Force of a jet on an inclined flat plate (stationary) Ratio of the normal (perpendicular-to-plate) force when the plate is at 30° to the jet, to the force when the plate is normal to the jet, equals:

Introduction / Context:
When a water jet strikes a stationary flat plate, the force depends on how much of the jet's momentum is stopped in the direction normal to the plate surface. Tilting the plate reduces the effective normal component of velocity, hence reducing the normal force compared to the case when the plate is perpendicular to the jet.

Given Data / Assumptions:

• Jet of uniform velocity V and area a strikes a stationary, smooth, flat plate.
• The plate is oriented at θ = 30° to the jet direction (0° = parallel, 90° = normal).
• Flow follows the plate with negligible splashing loss for the basic estimate.

Concept / Approach:
The normal component of velocity is V_n = V * sin θ. Only this component is brought to rest normal to the plate, so the normal momentum flux is ρ * a * V_n^2. Thus, normal force F_n ∝ ρ * a * V^2 * sin^2 θ. For θ = 90° (plate normal to jet), sin θ = 1 and F_n(normal) ∝ ρ * a * V^2.

Step-by-Step Solution:
For θ = 30°: V_n = V * sin 30° = V * 1/2.Normal momentum flux ∝ V_n^2 = (V/2)^2 = V^2 / 4.Force ratio = F_n(30°) / F_n(90°) = sin^2 30° / sin^2 90° = (1/2)^2 / 1^2 = 1/4.Hence the normal force at 30° is one-quarter of that at 90°.

Verification / Alternative check:
Dimensional reasoning and momentum balance lead to the same sin^2 θ dependence. At θ → 0°, ratio → 0 (grazing flow). At θ = 90°, ratio = 1, matching limits.

Why Other Options Are Wrong:
1/2 and 3/4 underestimate the reduction; 1 implies no change with angle; 2 is impossible because tilting cannot increase the normal force beyond the 90° case.

Common Pitfalls:
Confusing force along the jet direction with force normal to the plate; mixing sin θ and cos θ depending on chosen angle definition.

Final Answer:
1/4