Work, Hydrostatic Force, Centroids, and Average Value
Work, pumping work, hydrostatic pressure and force, center of mass, centroids, and average value formulas.
Work, pumping work, hydrostatic pressure and force, center of mass, centroids, and average value formulas.
It takes a force of $130$ N to stretch a spring $50$ cm. Find the work done in stretching the spring $20$ cm beyond its natural length.
Use $F=kx$. Since $130=k(0.50)$, $k=260$ N/m.
$$W=\int_0^{0.20}260x\,dx=\left[130x^2\right]_0^{0.20}$$
$$\boxed{5.2\text{ J}}$$
A bucket weighing $2$ lb contains $30$ lb of liquid. As it is lifted, liquid leaks at $\frac{1}{4}$ lb/ft. Find the work done in lifting it $5$ ft.
After the bucket has been lifted $x$ ft, its weight is $32-\frac{x}{4}$ lb.
$$W=\int_0^5\left(32-\frac{x}{4}\right)dx=\left[32x-\frac{x^2}{8}\right]_0^5$$
$$\boxed{156.875\text{ ft-lb}}$$
A hemispherical tank of radius $4$ m is full of water. Find the work done if the water is pumped to a vertical distance $6$ m above the top of the hemisphere.
The centroid of a solid hemisphere is $\frac{3r}{8}$ from its flat face, so the water's centroid is $1.5$ m below the top.
The volume is $\frac{2}{3}\pi r^3$, so the water weight is
$$\gamma V=9.81\left(\frac{2}{3}\pi(4)^3\right)=1314.9\text{ kN}.$$
The centroid is lifted $6+1.5=7.5$ m.
$$W=1314.9(7.5)=\boxed{9862\text{ kN-m}}$$
Find the average ordinate of $y=4-x^2$ from $x=-2$ to $x=2$.
$$y_{\text{avg}}=\frac{1}{2-(-2)}\int_{-2}^{2}(4-x^2)\,dx$$
$$=\frac{1}{4}\left[4x-\frac{x^3}{3}\right]_{-2}^{2}=\boxed{\frac{8}{3}}$$
Traffic speed is modeled by $S(t)=t^3-10.5t^2+30t+20$ mph, where $t$ is hours after noon. Find the average speed from 1:00 PM to 6:00 PM.
Use the average value formula on $1\le t\le 6$.
$$S_{\text{avg}}=\frac{1}{5}\int_1^6(t^3-10.5t^2+30t+20)\,dt$$
$$=\frac{1}{5}\left[\frac{t^4}{4}-3.5t^3+15t^2+20t\right]_1^6$$
$$\boxed{39.25\text{ mph}}$$
A vertical rectangular flood gate is 3 m wide and 2 m tall. Its top edge is 1 m below the water surface. Find the total hydrostatic force on the gate. (Use $\gamma_w=9810\text{ N/m}^3$.)
Set $y=0$ at the water surface (positive downward). The gate extends from $y=1$ m to $y=3$ m. Pressure at depth $y$ is $p=\gamma_w y$. Strip area: $dA=3\,dy$.
$$F=\int_1^3\gamma_w y\cdot3\,dy=3(9810)\int_1^3 y\,dy=29430\left[\frac{y^2}{2}\right]_1^3$$
$$=29430\left(\frac{9-1}{2}\right)=29430(4)=\boxed{117{,}720\text{ N}=117.72\text{ kN}}$$
Find the centroid $(\bar{x},\bar{y})$ of the upper semicircular region bounded by $x^2+y^2=r^2$ and the $x$-axis. (By symmetry, $\bar{x}=0$.)
Area of semicircle: $A=\frac{\pi r^2}{2}$. To find $\bar{y}$, use horizontal strips of width $2\sqrt{r^2-y^2}$ and height $dy$.
$$M_x=\int_0^r y\cdot 2\sqrt{r^2-y^2}\,dy$$
Let $u=r^2-y^2$, $du=-2y\,dy$:
$$M_x=\int_{r^2}^0\sqrt{u}\cdot(-du)=\int_0^{r^2}\sqrt{u}\,du=\left[\frac{2}{3}u^{3/2}\right]_0^{r^2}=\frac{2r^3}{3}$$
$$\bar{y}=\frac{M_x}{A}=\frac{2r^3/3}{\pi r^2/2}=\boxed{\frac{4r}{3\pi}}$$
For a unit semicircle ($r=1$): $\bar{y}=4/(3\pi)\approx0.424$. The centroid is about 42.4% of the radius above the diameter.
A conical tank (vertex down) has radius 3 m at the top and height 6 m. It is full of water ($\gamma=9810\text{ N/m}^3$). Find the work needed to pump all the water out over the rim.
Set $y=0$ at the vertex (bottom). At height $y$, the radius of the water cross-section is $r=\frac{3}{6}y=\frac{y}{2}$ by similar triangles. Slice area: $A(y)=\pi(y/2)^2=\frac{\pi y^2}{4}$.
Each slice must be lifted a distance $(6-y)$ to the rim.
$$W=\int_0^6\gamma\cdot A(y)\cdot(6-y)\,dy=9810\int_0^6\frac{\pi y^2}{4}(6-y)\,dy$$
$$=\frac{9810\pi}{4}\int_0^6(6y^2-y^3)\,dy=\frac{9810\pi}{4}\left[2y^3-\frac{y^4}{4}\right]_0^6$$
$$=\frac{9810\pi}{4}(432-324)=\frac{9810\pi}{4}(108)=\frac{9810\pi(108)}{4}$$
$$\boxed{W=831{,}794\text{ J}\approx832\text{ kJ}}$$
A hemispherical tank of radius $20$ ft is filled with water to a depth of $15$ ft. Determine the work done in pumping all the water to the top of the tank. Use $62.4$ lb/ft$^3$ for water.
Let $y$ be the vertical distance below the top rim. The hemisphere runs from $y=0$ at the rim to $y=20$ at the bottom.
At depth $y$, the slice radius satisfies $r^2=400-y^2$, so
$$dV=\pi(400-y^2)\,dy.$$
The tank is filled to a depth of $15$ ft from the bottom, so the water occupies $5\le y\le20$. Each slice is lifted $y$ ft to the rim.
$$W=62.4\pi\int_5^{20}y(400-y^2)\,dy$$
$$=62.4\pi\left[200y^2-\frac{y^4}{4}\right]_5^{20}$$
$$=\boxed{2{,}193{,}750\pi\text{ ft-lb}\approx6.89\times10^6\text{ ft-lb}}$$