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Vector \(\vec{A}\) is in the direction \(34.0^{\circ}\) clockwise from the \(-y\) -axis. The \(x\) -component of \(\vec{A}\) is \(A_{x}=-16.0 \mathrm{m} .\) (a) What is the \(y\) -component of \(\vec{A} ?\) (b) What is the magnitude of \(\vec{A} ?\)

Short Answer

Expert verified
The y-component is approximately 21.6 m, and the magnitude of \(\vec{A}\) is about 26.9 m.

Step by step solution

01

Understanding the Direction

The given vector \(\vec{A}\) is \(34.0^{\circ}\) clockwise from the \(-y\)-axis. This means that the angle \(\theta\) for calculating the vector components is \(34.0^{\circ} + 90.0^{\circ} = 124.0^{\circ}\) with the positive \(x\)-axis in the coordinate plane since the angles are usually measured counterclockwise from the positive \(x\)-axis.
02

Formula for Components

The formula to find the components of a vector \(\vec{A}\) with magnitude \(A\) and angle \(\theta\) is:* \(A_x = A \cos(\theta)\)* \(A_y = A \sin(\theta)\)Here, we know \(A_x = -16.0 \, \text{m}\).
03

Calculating the y-component

To find \(A_y\), use the relation between the x and y components, given the angle \(\theta = 124.0^{\circ}\):Since \(A_x = A \cos(124^{\circ}) = -16.0 \, \text{m}\), determine \(A \sin(124^{\circ})\) using:\[ \tan(\theta) = \frac{A_y}{A_x} \Longrightarrow A_y = A_x \tan(124^{\circ}) \]Calculate \(A_y\):\[ A_y = -16.0 \tan(124^{\circ}) \approx 21.6 \, \text{m} \]
04

Magnitude of the Vector

To find the magnitude \(A\) of vector \(\vec{A}\), use the formula:\[ A = \sqrt{A_x^2 + A_y^2} \]Substitute \(A_x = -16.0 \, \text{m}\) and \(A_y = 21.6 \, \text{m}\) to find \(A\):\[ A = \sqrt{(-16.0)^2 + (21.6)^2} \approx 26.9 \, \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Trigonometry in Vector Components
Trigonometry plays a crucial role in understanding vector components. Vectors are quantities that have both magnitude and direction. When breaking down vectors into their components, trigonometry helps us determine these separate parts along the coordinate axes.
To break a vector down into its components using trigonometry, an angle and a magnitude are necessary. In our case, the angle is given as relative to the axis directions, and we often work with
  • Sine, which gives the ratio of the opposite side to the hypotenuse
  • Cosine, which gives the ratio of the adjacent side to the hypotenuse
  • Tangent, which is the ratio of the opposite side to the adjacent side

These trigonometric functions allow us to project the vector onto the x and y axes, especially vital in solving problems where the vector does not align with these axes. By calculating both components using a combination of sine and cosine functions, we can figure out how much of the vector points in each direction.
Calculating the Magnitude of a Vector
The magnitude of a vector is essentially its length and provides a scalar value that denotes the strength or size of the vector. To determine the magnitude of a vector with known components, we use the Pythagorean theorem in a slightly adapted form.
Imagine the vector components as the two sides of a right triangle, with the vector's magnitude being the hypotenuse. This relationship is expressed by:\[ \text{Magnitude} \, A = \sqrt{A_x^2 + A_y^2} \]
Here, \(A_x\) and \(A_y\) represent the x and y components respectively. These comes directly from the vector formed by these two components.
This formula is incredibly useful for translating the vector's directional components into a single scalar that can be more easily worked with in calculations involving vector addition, kinematics, and dynamics.
For example, in the problem described, \( A_x = -16.0 \, \text{m} \) and \( A_y = 21.6 \, \text{m} \), leading to a magnitude calculated as:\[ A = \sqrt{(-16.0)^2 + (21.6)^2} \approx 26.9 \, \text{m} \] This shows how vector properties tie together to provide comprehensive insights into physical dynamics.
Exploring Coordinate Systems
Coordinate systems allow us to map vectors in readable and calculable formats. The most common is the Cartesian coordinate system, which uses perpendicular axes to describe locations and directions in a plane or space.
In these systems, the x and y axes are usually perpendicular, creating a framework where any vector can be expressed in terms of its individual x and y components.
Understanding how angles are measured is significant in solving vector problems. Angles are typically measured counterclockwise from the positive x-axis. However, knowing how to adjust for given conditions—like angles described as clockwise from the negative y-axis as in the problem—ensures that components are correctly calculated.
  • Angles in vector problems are reinterpreted to fit this standard direction system.
  • Adjusting initial angles to the coordinate plane's rules is essential for accuracy.
With the exercise, the provided angle needed conversion to align correctly with the coordinate system and apply trigonometric calculations accurately. This adjustment is crucial in providing exact x and y vector components that describe the situation accurately. A keen understanding of how coordinate systems work ensures problem-solving approaches seamlessly bridge from abstract to tangible outcomes.

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Most popular questions from this chapter

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