Problem 43 - Trigonometry

Question

Find the work performed when the given force \(\mathbf{F}\) is applied to an object, whose resulting motion is represented by the displacement vector \(d\). Assume the force is in pounds and the displacement is measured in feet. \(\mathbf{F}=-54 \mathrm{i}, \mathrm{d}=20 \mathrm{i}\)

Step-by-Step Solution

Answer

The work performed is -1080 foot-pounds.

Step 1: Understand the formula for Work

The work done by a force on an object is calculated using the dot product of the force vector \( \mathbf{F} \) and the displacement vector \( \mathbf{d} \). This can be expressed as \( W = \mathbf{F} \cdot \mathbf{d} \).

Step 2: Determine the components of the vectors

Given \( \mathbf{F} = -54 \mathrm{i} \) and \( \mathbf{d} = 20 \mathrm{i} \), these vectors only have components in the \( \mathrm{i} \)-direction. So, \( F = -54 \) and \( d = 20 \).

Step 3: Compute the dot product

The dot product of \( \mathbf{F} \) and \( \mathbf{d} \) is simply the product of their respective components in the same direction. Thus, \( W = -54 \times 20 \).

Step 4: Calculate the work

Perform the multiplication, \( -54 \times 20 = -1080 \), to find the work done.

Key Concepts

Understanding the Dot Product

The dot product is a vital concept in trigonometry, particularly when dealing with vectors. It allows us to calculate the work done by a force when acting along a certain displacement. To put it simply, the dot product of two vectors is a scalar value. It quantifies how much of one vector goes in the direction of another.
In mathematical terms, if we have two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), their dot product is calculated as:
\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \]
This shows us that the dot product is simply the sum of the products of their corresponding components. This method turns lengthy vector calculations into straightforward arithmetic.

• The result of the dot product is a single number, not a vector.
• The dot product can reveal information like the angle between vectors or indicate perpendicularity when it equals zero.

In our exercise, the dot product \( \mathbf{F} \cdot \mathbf{d} \) gives us the work done when the force is applied along the displacement. When both vectors are in the same line, as in our case, this calculation simplifies significantly.

Breaking Down Vector Components

Vectors are fundamental in describing quantities that have both magnitude and direction, such as force or velocity. Each vector can be decomposed into components to simplify calculations. In 2D, a vector can be broken into its components along the \( x \)-axis and the \( y \)-axis.
For a general vector \( \mathbf{V} \), represented as \( v_x \mathrm{i} + v_y \mathrm{j} \), it is essential to understand these components:

• \( v_x \) is the horizontal component, determining the vector's effect along the \( x \)-axis.
• \( v_y \) is the vertical component, affecting the vector's behavior along the \( y \)-axis.

Dissecting vectors helps in applying operations like dot products by focusing on individual components. This is especially straightforward when all the action is in one direction, as in our task.

In the given exercise, both the force \( \mathbf{F} \) and the displacement \( \mathbf{d} \) vectors have components solely in the \( \mathrm{i} \)-direction. This simplification means we only work with the scalar magnitudes \( F = -54 \) and \( d = 20 \), making it easier to compute their dot product directly.

Exploring Force and Displacement

Understanding force and displacement is crucial in physics and engineering. Force, often represented as a vector, indicates how much an object is pushed or pulled. Its magnitude is measured in units like pounds or newtons, while its direction matters because it dictates where the force is applied.
Displacement, also a vector, tells us about the change in position of an object. This is different from distance, as it considers the shortest path between starting and ending points. Measured in units such as feet or meters, displacement indicates both size and direction.

The relationship between force and displacement, especially when they align, determines the work performed. When the force is applied in the same direction as the displacement, the full magnitude contributes to the work done. In contrast, if the force is applied perpendicularly, no work is done.

• When analyzing problems, consider both the direction of force and the path of displacement.
• Work is maximized when force and displacement vectors are collinear.

In our exercise, the negative force component \( -54 \mathrm{i} \) acts along the same straight line as the positive displacement \( 20 \mathrm{i} \), leading to a straightforward calculation for work: \(-54 \times 20 = -1080\). Thus, the negative sign indicates that the force is opposing the direction of displacement, which aligns with the concept of doing negative work.