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Chapter 1: Problem 29

Vector \(\overrightarrow{A}\) has \(y\)-component A\(_y\) = \(+\)9.60 m. \(\overrightarrow{A}\) makes anangle of 32.0\(^{\circ}\) counterclockwise from the \(+$$y\)-axis. (a) What is the \(x\)-component of \(\overrightarrow{A}\)? (b) What is the magnitude of \(\overrightarrow{A}\)?

Short Answer

Expert verified
(a) The x-component is approximately 6.0 m. (b) The magnitude of \(\overrightarrow{A}\) is approximately 11.32 m.

Step by step solution

01

Understanding Vector Components

Vectors can be decomposed into parts, or components, in the x-direction and y-direction. Given that \(\overrightarrow{A}\) has a known y-component \(A_y = +9.60 \text{ m}\) and makes an angle \(\theta = 32.0^{\circ}\) with the y-axis, we will use trigonometry to find the unknown x-component and magnitude.
02

Find the x-component Using Trigonometry

Since the vector \(\overrightarrow{A}\) has an angle of 32.0掳 from the positive y-axis, we use the sine function to find the x-component. Recall, \( \sin(\theta) = \frac{A_x}{A} \), and \( \cos(\theta) = \frac{A_y}{A} \).Rearrange the sine formula to solve for the x-component:\[ A_x = A \cdot \sin(\theta) \]First calculate: \[ A_x = A_y \cdot \tan(\theta) \]Substitute the given values: \[ A_x = 9.60 \cdot \tan(32.0^{\circ}) \]
03

Calculate the x-component

Using the tangent formula we calculated in the previous step: \[ A_x = 9.60 \times \tan(32.0^{\circ}) \] \[ A_x \approx 9.60 \times 0.6249 \] \[ A_x \approx 5.999 \text{ m} \]
04

Calculate the Magnitude of Vector \(\overrightarrow{A}\)

We now use the Pythagorean theorem to find the magnitude of the vector \(\overrightarrow{A}\):\[ A = \sqrt{A_x^2 + A_y^2} \]Substitute the values:\[ A = \sqrt{(5.999)^2 + (9.60)^2} \]Calculate the magnitude:\[ A = \sqrt{35.988 + 92.16} \]\[ A \approx \sqrt{128.148} \]\[ A \approx 11.32 \text{ m} \]

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Key Concepts

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

Trigonometry in Physics
Trigonometry is a fundamental concept in physics for analyzing movements and forces. When dealing with vectors, understanding how they relate to trigonometric functions is crucial. Vectors have both magnitude and direction, which can be broken down into horizontal and vertical components using angles. Let's consider a vector \(\overrightarrow{A}\) that makes an angle with one of the coordinate axes.

When you know one component of a vector and the angle it makes with an axis, you can use trigonometric functions like sine, cosine, and tangent to calculate other components. In the given problem, we used the tangent function because we knew the \(y\)-component of the vector and needed to find the \(x\)-component:

  • Sine (\
Vector Magnitude Calculation
The magnitude of a vector gives us a "size" or "length" for the vector, independent of its direction. After calculating the components of the vector, the next step is often to find its magnitude. Imagine it as the diagonal of a right triangle where the legs are the vector's components along the x and y axes.

To find the magnitude of a vector \(\overrightarrow{A}\), use the Pythagorean theorem in the context of vector components, which are treated like the sides of a right triangle. The formula applied is:

\[ A = \sqrt{A_x^2 + A_y^2} \]

This principle stems from the idea that the hypotenuse (magnitude) of the triangle can be solved by calculating the square root of the sum of the squares of its legs (components).

  • The x-component is squared and then added to the square of the y-component.
  • Take the square root of this sum to find the magnitude of the vector.
By doing this, we transformed the vector into a singular value that describes its overall size, which in our example problem was computed as 11.32 meters.
Pythagorean Theorem in Vectors
The Pythagorean theorem is not just crucial in geometry but is also widely used in physics, especially for vector analysis. When dealing with vectors, it helps us relate the components of vectors to the overall magnitude.

Consider that any vector in a two-dimensional plane forms a right triangle with its x and y components. The Pythagorean theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. For vectors, this means:

\[ A = \sqrt{A_x^2 + A_y^2} \]

Where:
  • \(A_x\) is the horizontal component
  • \(A_y\) is the vertical component
  • \(A\) is the hypotenuse, representing the vector's magnitude
This approach is especially useful because it allows us to calculate one vector's overall impact or size using just its directional components.

By applying the Pythagorean theorem, you can effectively translate between component form and magnitude, which is a critical skill in physics problems involving motion and force. In our exercise example, using the theorem allowed us to correctly compute the magnitude of vector \(\overrightarrow{A}\) based on its components.

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Most popular questions from this chapter

As noted in Exercise 1.26, a spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction 45\(^{\circ}\) east of south, and then 280 m at 30\(^{\circ}\) east of north. After a fourth displacement, she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

In the methane molecule, CH\(_4\), each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the center. In coordinates for which one of the C\(-\)H bonds is in the direction of \(\hat{\imath}\) + \(\hat{\jmath}\) + \(\hat{k}\), an adjacent C\(-\)H bond is in the \(\hat{\imath}\) \(-\) \(\hat{\jmath}\) \(-\) \(\hat{k}\) direction. Calculate the angle between these two bonds.

A spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction \(45^{\circ}\) east of south. and then \(280 \mathrm{~m}\) at \(30^{\circ}\) east of north. After a fourth displacement,she finds herself back where she started. Use a scale drawing to determine the magnitude and direction of the fourth displacement. (See also Problem 1.61 for a different approach.)

Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 26.0 m in a direction 60.0\(^{\circ}\) west of north. Jane walks 16.0 m in a direction 30.0\(^{\circ}\) south of west. They then stop and turn to face each other. (a) What is the distance between them? (b) In what direction should Ricardo walk to go directly toward Jane?

You are camping with Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 21.0 m from yours, in the direction 23.0\(^{\circ}\) south of east. Karl's tent is 32.0 m from yours, in the direction 37.0\(^{\circ}\) north of east. What is the distance between Karl's tent and Joe's tent?
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