Question

1 a feed silo is made out of sheet steel 3 mm thick using a hemisphere, a cylinder, and a cone. a explain why the height of the cone must be 70 cm, and hence find the slant height of the conical section. b calculate the surface area of each of the three sections, and hence show that the total amount of steel used is about 15.7 square metres. c show that the silo would hold about 5.2 cubic metres of grain when completely full. d grain has a density of 0.68 g cm⁻³ and the density of steel is

Explanation:

Part (a)

Step 1: Determine the radius of the components

The diameter of the cylinder, hemisphere, and cone is \(1.6\) m, so the radius \(r=\frac{1.6}{2} = 0.8\) m \(= 80\) cm. The total height of the silo is \(3.3\) m and the height of the cylinder is \(1.8\) m. Let the height of the cone be \(h_{cone}\) and the radius of the hemisphere is also \(r = 0.8\) m (since it's attached to the cylinder, so the radius of the hemisphere is equal to the radius of the cylinder). The height of the hemisphere is its radius \(r=0.8\) m \( = 80\) cm. The total height of the silo is the sum of the height of the hemisphere, the height of the cylinder, and the height of the cone. Wait, actually, the hemisphere is on top, the cylinder in the middle, and the cone at the bottom. So total height \(H=\) height of hemisphere \(+\) height of cylinder \(+\) height of cone. Wait, no, the hemisphere's height (from base to top) is \(r\) (since it's a hemisphere, the distance from the flat circular face to the top is \(r\)). So total height \(H= r + h_{cylinder}+h_{cone}\). We know \(H = 3.3\) m \(= 330\) cm, \(h_{cylinder}=1.8\) m \( = 180\) cm, \(r = 80\) cm. So \(h_{cone}=H - r - h_{cylinder}\). Wait, \(330-80 - 180=70\) cm. So that's why the height of the cone is \(70\) cm.

Step 2: Calculate the slant height of the cone

For a cone, the slant height \(l\) is given by the Pythagorean theorem \(l=\sqrt{r^{2}+h_{cone}^{2}}\), where \(r = 80\) cm and \(h_{cone}=70\) cm. So \(l=\sqrt{80^{2}+70^{2}}=\sqrt{6400 + 4900}=\sqrt{11300}\approx106.3\) cm \( = 1.063\) m.

Part (b)

Step 1: Surface area of the hemisphere

The surface area of a hemisphere (excluding the circular base, since it's attached to the cylinder) is \(2\pi r^{2}\). \(r = 0.8\) m. So \(A_{hemisphere}=2\pi(0.8)^{2}=2\pi\times0.64 = 1.28\pi\) m².

Step 2: Surface area of the cylinder

The lateral (curved) surface area of a cylinder is \(2\pi rh_{cylinder}\), where \(r = 0.8\) m and \(h_{cylinder}=1.8\) m. So \(A_{cylinder}=2\pi\times0.8\times1.8=2.88\pi\) m².

Step 3: Surface area of the cone

The lateral (curved) surface area of a cone is \(\pi rl\), where \(r = 0.8\) m and \(l\approx1.063\) m (from part a). So \(A_{cone}=\pi\times0.8\times1.063\approx0.8504\pi\) m².

Step 4: Total surface area

Total surface area \(A = A_{hemisphere}+A_{cylinder}+A_{cone}=1.28\pi+2.88\pi + 0.8504\pi=(1.28 + 2.88+0.8504)\pi=5.0104\pi\approx5.0104\times3.1416\approx15.7\) m².

Part (c)

Step 1: Volume of the hemisphere

The volume of a hemisphere is \(\frac{2}{3}\pi r^{3}\), with \(r = 0.8\) m. So \(V_{hemisphere}=\frac{2}{3}\pi(0.8)^{3}=\frac{2}{3}\pi\times0.512\approx0.3413\pi\) m³.

Step 2: Volume of the cylinder

The volume of a cylinder is \(\pi r^{2}h_{cylinder}\), with \(r = 0.8\) m and \(h_{cylinder}=1.8\) m. So \(V_{cylinder}=\pi\times(0.8)^{2}\times1.8=\pi\times0.64\times1.8 = 1.152\pi\) m³.

Step 3: Volume of the cone

The volume of a cone is \(\frac{1}{3}\pi r^{2}h_{cone}\), with \(r = 0.8\) m and \(h_{cone}=0.7\) m (since \(70\) cm \( = 0.7\) m). So \(V_{cone}=\frac{1}{3}\pi\times(0.8)^{2}\times0.7=\frac{1}{3}\pi\times0.64\times0.7\approx0.1493\pi\) m³.

Step 4: Total volume

Total volume \(V=V_{hemisphere}+V_{cylinder}+V_{cone}=0.3413\pi + 1.152\pi+0.1493\pi=(0.3413 + 1.152+0.1493)\pi=1.6426\pi\approx1.6426\times3.1416\approx5.16\approx5.2\) m³.

Answer:

(a) The height of the cone is found by subtracting the height of the hemisphere (\(0.8\) m) and the height of the cylinder (\(1.8\) m) from the total height (\(3.3\) m): \(3.3-(0.8 + 1.8)=0.7\) m \( = 70\) cm. The slant height of the cone is \(\sqrt{0.8^{2}+0.7^{2}}=\sqrt{0.64 + 0.49}=\sqrt{1.13}\approx1.063\) m.

(b) Surface area of hemisphere: \(2\pi(0.8)^{2}=1.28\pi\) m², cylinder: \(2\pi(0.8)(1.8)=2.88\pi\) m², cone: \(\pi(0.8)(1.063)\approx0.8504\pi\) m². Total: \(1.28\pi+2.88\pi + 0.8504\pi\approx5.0104\times3.1416\approx15.7\) m².

(c) Volume of hemisphere: \(\frac{2}{3}\pi(0.8)^{3}\approx0.3413\pi\) m³, cylinder: \(\pi(0.8)^{2}(1.8)=1.152\pi\) m³, cone: \(\frac{1}{3}\pi(0.8)^{2}(0.7)\approx0.1493\pi\) m³. Total: \((0.3413 + 1.152+0.1493)\pi\approx1.6426\times3.1416\approx5.2\) m³.