A cylindrical roller gate of diameter 6.0 m and length 10.0 m is placed on a dam to store water behind it, as shown in figure.
The magnitude of the resultant force due to water acting on that gate when the water is about to spill is ______ × 10⁶ N (rounded off to two decimal places).
Consider acceleration due to gravity (g) = 9.81 m/s²;
density of water (ρ) = 1000 kg/m³; and π = 3.14.

Question

Which one of the following gases is primarily responsible for the formation of photochemical smog?

Answer 1

Instead of separately deriving the vertical force from a volume argument, it can be obtained directly by integrating the pressure around the wetted half of the cylinder, which also confirms the volume-based shortcut used in the other method.

Place the origin at the centre of the circle, and describe a point on the wetted (water-side) surface by the angle $\theta$ measured from the top point, so its depth below the free surface is $h(\theta) = R(1-\cos\theta)$ for $\theta$ from $0$ (top) to $\pi$ (bottom).

The pressure there is $p(\theta) = \rho g R(1-\cos\theta)$, and it acts radially toward the centre. Its vertical component, integrated over the whole wetted arc length $R\,d\theta$ and length $L$, works out to $$F_v = \rho g R^2 L\int_0^{\pi}(1-\cos\theta)\cos\theta\,d\theta$$

Evaluating the integral, $\int_0^\pi(1-\cos\theta)\cos\theta\,d\theta = -\pi/2$, and after accounting for the sign convention (the lower quadrant pushes up more than the upper quadrant pushes down), this reduces to the clean closed form $$F_v = \rho g \frac{\pi R^2}{2} L = 1000\times9.81\times\frac{3.14\times9}{2}\times10 = 1{,}386{,}153\ N$$

This is exactly the weight of a half-cylinder of water of the same radius and length, confirming the shortcut. Combining with the horizontal force $F_h = \rho g\cdot2R^2L = 1{,}765{,}800$ N found from the projected area: $$F = \sqrt{F_h^2+F_v^2} = \sqrt{(1{,}765{,}800)^2+(1{,}386{,}153)^2} = 2{,}244{,}876\ N$$ \[\boxed{F \approx 2.24\times10^6\ N}\]

Answer 2

Instead of separately deriving the vertical force from a volume argument, it can be obtained directly by integrating the pressure around the wetted half of the cylinder, which also confirms the volume-based shortcut used in the other method.

Place the origin at the centre of the circle, and describe a point on the wetted (water-side) surface by the angle $\theta$ measured from the top point, so its depth below the free surface is $h(\theta) = R(1-\cos\theta)$ for $\theta$ from $0$ (top) to $\pi$ (bottom).

The pressure there is $p(\theta) = \rho g R(1-\cos\theta)$, and it acts radially toward the centre. Its vertical component, integrated over the whole wetted arc length $R\,d\theta$ and length $L$, works out to $$F_v = \rho g R^2 L\int_0^{\pi}(1-\cos\theta)\cos\theta\,d\theta$$

Evaluating the integral, $\int_0^\pi(1-\cos\theta)\cos\theta\,d\theta = -\pi/2$, and after accounting for the sign convention (the lower quadrant pushes up more than the upper quadrant pushes down), this reduces to the clean closed form $$F_v = \rho g \frac{\pi R^2}{2} L = 1000\times9.81\times\frac{3.14\times9}{2}\times10 = 1{,}386{,}153\ N$$

This is exactly the weight of a half-cylinder of water of the same radius and length, confirming the shortcut. Combining with the horizontal force $F_h = \rho g\cdot2R^2L = 1{,}765{,}800$ N found from the projected area: $$F = \sqrt{F_h^2+F_v^2} = \sqrt{(1{,}765{,}800)^2+(1{,}386{,}153)^2} = 2{,}244{,}876\ N$$ \[\boxed{F \approx 2.24\times10^6\ N}\]

Show Hint

Correct Answer: 2.24

Solution and Explanation

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