Question

A cone clutch has a contact surface with half angle of 62° and a normal force of 800 N.
The friction coefficient of the contact surface is 0,36. Determine the axial force required to engage the clutch.

Answer 1

Resolve the normal force into axial components. The normal N on the conical surface makes angle (90° − α) with the axis, so its axial component is N sin α. The friction f = μN acts along the surface; its axial component is μN cos α and opposes engagement, so the axial thrust required is

Fa = N(sin α + μ cos α).

With α = 62°, N = 800 N, μ = 0.36:

sin 62° = 0.8829, cos 62° = 0.4695
Fa = 800(0.8829 + 0.36·0.4695) = 800(0.8829 + 0.1690) ≈ 800·1.0519 ≈ 842 N.

So the axial force required is about 8.42 × 10^2 N (≈842 N).