Question

a conical container, oriented such that its vertex is at the bottom, has radius 5 ft and height 20 ft. it is filled to a height of 16 ft of a liquid weighing 51.8 lb/ft³. how much work will it take to pump the contents to the rim? how much work will it take to pump the liquid to a level of 4 ft above the cone’s rim? the amount of work required to pump the liquid to the rim of the tank is 111093 ft-lb. (round to the nearest whole number as needed.) the amount of work required to pump the liquid to a level 4 ft above the rim of the tank is □ ft-lb. (round to the nearest whole number as needed.)

Explanation:

Step1: Understand the Problem

We know the work to pump liquid to the rim is \( W_1 = 111093 \) ft - lb. Now we need to find the work to pump it 4 ft above the rim. First, we need to find the relationship between the height and the work. For a cone, the radius \( r \) at height \( h \) from the vertex is given by similar triangles. The cone has radius \( R = 5 \) ft and height \( H = 20 \) ft, so \( \frac{r}{h}=\frac{5}{20}=\frac{1}{4} \), so \( r=\frac{h}{4} \). The volume of a thin slice of liquid at height \( h \) with thickness \( dh \) is \( dV=\pi r^{2}dh=\pi(\frac{h}{4})^{2}dh \). The weight of the liquid is \( ho = 51.8 \) lb/ft³, so the weight of the slice is \( dF= ho dV = 51.8\pi(\frac{h}{4})^{2}dh \). The distance to pump this slice to the rim (when \( h \) is from 0 to 16 ft, since it's filled to 16 ft) is \( (20 - h) \) ft. But we already know the work to pump to the rim. Now, to pump 4 ft above the rim, the distance for each slice (originally from 0 to 16 ft) will be \( (20 - h)+4=(24 - h) \) ft. But we can also think in terms of the additional work. Wait, actually, the work to pump to a height \( H + 4 \) (where \( H = 20 \) is the rim height) from the original height (filled to 16 ft) can be found by considering the work done to lift the liquid from the rim level to 4 ft above. Wait, no. Wait, the liquid is filled to 16 ft, so the height from the vertex is 16 ft. The rim is at 20 ft, so the distance from the liquid surface to the rim is \( 20 - 16 = 4 \) ft? Wait, no, the cone has height 20 ft (vertex at bottom, rim at top). So the liquid is filled to height 16 ft (from vertex), so the depth from the rim is \( 20 - 16 = 4 \) ft? Wait, no, height from vertex is 16 ft, so the distance from the liquid surface to the rim is \( 20 - 16 = 4 \) ft. Wait, but the first part says the work to pump to the rim is 111093 ft - lb. Now, to pump it 4 ft above the rim, we need to lift each particle of liquid an additional 4 ft. So the work done will be the original work to pump to the rim plus the work to lift the entire liquid 4 ft. Wait, is that correct? Let's check. The work to pump to the rim is \( W_1=\int_{0}^{16} dF\times(20 - h) \). The work to pump to 4 ft above the rim is \( W_2=\int_{0}^{16} dF\times(20 - h + 4)=\int_{0}^{16} dF\times(24 - h)=\int_{0}^{16} dF\times((20 - h)+4)=W_1 + 4\int_{0}^{16} dF \). Now, \( \int_{0}^{16} dF \) is the total weight of the liquid. Let's find the total weight. The volume of the liquid is \( V=\frac{1}{3}\pi r^{2}h \), where \( r=\frac{16}{4}=4 \) ft (since \( r=\frac{h}{4} \), \( h = 16 \)), so \( V=\frac{1}{3}\pi(4)^{2}(16)=\frac{256}{3}\pi \) ft³. The weight is \( F= ho V=51.8\times\frac{256}{3}\pi \). Let's calculate that: \( 51.8\times\frac{256}{3}\times3.1416\approx51.8\times\frac{804.2496}{3}\approx51.8\times268.0832\approx13906.71 \) lb. Then the additional work is \( 4\times13906.71\approx55626.84 \) ft - lb. Then total