For the screw to stay put under the load alone, the resisting friction force must be at least as large as the force trying to slide it back down: \(\mu W\cos\alpha \geq W\sin\alpha\).
4. Simplifying: Dividing both sides by \(W\cos\alpha\) (positive, so the inequality direction is unchanged) gives \(\mu \geq \tan\alpha\), i.e. the coefficient of friction must be equal to or greater than the tangent of the load (helix) angle.

This force-balance derivation confirms, from the ground up, that self-locking depends purely on how the friction coefficient compares to the tangent of the load angle, matching none of the other three options about pins, latches, or pitch.

Therefore, the correct answer is Has coefficient of friction equal to or greater than the tangent of the load angle.