Work, Fluid Force, and Variable Density

Why a single multiplication may fail

In elementary physics, work is often written as force times displacement: $W=Fd$. That formula is exactly right when the force is constant and acts along the direction of motion. Many physical systems are not so accommodating. A spring becomes harder to stretch as it lengthens, the weight of a hanging cable decreases as more cable reaches the floor, and water pressure increases with depth. In each case, the useful strategy is the same: describe one tiny piece whose force is nearly constant, find its small contribution, and add all such contributions.

The word work has a precise unit. In SI units, force is measured in newtons, abbreviated $mathrm{N}$, and displacement is measured in meters, abbreviated $mathrm{m}$. Their product is $mathrm{N}\cdot\mathrm{m}$, called a joule, abbreviated $mathrm{J}$. Writing units through a setup is more than bookkeeping. It distinguishes force from work and exposes a missing distance factor immediately.

Work as accumulated force-distance

Suppose a spring obeys Hooke’s law $F(x)=kx$, where $x$ is extension from its relaxed length and $k$ is the spring constant. The constant $k$ has units $mathrm{N}/\mathrm{m}$, so $kx$ has units of force. To stretch the spring from $x=0$ to $x=a$, the small work needed to increase its extension by $dx$ is

$$dW=F(x)\,dx=kx\,dx.$$

Adding those small contributions gives

$$W=\int_0^a kx\,dx=\frac{1}{2}ka^2.$$

For example, if $k=200\,\mathrm{N}/\mathrm{m}$ and $a=0.10\,\mathrm{m}$, then the work is $1.0\,\mathrm{J}$. The result is not $F(a)a=2.0\,\mathrm{J}$ because that calculation incorrectly uses the final, largest force throughout the whole stretch. The graph makes the correction visible: the work is the triangular area under an increasing line, half of the rectangle with the same final height and width.

When a force points opposite the displacement, its work is negative. This sign is meaningful: it records energy transferred out of the moving object by the force. A problem must therefore state or imply a direction. In the examples here, $F(x)$ is the magnitude of the applied force in the direction of motion, so each small contribution is positive. Never insert a sign only because an integral looks unfamiliar; derive it from the chosen coordinate direction.

Fluid force begins with pressure

Pressure is force per area. In a fluid at rest, pressure increases with depth because deeper material supports more fluid above it. If $h$ is depth below the surface, the pressure can be modeled as $p(h)=\rho gh$, where $\rho$ is mass density and $g$ is gravitational acceleration. In SI units, $ho$ has units $mathrm{kg}/\mathrm{m}^3$ and $g$ has units $mathrm{m}/\mathrm{s}^2$, giving pressure units of $mathrm{N}/\mathrm{m}^2$.

Take a horizontal strip of a vertical plate at depth $h$ with width $w(h)$ and small height $dh$. Its area is $w(h)\,dh$, so the small fluid force is

$$dF=p(h)w(h)\,dh=\rho gh\,w(h)\,dh.$$

The total force is found by integrating over the plate’s depth interval. Notice that this is not a volume calculation. The strip’s thickness is a vertical distance used to make an area strip, then pressure times that area gives force. The units confirm it: $(\mathrm{N}/\mathrm{m}^2)(\mathrm{m}^2)=\mathrm{N}$. Calling every $dx$ or $dh$ “volume” obscures the physical model.

Variable density makes the same idea portable

Density can vary along an object. A rod on the $x$-axis with linear density $\lambda(x)$ measured in $mathrm{kg}/\mathrm{m}$ has a small mass $dm=\lambda(x)\,dx$. Its weight contribution is $g\lambda(x)\,dx$, measured in newtons. If that piece must be lifted a distance $D(x)$, then its small work contribution is

$$dW=g\lambda(x)D(x)\,dx.$$

This expression is worth reading in words: gravity times mass-per-length times a thin length gives the piece’s weight, and multiplying by lift distance gives its work. Different contexts merely change the factor that varies. In a cable-lifting problem, $D(x)$ may depend on where the piece starts. In a pumping problem, a fluid slice’s weight is determined by its volume and density, while its lift distance is determined by its depth. The integral is a concise record of this local accounting.

Before integrating, make a table with four columns: the small piece, its amount or area, the force on it, and the distance through which that force acts. This is more reliable than hunting for a memorized formula. It makes assumptions testable and keeps dimensions attached to every factor. If two terms describe the same physical feature, the table will often reveal that you counted it twice.

Retrieval check

A cable of length $12\,\mathrm{m}$ has constant linear density $3\,\mathrm{kg}/\mathrm{m}$ and hangs vertically from a platform. Set up, but do not yet evaluate, the work required to pull the entire cable onto the platform. Choose a coordinate, describe the representative piece, state its weight and its lifting distance, and check the units of your integrand. Then explain why multiplying the total cable weight by $12\,\mathrm{m}$ would overestimate the work. The next lesson uses the same representative-piece discipline to locate centers, moments, and probability-weighted averages.