Jet on a fixed inclined plate: force component normal to the original flow A water jet of velocity V and area a strikes a fixed flat plate inclined at angle θ to the jet direction. Find the force exerted by the jet on the plate in the direction perpendicular to the initial jet (normal to flow). Choose the correct expression.
Mechanical Engineering
Hydraulic Machines
Difficulty: Medium
Choose an option
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AF = ρ a V^2 (1 − cos θ)
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BF = ρ a V^2 sin θ
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CF = ρ a V^2 cos θ
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DF = ρ a V^2 (1 + cos θ)
Answer
Correct Answer: F = ρ a V^2 sin θ
Explanation
Given
- Jet: speed V, area a, density ρ → mass flow rate ṁ = ρ a V.
- Plate: fixed, inclined at θ to jet.
ApproachTake x along the jet and y normal to the jet. The plate eliminates the normal-to-plate velocity component; in x–y, the outgoing flow has zero y-component.
Step-by-stepIncoming momentum flux in y = ṁ (V sin θ) = ρ a V · V sin θ = ρ a V^2 sin θ.Outgoing momentum flux in y = 0 (fixed plate, flow slides along the plate).Force on plate in +y (on fluid in −y) = Δ(momentum flux in y) = ρ a V^2 sin θ.
Checksθ = 0° ⇒ F = 0 (plate parallel to flow) ✔θ = 90° ⇒ F = ρ a V^2 (plate normal to flow) ✔
Final AnswerF = ρ a V^2 sin θ