Problem 35 - Advanced Engineering Mathematics

Question

Find the work done by a constant force \(\mathbf{F}(x, y)=a \mathbf{i}+b \mathbf{j}\) acting counterclockwise once around the circle \(x^{2}+y^{2}=9\).

Step-by-Step Solution

Answer

The work done is zero.

Step 1: Understand the Problem Statement

We need to find the work done by a constant force as it acts around the circle described by the equation \(x^2 + y^2 = 9\). The force vector given is \( \mathbf{F}(x, y) = a \mathbf{i} + b \mathbf{j} \).

Step 2: Identify the Circle's Equation

The equation \(x^2 + y^2 = 9\) represents a circle centered at the origin with a radius of 3. This serves as the path along which the force is acting.

Step 3: Recognize the Nature of the Force

The force \( \mathbf{F}(x, y) = a \mathbf{i} + b \mathbf{j} \) is constant, meaning it doesn't change with respect to position \(x\) or \(y\).

Step 4: Recall the Work Formula for Closed Paths

The work done by a force \( \mathbf{F} \) around a closed path \( C \) is given by the line integral \( W = \oint_C \mathbf{F} \cdot d\mathbf{r} \).

Step 5: Consider the Vector Field's Properties

For a vector field \( \mathbf{F}(x, y) = a \mathbf{i} + b \mathbf{j} \), the line integral around a closed path like a circle can be simplified if the path is closed, as for a conservative field or when dealing with potential functions.

Step 6: Apply Green's Theorem

Green's Theorem states \( \oint_C \, \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA\), where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \). Here, \( M = a \) and \( N = b \). The partial derivatives are zero since \( a \) and \( b \) are constants.

Step 7: Calculate the Work Done

By Green's Theorem, we evaluate \( \iint_D (0 - 0) \, dA = 0 \). Thus, the work done by the force around the circle is zero.

Key Concepts

work done by force

When we talk about the work done by a force in physics, it is a measure of energy transfer. In a mathematical sense, work is calculated as the dot product of force and displacement vectors. However, this concept becomes a bit more intriguing when the force acts along a curve or a path.
The work done by a force \( \mathbf{F} \) moving along a path is computed as the line integral of \( \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is an infinitesimal displacement along the path.
For example, if a force acts in a uniform direction along the path of a circle or any closed loop, we explore how much work is done over that cycle. Interestingly, for constant forces around closed paths, the work done can often be zero, especially when the path brings everything back to its starting point without net displacement.

line integrals

Line integrals are a powerful tool in vector calculus for integrating functions along a curve. Instead of integrating functions with respect to variables alone, line integrals consider the path taken by a curve in space.
Imagine you walk along a path that curves through a field of forces. While walking, you experience different levels of force, which can do varying amounts of work on you, depending on the curve's direction and force field orientation.
Mathematically, the line integral over a force field \( \mathbf{F}(x, y) \) along a path can be expressed as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This represents the total work done by the force along the path \( C \). Important applications include calculating work, flux, and circulation in fields like physics, engineering, and computer graphics.

Green's theorem

Green's Theorem is a foundational theorem in vector calculus, serving as a bridge between line integrals and double integrals. It allows us to transform a line integral around a closed curve into a double integral over the region bounded by the curve.
For a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), Green's Theorem tells us that \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \) dA \). This conversion simplifies many calculations, especially when a line integral seems complex to evaluate directly.
In cases like the constant force problem around the circle, utilizing Green's theorem reveals that if the force has no swirliness or doesn't lead to rotation (often the derivatives are zero), the work done around the closed path is zero. This highlights why such a powerful theoretical concept is frequently used in physics and engineering.