Question
Download Solution PDFThe force exerted by a jet of water on a stationary vertical plate in the direction of the jet is ________.
where, ρ = mass density of the liquid, a = cross-sectional area of the jet, V = velocity of the jet.Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Jet Striking a stationary vertical plate:
From Newton’s second law
F = dp/dt; where; p = momentum = mv
∴ \(F = \frac{{\left( {initial\;momentum\; - \;final\;momentum} \right)}}{{time}}\)
∴ \(F = \frac{{\left( {mass\; \times \;initial\;velocity\; - \;mass\; \times \;final\;velocity} \right)}}{{time}}\)
∴ \(F = \frac{{m\;\left( {velocity\;of\;jet\;before\;striking - velocity\;of\;jet\;after\;striking} \right)}}{{time}}\)
∴ \(F = \frac{{m\left( {V - 0} \right)}}{{time}}\) ----(1)
Now
Mass = ρ × volume ; volume = area × length
∴ [mass (m)]/time = [ρ × area × length]/time ∵ length/time = velocity
∴ \(\frac{m}{{time}} = \rho \times area \times velocity\) ----(2)
Substitute equation (2) in equation (1)
∴ F = ρ × a × v2
Note: Jet Striking a Stationary incline plate
Let us apply the impulse-momentum equation in the direction normal to the plate
\(\begin{array}{l} {F_n} = \rho aV\left( {Vsin\theta - 0} \right) = \rho a{V^2}\sin \theta \\ {F_x} = {F_n}\sin \theta = \rho {V^2}\sin \theta \times \sin \theta = \rho a{V^2}{\sin ^2}\theta \\ {F_y} = {F_n}\cos \theta = \rho {V^2}\sin \theta \times \cos \theta = \rho a{V^2}{\sin }\theta \cos \theta \end{array}\)Last updated on Jul 1, 2025
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