일

$36$ J

a. $3,000,000$ lb, b. $750,000$ lbs

$23.25\pi$ million ft-lb

$\frac{A\rho {H}^{2}}{2}$

11,100 ft–lb

$953,284$ J

(a) 52,929.6 ft–lb (b) 18,525.3 ft–lb (c) When 3.83 ft of water have been pumped from the tank, leaving about 2.17 ft in the tank.

192,767 ft–lb. Note that the tank is oriented horizontally. Let the origin be the center of one of the circular ends of the tank. Since the radius is 3.75 ft, the fluid is being pumped to $y=4.75$; thus the distance the gas travels is $h(y)=4.75-y$. A differential element of water is a rectangle, with length 20 and width $2\sqrt{3.75^2-y^2}$. Thus the force required to move that slab of gas is $F(y) = 40\cdot45.93\cdot\sqrt{3.75^2-y^2}dy$. Total work is $\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75-y)\sqrt{3.75^2-y^2}\ dy$. This can be evaluated without actual integration; split the integral into $\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75)\sqrt{3.75^2-y^2}\ dy + \int_{-3.75}^{3.75} 40\cdot45.93\cdot(-y)\sqrt{3.75^2-y^2}\ dy$. The first integral can be evaluated as measuring half the area of a circle; the latter integral can be shown to be 0 without much difficulty. (Use substitution and realize the bounds are both 0.)

(a) We defined $\Delta x = \frac{b-a}{n}$: that is, the length of one interval, when we chop $[a,b]$ into $n$ of them. If $b$ and $a$ are measured in metres, then $\Delta x$ is measured in metres as well. So, the units of $\Delta x$ are metres.

Put another way, since $a$ and $b$ both describe a quantity in metres, $b-a$ describes a quantity in metres as well. (When we add or subtract quantities of the same units, their sum or difference is given in the same units.) Since $n$ is a unitless quantity (simply a number: not “$n$ kg" or “$n$ m"), $\frac{b-a}{n}$ still describes a quantity in metres. (If I have 6 metres of cloth, and I cut it into 3 pieces, each piece has $\frac{6}{3}=2$ metres–not 2 kilograms, or 2 metres per second.) (b) Since $F(x)$ is measured in kilogram-metres per second squared (newtons), the units of $F(x_i)$ are kilogram-metres per second squared (newtons). (c) $W$ is calculated by adding up summands of the form $F(x_i)\Delta x$. The units of $F(x_i)\Delta x$ are the products of the units of $F(x_i)$ with the units of $\Delta x$. That is, the units of $F(x_i)\Delta x$ are $\left(\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\right)\left(\text{m}\right) = \frac{\text{kg}\cdot\text{m}^2}{\text{sec}^2} = J$. The sum of terms given in joules is itself given in joules, so the units of $W$ are joules.

Let's start by converting from time spent pulling to amount pulled. When $y$ metres of rope have been pulled up, $2y$ seconds have passed, so $\frac{1}{5}y$ litres of water have leaked out of the bucket, leaving $5-\frac{1}{5}y$ litres. (This only makes sense when $\frac{1}{5}y \leq 5$, but we only consider values of $y$ from 0 to 5, so it's not a problem. That is, we're never hauling up an empty bucket that can't leak any more.)

When we've pulled up $y$ metres of rope, the mass in the bucket is $(5-\frac{1}{5}y)$ kg, so the force of gravity acting on it is $9.8(5-\frac{1}{5}y)$ N. Since we pull up 5 metres of rope, the work done is: $$W = \int_0^5 9.8\left(5-\frac{1}{5}y\right) \,\mathrm{d}y = 9.8\left[5y-\frac{1}{10}y^2\right]_0^5 = 9.8\left[25-2.5\right]=220.5\text{ J}$$

(a) First, we set up axes and a sketch for one of the ends of the cylindrical tank. It is simplest to let ground level correspond to the (horizontal) $y$-axis with the positive $x$-axis pointing straight down, so that $x = 0$ corresponds to ground level and the value of $x$ tells us the depth of a particular location. The circular end of the tank has radius $3$ and is centered at the point $(5,0)$ since its top is $2$ feet underground. Thus, the equation of that circle is $$(x-5)^2 + y^2 = 3^2\,.$$ Solving for $y$, we find the equation of one half of the circle, the half that is pictured at right in the figure below: $y = f(x) = \sqrt{9 - (x-5)^2}$. Next, for a slice at depth $x$, its shape is a rectangular slab with thickness $\triangle x$. Its length is $10$, and its width is $2 f(x) = 2 \sqrt{9 - (x-5)^2}$. Thus, the volume of such a slice is $$V_{\text{slice}} \approx 10 \cdot 2 \sqrt{9 - (x-5)^2} \cdot \triangle x\,.$$ With the water weighing $62.4$ pounds per cubic foot, the weight of (and thus force to be exerted upon) a typical slice is $$F_{\text{slice}} \approx 62.4 \cdot 20 \sqrt{9 - (x-5)^2} \triangle x\,.$$ To move such a slice from its depth $x$ feet below the ground to a tank on a truck that's $4.5$ feet above ground, each slice must travel $x + 4.5$ feet, and thus the work to move the slice is $$W_{\text{slice}} \approx (x+4.5) 1248 \sqrt{9 - (x-5)^2} \triangle x\,.$$ Finally, to empty half the tank, we need to consider slices from $x = 2$ to $x = 5$, and thus using a definite integral to sum the work required for each slice, we get $$W = \int_2^5 1248 (x+4.5) \sqrt{9 - (x-5)^2} \, dx = 1404(19\pi - 8) \approx 72573.13$$ foot-pounds. (b) To find the hydrostatic force due to water pressure that is exerted on one end of the tank by the remaining water, we first find the area of a rectangular slice at a location $x$, which is $$A_{\text{slice}} \approx 2 f(x) \triangle x = 2 \sqrt{9 - (x-5)^2} \triangle x\,.$$ We note that such a slice lies between $x = 5$ and $x = 8$, which is the values of $x$ that represent the depths of remaining water from top to bottom. At an $x$ in this interval, the slice's depth below the surface of the water is $x - 5$. Thus, since pressure is weight-density times depth, it follows that $$P_{\text{slice}} \approx 62.4 (x-5)\,.$$ Finally, since $F = P \cdot A$, the total force water is exerting on this slice is $$F_{\text{slice}} = P_{\text{slice}} \cdot A_{\text{slice}} \approx 62.4 (x-5) \cdot 2 \sqrt{9 - (x-5)^2} \triangle x\,.$$ Integrating to sum the total force, we find $$F = \int_5^8 124.8 (x-5) \sqrt{9 - (x-5)^2} \, dx \approx 1123.2$$ pounds.