Roll crusher nip condition: which inequality correctly relates the coefficient of friction (μ) between rock and roll and the half-angle of nip (α) required to draw particles into the rolls?

Introduction:
For a roll crusher to “nip” a particle and pull it into the crushing zone, friction must be sufficient to overcome the particle’s tendency to slide. This basic force balance produces a design inequality relating friction to the nip angle. Recognizing the correct form helps in selecting roll diameter, surface, and operating conditions.

Given Data / Assumptions:

• Two smooth rolls; half-angle of nip is α.
• Coefficient of friction between particle and roll is μ.
• Quasi-static force balance at the point of contact.

Concept / Approach:
Resolving normal and tangential components on the particle shows that the tangential friction force must at least equal the component of normal force trying to eject the particle. The well-known criterion is μ ≥ tan α (strictly μ must not be less than tan α). If μ is lower, the particle will slip instead of being nipped.

Step-by-Step Solution:
At contact, let N be normal force; friction available = μN.Tangential component of the geometry at angle α is N sin α; normal component is N cos α.Condition to avoid slip: μN ≥ N tan α ⇒ μ ≥ tan α.Hence the correct inequality is μ ≥ tan α.

Verification / Alternative check:
Textbook derivations for smooth rolls produce the same inequality and relate maximum nip angle to friction (larger μ allows larger nip angles). Surface roughening effectively increases the usable μ.

Why Other Options Are Wrong:

• μ > tan α: sometimes stated, but the minimal requirement includes equality; ≥ is the more correct design condition.
• μ > tan 2α: dimensionally wrong; angle should not be doubled.
• μ ≤ tan α: would predict slipping at or above nip condition.
• μ = cot α: unrelated to the nip balance.

Common Pitfalls:
Confusing α (half-angle) with the full nip angle, or forgetting that surface texturing changes effective friction.

Final Answer:
μ ≥ tan α