Lock gates – reaction between the two leaves In a lock gate, the resultant hydrostatic pressure on the gate is P, and each gate leaf is inclined at angle α to the normal to the lock side. The reaction between the two gates equals:

Introduction:Mitre gates in navigation locks split the hydrostatic resultant into reactions along each leaf. This item checks your ability to resolve forces based on the gate geometry.

Given Data / Assumptions:

• Total hydrostatic resultant on the mitre is P acting normal to the lock wall plane.
• Two identical leaves meet at the mitre with each leaf at angle α to the wall normal.
• Friction and weight components are neglected for the force-resolution concept.

Concept / Approach:The reaction R between the leaves acts along the bisector at the mitre. By symmetry, each leaf carries the same axial thrust. Resolving P into components along the two leaves gives the relation for R in terms of P and α.

Step-by-Step Solution:1) The axial force along each leaf is R.2) Horizontal components of the two leaf forces must balance P.3) Each contributes R * 2 * sin α in total toward balancing P.4) Therefore P = 2 * R * sin α → R = P / (2 * sin α).

Verification / Alternative check:As α → 90°, sin α → 1 and R → P / 2, matching intuition that when gates are nearly perpendicular, each carries half the load axially.

Why Other Options Are Wrong:

• P * sin α or P * cos α: incorrect dimensionally for axial reaction.
• P / (2 * cos α): arises from wrong component resolution.
• P / 2: valid only for α = 90°, not general.

Common Pitfalls:Confusing angles measured to the gate face vs the lock wall; a wrong reference angle flips sine/cosine.

Final Answer:P / (2 * sin α)