maths.free › Calculus › 6. Using Definite Integrals › Physics applications: work, force, and pressure

Physics applications: work, force, and pressure

Introduction

We have seen several different circumstances where the definite integral enables us to measure the accumulation of a quantity that varies, provided the quantity is approximately constant over small intervals. For instance, to find the area bounded by a nonnegative curve \(y = f(x)\) and the \(x\)-axis on an interval \([a,b]\), we take a representative slice of width \(\Delta x\) that has area \(A_{\text{slice} } = f(x) \Delta x\). As we let the width of the representative slice tend to zero, we find that the exact area of the region is \[\begin{aligned}\end{aligned}\].

In a similar way, if we know the velocity \(v(t)\) of a moving object and we wish to know the distance the object travels on an interval \([a,b]\) where \(v(t)\) is nonnegative, we can use a definite integral to generalize the fact that \(d = r \cdot t\) when the rate, \(r\), is constant. On a short time interval \(\Delta t\), \(v(t)\) is roughly constant, so for a small slice of time, \(d_{\text{slice} } = v(t) \Delta t\). As the width of the time interval \(\Delta t\) tends to zero, the exact distance traveled is given by the definite integral \[\begin{aligned}\end{aligned}\].

Finally, if we want to determine the mass of an object of non-constant density, because \(M = D \cdot V\) (mass equals density times volume, provided that density is constant), we can consider a small slice of an object on which the density is approximately constant, and a definite integral may be used to determine the exact mass of the object. For instance, if we have a thin rod whose cross sections have constant density, but whose density is distributed along the \(x\) axis according to the function \(y = \rho(x)\), it follows that for a small slice of the rod that is \(\Delta x\) thick, \(M_{\text{slice} } = \rho(x) \Delta x\). In the limit as \(\Delta x \to 0\), we then find that the total mass is given by \[\begin{aligned}\end{aligned}\].

We next turn to the notion of work: from physics, a basic principle is that work is the product of force and distance. For example, if a person exerts a force of 20 pounds to lift a 20-pound weight 4 feet off the ground, the total work accomplished is \[\begin{aligned}\end{aligned}\].

If force and distance are measured in English units (pounds and feet), then the units of work are foot-pounds. If we work in metric units, where forces are measured in Newtons and distances in meters, the units of work are Newton-meters.

Condensed — the full section is in Boelkins, Active Calculus.

Work

Because work is calculated by the rule \(W = F \cdot d\) whenever the force \(F\) is constant, it follows that we can use a definite integral to compute the work accomplished by a varying force. For example, suppose that a bucket whose weight at height \(h\) is given by \(B(h) = 12 + 8e^{-0.1h}\) is being lifted in a 50-foot well.

In contrast to the problem in the preview activity, this bucket is not leaking at a constant rate; but because the weight of the bucket and water is not constant, we have to use a definite integral to determine the total work done in lifting the bucket. At a height \(h\) above the water, the approximate work to move the bucket a small distance \(\Delta h\) is \[\begin{aligned}\end{aligned}\].

Hence, if we let \(\Delta h\) tend to 0 and take the sum of all of the slices of work accomplished on these small intervals, it follows that the total work is given by \[\begin{aligned}\end{aligned}\].

While we could evaluate this integral exactly using the First Fundamental Theorem of Calculus, in applied settings such as this one we will typically use computing technology. Here, it turns out that \(W = \int_0^{50} (12 + 8e^{-0.1h}) \, dh \approx 679.461\) foot-pounds.

Our work in Preview Activity and in the most recent discussion above employs the following important general principle.

For an object being moved in the positive direction along an axis with location \(x\) by a force \(F(x)\), the total work to move the object from \(a\) to \(b\) is given by \[\begin{aligned}\end{aligned}\].

Work: Pumping Liquid from a Tank

In certain geographic locations where the water table is high, residential homes with basements have a peculiar feature: in the basement, one finds a large hole in the floor, and in the hole, there is water. For example, in Figure we see a sump crock Image credit to www.warreninspect.com/basement-moisture. . A sump crock provides an outlet for water that may build up beneath the basement floor to prevent flooding the basement.

