Science:Math Exam Resources/Courses/MATH101/April 2005/Question 05 (a)

MATH101 April 2005

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Question 05 (a)

A water tank with depth 3 feet is in the shape of the trough depicted below. The bottom edge of the tank is 1 foot above the ground, and the ends of the tank are equilateral triangles of side length $2{\sqrt {3}}$ feet; the top of the tank is a rectangle of length 12 feet and width $2{\sqrt {3}}$ feet. (a) The tank is filled with water pumped up from the ground level. How much work is done in filling the tank? Express your answer in foot-pounds, and use the value 62 lb/ft$^{3}$ for the density of water. (Note that in the Imperial System, pounds are a unit for force.) (Do not simplify your answer.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Imagine looking at the trough face on to the equilateral triangle. Choose a coordinate system so that 0 is on the top. Choose a sample point and consider the volume that the small rectangular prism creates.

Hint 2
The following formulas will probably be useful Volume of a rectangular prism = length x width x height Imperial Force = Imperial density x volume = gravity x mass = 9.8 x mass Work = Imperial Force x displacement

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Consider the following diagram: In the picture, we take the perpendicular bisector of the top base and draw the line. This splits our equilateral triangle into two equal parts. The base is ${\sqrt {3}}$ and the height is 3 and the other side length is $2{\sqrt {3}}$. Choose a sample point say $\displaystyle x_{i}^{*}$. Let's call the length of the small triangle point downward l. By similar triangles, we have $\displaystyle {\frac {\sqrt {3}}{3}}={\frac {l/2}{3-x_{i}^{*}}}$ and hence $\displaystyle l={\frac {2(3-x_{i}^{*}){\sqrt {3}}}{3}}$ The volume of the thin prism at this height is $\displaystyle V_{i}=l(12)\Delta x={\frac {2(3-x_{i}^{*}){\sqrt {3}}}{3}}(12\Delta x)=8(3-x_{i}^{*}){\sqrt {3}}\Delta x$ Next, the force acting on this prism is $\displaystyle F_{i}=V_{i}(density)=(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x$ Lastly, the work done on this piece, noting that the displacement is 1 added to the distance to the bottom of the triangle (which is $\displaystyle (3-x_{i}^{*})$) $\displaystyle W_{i}=F_{i}(displacement)=(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x(4-x_{i}^{*})$ Summing over all pieces, we have $\displaystyle W=\lim _{n\to \infty }\sum _{i=1}^{n}W_{i}=\lim _{n\to \infty }\sum _{i=1}^{n}(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x(4-x_{i}^{*})$ and this last value is an integral given by ${\begin{aligned}W&=\int _{0}^{3}(62)8(3-x){\sqrt {3}}(4-x)\,dx\\&=62(8){\sqrt {3}}\int _{0}^{3}(12-7x+x^{2})\,dx\\&=62(8){\sqrt {3}}(12x-7x^{2}/2+x^{3}/3){\bigg |}_{0}^{3}\\&=62(8){\sqrt {3}}(12(3)-7(3)^{2}/2+(3)^{3}/3-0)\\&=62(8){\sqrt {3}}(36-63/2+9)\\&=62(4)(27){\sqrt {3}}\end{aligned}}$

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