Question

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.

Hint:

Split the force into horizontal (projected vertical area) and vertical (weight of the half-cylinder volume of water) components, then combine them as a vector sum to get the resultant.

Answer 1

Correct Answer: 2.24

Step 1: Set up the geometry.
The gate is a full cylinder of radius \(R = 3.0\ m\) and length \(L = 10.0\ m\), resting on the dam crest with the water surface exactly at the top of the cylinder (about to spill). The water is held back against one half of the curved surface, from the top point around to the bottom tangent point.

Step 2: Find the horizontal component of the force.
The horizontal force on a curved surface equals the hydrostatic force on its vertical projection, which here is a rectangle of height equal to the full diameter \(D = 2R\) and width \(L\). The centroid of this projected area lies at depth \(D/2\) below the free surface, so \[F_h = \rho g \left(\frac{D}{2}\right)(D \times L) = \rho g \cdot 2R^2 \cdot L\] Substituting values: \[F_h = 1000 \times 9.81 \times 2 \times 3^2 \times 10 = 1{,}765{,}800\ N = 1.7658\times10^6\ N\]

Step 3: Find the vertical component of the force.
The vertical force on this curved (half-cylinder) surface equals the weight of the volume of water that would occupy a half-cylinder of the same radius and length (a standard result for a semicircular wetted surface spanning from the free surface down to the base): \[F_v = \rho g \left(\frac{\pi R^2}{2}\right) L\] Substituting values: \[F_v = 1000 \times 9.81 \times \left(\frac{3.14\times3^2}{2}\right) \times 10 = 1000\times9.81\times14.13\times10 = 1{,}386{,}153\ N = 1.386153\times10^6\ N\]

Step 4: Combine the components to get the resultant. \[F = \sqrt{F_h^2 + F_v^2} = \sqrt{(1.7658)^2 + (1.386153)^2}\times10^6\ N\] \[F = \sqrt{3.1180 + 1.9214}\times10^6 = \sqrt{5.0395}\times10^6 = 2.2449\times10^6\ N\]

Step 5: Round the answer.
Rounding to two decimal places gives \(F \approx 2.24\times10^6\ N\), well within the expected 2.00-2.50 range.