In the crock we see a floating pump. This pump is activated by elevation, so when the water level reaches a particular height, the pump turns on and pumps water out of the crock, hence relieving the water buildup beneath the foundation. One of the questions we'd like to answer is: how much work does a sump pump accomplish?

The preceding example demonstrates the standard approach to finding the work required to empty a tank filled with liquid. The main task in each such problem is to determine the volume of a representative slice, followed by the force exerted on the slice, as well as the distance such a slice moves. In the case where the units are metric, there is one key difference: in the metric setting, rather than weight, we normally first find the mass of a slice. For instance, if distance is measured in meters, the mass density of water is 1000 kg/m\(^3\). In that setting, we can find the mass of a typical slice (in kg). To determine the force required to move it, we use \(F = ma\), where \(m\) is the object's mass and \(a\) is the gravitational constant \(a=9.81\) N/kg. That is, in metric units, the weight density of water is 9810 N/m\(^3\).

Condensed — the full section is in Boelkins, Active Calculus.

Force due to Hydrostatic Pressure

When building a dam, engineers need to know how much force water will exert against the face of the dam. This force comes from water pressure. The pressure a force exerts on a region is measured in units of force per unit of area: for example, the air pressure in a tire is often measured in pounds per square inch (PSI). Hence, we see that the general relationship is given by \[\begin{aligned}\end{aligned}\], where \(P\) represents pressure, \(F\) represents force, and \(A\) the area of the region being considered. Of course, in the equation \(F = PA\), we are assuming that the pressure is constant over the entire region \(A\).

We know from experience that the deeper one dives underwater while swimming, the greater the pressure exerted by the water. This is because at a greater depth, there is more water right on top of the swimmer: it is the force that column of water exerts that determines the pressure the swimmer experiences. The total water pressure is found by computing the total weight of the column of water that lies above a region of area 1 square foot at a fixed depth. At a depth of \(d\) feet, a rectangular column has volume \(V = 1 \cdot 1 \cdot d\) ft\(^3\), so the corresponding weight of the water overhead is \(62.4d\). This is the amount of force being exerted on a 1 square foot region at a depth \(d\) feet underwater, so the pressure exerted by water at depth \(d\) is \(P = 62.4 d\) (lbs/ft\(^2\)).

Because pressure is force per unit area, or \(P = \frac{F}{A}\), we can compute the total force from a variable pressure by integrating \(F = PA\).

Although there are many different formulas involving work, force, and pressure, the fundamental ideas behind these problems are similar to others we've encountered in applications of the definite integral. We slice the quantity of interest into more manageable pieces and then use a definite integral to add them up.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

To measure the work done by a varying force in moving an object, we divide the problem into pieces on which we can use the formula \(W = F \cdot d\), and then use a definite integral to sum the work done on each piece.
To find the total force exerted by water against a dam, we use the formula \(F = P \cdot A\) to measure the force exerted on a slice that lies at a fixed depth, and then use a definite integral to sum the forces across the appropriate range of depths.
Because work is computed as the product of force and distance (provided force is constant), and the force water exerts on a dam can be computed as the product of pressure and area (provided pressure is constant), problems involving these concepts are similar to earlier problems we did using definite integrals to find distance (via distance equals rate times time) and mass (mass equals density times volume).