work \( W_2=111093 + 55626.84\approx166719.84 \), which rounds to 166720. Wait, but maybe there's a simpler way. Wait, the work to pump to the rim is \( W_1 \), and the work to pump an additional distance \( d \) is \( W_{additional}=F\times d \), where \( F \) is the total weight of the liquid. So first, find \( F \). From the cone, \( r=\frac{h}{4} \), so at \( h = 16 \), \( r = 4 \). Volume \( V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(4)^{2}(16)=\frac{256\pi}{3} \). Weight \( F= ho V=51.8\times\frac{256\pi}{3} \). Let's compute \( F \): \( 51.8\times256 = 13260.8 \), \( 13260.8\times\pi\approx13260.8\times3.1416\approx41660.5 \), then divide by 3: \( 41660.5\div3\approx13886.83 \) lb. Then additional work is \( 4\times13886.83\approx55547.32 \). Then total work \( W_2=111093 + 55547.32\approx166640.32 \). Wait, maybe my first calculation of \( F \) was wrong. Wait, let's recalculate \( V \): \( \frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(4)^{2}(16)=\frac{1}{3}\pi\times16\times16=\frac{256\pi}{3}\approx\frac{256\times3.1416}{3}\approx\frac{804.2496}{3}\approx268.0832 \) ft³. Then weight \( F = 51.8\times268.0832\approx51.8\times268.0832 \). Let's compute 50×268.0832 = 13404.16, 1.8×268.0832 = 482.54976, so total \( 13404.16 + 482.54976 = 13886.70976 \) lb. So additional work is \( 4\times13886.70976\approx55546.839 \). Then total work \( W_2=111093 + 55546.839\approx166639.839 \), which rounds to 166640. Wait, but maybe the problem is that the work to pump to the rim is already calculated, and now we need to pump it 4 ft above, so the distance each particle is lifted is increased by 4 ft. So the work is the original work plus the work to lift the entire liquid 4 ft. So \( W_2 = W_1 + F\times4 \), where \( F \) is the total weight of the liquid. We can find \( F \) from the work to pump to the rim. The work to pump to the rim is \( W_1=\int_{0}^{16} ho\pi(\frac{h}{4})^{2}(20 - h)dh \). Let's compute that integral to check. \( \int_{0}^{16} ho\pi\frac{h^{2}}{16}(20 - h)dh=\frac{ ho\pi}{16}\int_{0}^{16}(20h^{2}-h^{3})dh=\frac{ ho\pi}{16}[20\times\frac{h^{3}}{3}-\frac{h^{4}}{4}]_{0}^{16}=\frac{ ho\pi}{16}[20\times\frac{4096}{3}-\frac{65536}{4}]=\frac{ ho\pi}{16}[\frac{81920}{3}-16384]=\frac{ ho\pi}{16}[\frac{81920 - 49152}{3}]=\frac{ ho\pi}{16}[\frac{32768}{3}]=\frac{ ho\pi\times32768}{48}=\frac{ ho\pi\times4096}{6} \). Plugging in \( ho = 51.8 \), \( \pi\approx3.1416 \): \( \frac{51.8\times3.1416\times4096}{6}\approx\frac{51.8\times12867.96}{6}\approx\frac{666560.328}{6}\approx111093.39 \), which matches the given \( W_1 = 111093 \) ft - lb. Good, so the integral is correct. Now, the total weight \( F=\int_{0}^{16} ho\pi(\frac{h}{4})^{2}dh=\frac{ ho\pi}{16}\int_{0}^{16}h^{2}dh=\frac{ ho\pi}{16}[\frac{h^{3}}{3}]_{0}^{16}=\frac{ ho\pi}{16}\times\frac{4096}{3}=\frac{ ho\pi\times256}{3} \), which is what we had before. So \( F=\frac{51.8\times\pi\times256}{3}\approx13886.71 \) lb. Then the additional work to lift 4 ft above the rim is \( F\times4\approx13886.71\times4 = 55546.84 \) ft - lb. Then total work \( W_2=W_1 + 55546.84=111093 + 55546.84 = 166639.84 \), which rounds to 166640.

Step2: Calculate the Additional Work

We found that the total weight \( F\approx13886.71 \) lb. The additional distance to pump is 4 ft, so the additional work is \( 13886.71\times4 = 55546.84 \) ft - lb.

Step3: Calculate the Total Work

Add the original work (to pump to the rim) and the additional work: \( 111093+55546.84 = 166639.84 \), which rounds to 166640.

Answer:

166640