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ECH 141 Problem Set #3 1. Whitaker problem 2-8. Substitute y=6m for y=6ft, and use β =1m 1/2 . Solution If L=6m, then the pressure in the water is, from derivations in class, p = p atm + ρ g L y ( ) . To calculate the force on the curved gate in the figure, we start from F = pn A G dA so that the y- and x-components of the force can be found from F x = e x F = pe x n A G dA and F y = e y F = pe y n A G dA . These expressions yield areas projected to the yz plane for the x-component and the zx plane for the y- component. Note that projection to the yz plane yields a negative sign (in calculation of the x- component of force), whereas projection to the xz plane does not yield a negative sign. Since atmospheric pressure operates on both sides of the gate it makes no net contribution. We find F x = ρ g 6m y ( ) dzdy 0 W 0 4m or F x = ρ gW 6my y 2 2 0 4m = 1000 kg m 3 9.8 m s 2 W 6my y 2 2 0 4m or F x = ρ gW 6my y 2 2 0 4m = 1000 kg m 3 9.8 m s 2 4m ( ) W 4m ( ) . This result shows clearly that the mean pressure is the pressure at L-y=4m, or y=2m, half way up the gate, and the projected area is W(4m). The numerical result is This study resource was
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