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Chapter 1: Problem 30

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

Short Answer

Expert verified
(a) The y-component is 23.7 m. (b) The magnitude is 28.6 m.

Step by step solution

01

Understand the Direction

The vector \(\overrightarrow{A}\) is directed 34.0掳 clockwise from the negative \(y\)-axis. This means it's in the fourth quadrant of the coordinate system.
02

Apply Trigonometric Functions

To find the \(y\)-component \(A_y\), use the tangent function. Because \(\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}\), and here \(\theta = 34.0^\circ\), the components can be described by:\[ \tan(-34.0^\circ) = \frac{A_x}{A_y} \] The angle is negative because it's clockwise.
03

Solve for the y-component

Substitute the known \(x\)-component \(A_x = -16.0\) m and solve for \(A_y\):\[ A_y = \frac{-16.0}{\tan(-34.0^\circ)} \]Calculate \(A_y\) using \(\tan(-34.0^\circ) \approx -0.6745\), so:\[ A_y = \frac{-16.0}{-0.6745} \approx 23.7 \text{ m} \]
04

Calculate the Magnitude of the Vector

Using the Pythagorean theorem for calculating the magnitude of the vector, \(\overrightarrow{A}\), with components \(A_x\) and \(A_y\):\[ |\overrightarrow{A}| = \sqrt{A_x^2 + A_y^2} \] Substitute the known values \(A_x = -16.0\) m and \(A_y \approx 23.7\) m.
05

Compute the Magnitude

Plug into the magnitude formula:\[ |\overrightarrow{A}| = \sqrt{(-16.0)^2 + (23.7)^2} = \sqrt{256 + 561.69} = \sqrt{817.69} \] Calculate the final result: \[ |\overrightarrow{A}| \approx 28.6 \text{ m} \]

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. They are especially useful in physics and engineering to find unknown components of vectors. In this exercise, you encounter the tangent function, which is defined as the ratio of the opposite side to the adjacent side of a right triangle:
  • The tangent function is noted as \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\).
  • When dealing with vectors, the opposite and adjacent sides are often the vector components in the coordinate system.

For the given vector \(\overrightarrow{A}\), by setting up \(\tan(-34.0^\circ) = \frac{-16.0}{A_y}\), it becomes straightforward to solve for the unknown component \(A_y\) by isolating it in the equation:
  • The angle is negative, reflecting its direction clockwise from the axis.
  • This manipulation highlights the power of trigonometric functions to extract vector information from angular data.
Coordinate System
A coordinate system is crucial when dissecting the direction and magnitude of vectors. In this exercise, the vector \(\overrightarrow{A}\) lies in a specific quadrant of the Cartesian coordinate plane. Here鈥檚 a breakdown:
  • The vector is described as being in the **fourth quadrant**. This quadrant is defined by positive \(x\)-values and negative \(y\)-values in a standard coordinate system.
  • Since \(A_x = -16.0\) m, it's clear the vector does not follow this usual quadrant placement, due to either specific direction description or problem context.

This problem involves considering how an angle measured clockwise from the negative \(y\)-axis affects the coordinates. Thus, a coordinate system is essential not just for plotting but also for resolving each component of \(\overrightarrow{A}\) accurately.
Pythagorean Theorem
The Pythagorean theorem is a fundamental tool in deriving vector magnitudes from their components in physics. It states that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
  • Expressed mathematically: \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse.
  • For vectors, the theorem translates to: \(|\overrightarrow{A}| = \sqrt{A_x^2 + A_y^2}\), linking component magnitudes to overall length.

In solving the exercise, the components of \(\overrightarrow{A}\) are plugged into this formula:
  • Using \(A_x = -16.0\) and \(A_y \approx 23.7\), the magnitude \(|\overrightarrow{A}|\) is computed as \(\sqrt{(-16.0)^2 + (23.7)^2}\).
  • This computation showcases how the Pythagorean theorem is indispensable for finding resultant vector lengths, ensuring precise measurement when components are known.

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Most popular questions from this chapter

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?

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm\(^3\) of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

(a) The recommended daily allowance (RDA) of the trace metal magnesium is 410 mg/day for males. Express this quantity in \(\mu\)g/day. (b) For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 75-kg adult receive? (c) A typical multivitamin tablet can contain 2.0 mg of vitamin B\(_2\) (riboflavin), and the RDA is 0.0030 g/day. How many such tablets should a person take each day to get the proper amount of this vitamin, if he gets none from other sources? (d) The RDA for the trace element selenium is 0.000070 g/day. Express this dose in mg/day.

Vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) have scalar product \(-\)6.00, and their vector product has magnitude \(+\)9.00. What is the angle between these two vectors?

While following a treasure map, you start at an old oak tree. You first walk 825 m directly south, then turn and walk 1.25 km at 30.0\(^{\circ}\) west of north, and finally walk 1.00 km at 32.0\(^{\circ}\) north of east, where you find the treasure: a biography of Isaac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, compare it with a graphical solution drawn roughly to scale.
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