Question

A child pulls a wagon with a force of 40 pounds. The handle of the wagon makes an angle of \(30^{\circ}\) with the ground. Express the force vector \(\mathbf{F}\) in terms of i and \(\mathbf{j}\).

Short answer

\[ \textbf{F} = 20\text{\textbf{i}} \text{sqrt}{3}} + 20 \text{\textbf{j}} \]

Step-by-step solution

Determine the Components of the Force Vector

The force vector \(\textbf{F}\) can be broken down into its horizontal (\textbf{i}) and vertical (\textbf{j}) components. The given force has a magnitude of 40 pounds and makes an angle of 30 degrees with the ground.

Calculate the Horizontal Component

Using the cosine function for the horizontal component, we have \[F_x = F \times \text{cos}(\theta) = 40 \times \text{cos}(30^{\text{o}}) \] We know that \(\text{cos}(30^{\text{o}}) = \frac{\sqrt{3}}{2} \), so \[F_x = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \]

Calculate the Vertical Component

Using the sine function for the vertical component, we have \[F_y = F \times \text{sin}(\theta) = 40 \times \text{sin}(30^{\text{o}}) \] We know that \(\text{sin}(30^{\text{o}}) = \frac{1}{2} \), so \[F_y = 40 \times \frac{1}{2} = 20 \]

Express the Force Vector in Terms of \textbf{i} and \textbf{j}

Combine the horizontal and vertical components to express the force vector as: \(\textbf{F} = 20\sqrt{3} \textbf{i} + 20 \textbf{j} \)

Key concepts

Force Vector

A force vector is a quantity that has both magnitude and direction. In this exercise, the child exerts a force of 40 pounds on the wagon, and the force is applied at an angle of 30 degrees to the horizontal ground. When dealing with vectors like this, it's important to break them down into their components to better understand the force's effect in different directions.

Trigonometric Functions

To break down the force vector into its horizontal and vertical components, we use trigonometric functions. These functions are essential for calculating the magnitude of vector components based on the given angle. The cosine function helps in finding the horizontal component, while the sine function helps in finding the vertical component. For this problem, using \(\text{cos}(30^{\text{o}}) = \frac{\textrm{\sqrt{3}}}{2}\) and \(\text{sin}(30^{\text{o}}) = \frac{1}{2}\) provides us with the accurate parts of the force.

Vector Notation

Vector notation allows us to concisely represent vector components along different axes. The standard notation for vectors in the plane uses \(\textbf{i}\) and \(\textbf{j}\) to denote the unit vectors in the horizontal (x) and vertical (y) directions, respectively. In this example, the force vector \(\textbf{F}\) expressed in terms of \(\textbf{i}\) and \(\textbf{j}\) is given by \(\textbf{F} = F_x \textbf{i} + F_y \textbf{j}\), where \({F_x} \) and \({F_y}\) are the magnitudes of the horizontal and vertical components.

Horizontal and Vertical Components

Identifying the horizontal and vertical components of a vector helps in understanding its direction and magnitude in different dimensions. For the force vector in this exercise:

• The horizontal component (\(F_x\)) is found using the cosine function: \[F_x = F \times \text{cos}(\theta) = 40 \times \frac{\text{\sqrt{3}}}{2} = 20\text{\sqrt{3}} \]
• The vertical component (\(F_y\)) is found using the sine function: \[F_y = F \times \text{sin}(\theta) = 40 \times \frac{1}{2} = 20 \]

Thus, the force vector is expressed as \(\textbf{F} = 20\text{\sqrt{3}} \textbf{i} + 20 \textbf{j}\). This representation helps us see how much force is applied in the horizontal direction versus the vertical direction.