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Question Number 116078 by Study last updated on 30/Sep/20

$$provethatFr=\frac{v^2}{gh}froudenumer$$

Answered by MrGaster last updated on 06/Jan/25

$$F_r=\frac{v^2}{gh}\left(froudenumer\right)$$$$\mathrm{Let}v\mathrm{be}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{object},g\mathrm{be}\mathrm{the}\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},\mathrm{and}h\mathrm{be}\mathrm{the}\mathrm{characteristic}\mathrm{length}$$$$\mathrm{From}\mathrm{dimensional}\mathrm{analysis}:$$$$\mid Fr\mid =\frac{\mid v{\mid }^2}{\mid g\mid \mid h\mid }$$$$\mid v\mid =\frac{L}{T},\left[g\right]=\frac{L}{T^2},\left[h\right]=L$$$$\left[Fr\right]=\frac{\left(\frac{L}{T}\right)}{\frac{L}{T^2}\cdot L}=\frac{\frac{L^2}{T^2}}{\frac{L^2}{T^2}}=1$$$$\mathrm{Thus}\mathrm{the}\mathrm{Froude}\mathrm{number}\mathrm{isidimensonless}.$$$$\mathrm{Consider}\mathrm{the}\mathrm{forces}\mathrm{acting}\mathrm{on}\mathrm{thejobect}:$$$$\mathrm{Gravitational}\mathrm{force}{F}_G\propto \rho {gL}^3$$$$\mathrm{The}\mathrm{Froude}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{ratio}\mathrm{ofn}$$$$\mathrm{iertial}\mathrm{force}\mathrm{to}\mathrm{gravitational}\mathrm{force}:$$$$F_r=\frac{F_1}{F_G}=\frac{\rho v^2{L}^2}{\rho {gL}^3}=\frac{v^2}{gL}$$$$\mathrm{Therefore}\mathrm{it}\mathrm{is}\mathrm{proven}\mathrm{that}{F}_r=\frac{v^2}{hh}.$$

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