§6.4 Physics applications: work, force, and pressure

Physics applications: work, force, and pressure

Introduction

Work

Work: Pumping Liquid from a Tank

Force due to Hydrostatic Pressure

Summary

A bucket leaving the water weighs $B(0) = 20$ lb and weighs $B(50) = 12$ lb at the top of a 50-foot well, losing water at a constant rate with height. Then

A 200-foot rope weighing 0.3 lb/ft hangs fully extended over a cliff, and $F(h) = 0.3(200-h)$ lb remains hanging after $h$ feet are hauled in. Compute the total work in foot-pounds using $W = \int_0^{200}F(h)\,dh$.

A bucket weighing 40 lb at the water's surface loses 0.1 lb per foot of height while being lifted 100 feet, so $B(h) = 40-0.1h$. Compute the total work in foot-pounds.

A bucket's weight at height $h$ is $B(h) = 25+15e^{-0.05h}$ lb over a 100-foot lift. The average force exerted over the lift is closest to

Hooke's Law says $F(x) = kx$ for a spring displaced $x$ from natural length. If holding the spring at $x = \frac{1}{3}$ ft takes 5 lb of force, then

For the spring with $k = 15$, compute the work to stretch it from 1 foot beyond natural length to 1.5 feet beyond, so $W = \int_1^{1.5}15x\,dx$. Report the value in foot-pounds.

A crock 4 feet deep has radius 1.5 ft at the surface tapering to 0.75 ft at the base, and water must be pumped 9 feet above the surface. With $y$ measured down from the surface, the work integral is

A vertical cylindrical tank of radius 2 m and depth 6 m holds 4 m of water, and only the top 1 m is pumped over the rim. With $x$ measured down from the top, the work integral is

A hemispherical tank of radius 10 ft holds 7 ft of water, and the top 5 ft are pumped to a truck 5 ft above the tank's rim. With $x$ measured down from the top of the tank, the work integral is

A trough 10 ft long with triangular ends (top width 5 ft, depth 4 ft) is full to within 1 ft of the top, and water is pumped out until 1 ft remains. With $x$ measured down from the top, the work integral is

A dam 25 ft tall is 90 ft wide at the top and 60 ft wide at the base, with water rising to within 5 ft of the top. With $x$ measured down from the top, the total force equals

For a rectangular dam 100 feet wide and 50 feet tall with water to the top, the force integral $\int_0^{50}6240x\,dx = 7800000$ lb grows with $x$ because deeper slices feel greater water pressure.

A semicircular dam of radius 30 ft holds water rising to within 10 ft of the top. With $x$ measured down from the top and the semicircle described by $y = \sqrt{900-x^2}$, the force integral is

A trough with triangular ends (top width 5 ft, depth 4 ft) is full to within 1 ft of the top. The force on one triangular end is $F = \int_1^4 62.4(x-1)(5-1.25x)\,dx$. Evaluate to find the force in pounds.

For the region under $f(x) = 3\cos(\frac{x^3}{4})$ rotated to form a 10-foot trough, the work to lower the full tank by $h$ feet uses slices $2f(x)\cdot 10$ wide and long. The work integral is

For the solid-of-revolution tank from $f(x) = 3\cos(\frac{x^3}{4})$, the work to empty the top foot is $W = \int_{0.34527}^{1.34527}561.6\pi x\cos^2(\frac{x^3}{4})\,dx$. Evaluate with technology, rounded to two decimals.

A cylindrical tank of radius 3 ft and length 10 ft lies buried with its top 2 ft underground, and the top half of the water goes to a truck 4.5 ft above ground. With $x$ measured down from ground level, the work integral is

With the same tank only half full, the force on one circular end is $F = \int_5^8 124.8(x-5)\sqrt{9-(x-5)^2}\,dx$. Evaluate to find the force in pounds, rounded to one decimal.