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Suppose we have a region bounded by a nonnegative function \(y = f(x)\) and the \(x\)-axis on an interval \([a, b]\) and that \(F(x)\) is an antiderivative of \(f(x)\). If we want take a representative thin rectangular slice of the bounded area that has width \(\Delta x\), what is the area of the representative slice?
In the same context where we have a region bounded by a nonnegative function \(y = f(x)\) and the \(x\)-axis on an interval \([a, b]\), and \(F(x)\) is an antiderivative of \(f(x)\), what calculus expression provides the exact area of the region?
Suppose we know the nonnegative (and changing) velocity \(v(t)\) of a moving object on the time interval \([a, b]\) and we wish to know the total distance the object travels. Recall that when the nonnegative velocity of a moving object is constant, \(d = r \cdot t\). In the usual way, let \(a(t)\) and \(s(t)\) be the corresponding acceleration and position functions that satisfy \(a(t) = v'(t)\) and \(s'(t) = v(t)\).
On a short time interval \(\Delta t\), \(v(t)\) is roughly constant, so for a small slice of time, the corresponding small amount of distance traveled will be
In the same context where we have a nonnegative function \(v(t)\) on an interval \([a, b]\), and \(s(t)\) is the corresponding position function, what calculus expression provides the distance traveled?
Recall that for an object with constant density, \(D\), mass equals density times volume, or \[\begin{aligned}\end{aligned}\]. If we want to determine the mass of an object that has non-constant density, we have to use calculus in order to find its mass. For example, suppose we have a thin rod of length \((b-a)\) lying on the \(x\)-axis from \(x=a\) to \(x=a\) whose cross sections have constant density, but whose density is distributed according to the function \(\rho(x)\).
On a short interval of length \(\Delta x\), \(\rho(x)\) is roughly constant, so for a small slice of the rod, the corresponding small amount of mass will be
In the same context where we have a nonconstant continuous density function \(\rho(x)\) on an interval \([a, b]\), what calculus expression provides the exact mass, \(M\), of the rod?
How are all three of these situations finding area, finding distance traveled, finding total mass similar? What general pattern(s) do you notice?
A cylindrical tank, buried on its side, has radius \(3\) feet and length \(10\) feet. It is filled completely with water whose weight density is \(62.4\) lbs/ft\(^3\), and the top of the tank is two feet underground.
1. Set up, but do not evaluate, an integral expression that represents the amount of work required to empty the top half of the water in the tank to a truck whose tank lies 4.5 feet above ground.
2. With the tank now only half-full, set up, but do not evaluate an integral expression that represents the total force due to hydrostatic pressure against one end of the tank.
Odhalte odpověď
1. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. With the water weighing \(62.4\) pounds per cubic foot, the weight of (and thus force to be exerted upon) a typical slice is \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\] foot-pounds.
2. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. Finally, since \(F = P \cdot A\), the total force water is exerting on this slice is \[\begin{aligned}\end{aligned}\]. Integrating to sum the total force, we find \[\begin{aligned}\end{aligned}\] pounds.
A cylindrical tank, buried on its side, has radius \(3\) feet and length \(10\) feet. It is filled completely with water whose weight density is \(62.4\) lbs/ft\(^3\), and the top of the tank is two feet underground.
1. Set up, but do not evaluate, an integral expression that represents the amount of work required to empty the top half of the water in the tank to a truck whose tank lies 4.5 feet above ground.
2. With the tank now only half-full, set up, but do not evaluate an integral expression that represents the total force due to hydrostatic pressure against one end of the tank.
Odhalte odpověď
1. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. With the water weighing \(62.4\) pounds per cubic foot, the weight of (and thus force to be exerted upon) a typical slice is \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\] foot-pounds.
2. 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 \[\begin{aligned}\end{aligned}\]. 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 \[\begin{aligned}\end{aligned}\]. Finally, since \(F = P \cdot A\), the total force water is exerting on this slice is \[\begin{aligned}\end{aligned}\]. Integrating to sum the total force, we find \[\begin{aligned}\end{aligned}\] pounds.

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

Euler's number
2.71828…, the base whose exponential is its own derivative.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Physics applications: work, force, and pressure

How do we measure the work accomplished by a varying force that moves an object a certain distance?
What is the total force exerted by water against a dam?
How are both of the above concepts and their corresponding use of definite integrals similar to problems we have encountered in the past involving formulas such as distance equals rate times time and mass equals density times volume?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Zkuste si vlastní.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Více v Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests