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Chapter 1: Problem 27

Vector \(\vec{A}\) has \(y\) -component \(A_{y}=+9.60 \mathrm{~m} . \vec{A}\) makes an angle of \(32.0^{\circ}\) counterclockwise from the \(+y\) -axis. (a) What is the \(x\) -component of \(\vec{A} ?\) (b) What is the magnitude of \(\vec{A} ?\)

Short Answer

Expert verified
The x-component of \(\vec{A}\) is \(A_x = +9.60 \times \sin(32.0^{\circ})\) m and the magnitude of \(\vec{A}\) is \(|\vec{A}| = \sqrt{(A_x)^2 + (9.60)^2}\) m. Calculation of these will yield the exact results.

Step by step solution

01

Calculate the x-component

Since the vector makes an angle 32.0 degrees with the y-axis, the x-component of the vector can be obtained using the sine of this angle (since in the counterclockwise direction from the \(+y\) -axis, we have to use sine instead of cosine). Hence, \(A_x = A_y \times \sin(32.0^{\circ})\). Substituting \(A_y = +9.60\) m in the equation yields \(A_x = +9.60 \times \sin(32.0^{\circ})\) m.
02

Calculate the magnitude of the vector

The magnitude of the vector can be calculated using the Pythagorean theorem, since \(\vec{A}\) and the components \(A_x\) and \(A_y\) form a right-angled triangle. So, \(|\vec{A}| = \sqrt{(A_x)^2 + (A_y)^2}\). Substituting the values of \(A_x\) and \(A_y\) from Step 1 and the given into the equation will give the magnitude of the vector.

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Key Concepts

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

Vector Magnitude
A vector's magnitude is a measure of its length or size in space, regardless of its direction. Think of it as the distance from the vector's origin to its tip.
Vectors like \( \vec{A} \) are often described in terms of their components along various axes, usually denoted as \( A_x \) and \( A_y \) in two-dimensional space. To find the magnitude, you'll examine both these components.

The calculation of the magnitude is akin to determining the hypotenuse of a right triangle where the legs are the vector's components.
This brings us to a very handy theorem discussed next.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle that relates the lengths of the sides of a right-angled triangle. It states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In the context of vectors, if you consider a vector with components \( A_x \) and \( A_y \), they form a right-angled triangle with the vector magnitude \( | \vec{A} | \) as the hypotenuse.

The formula for finding the magnitude \( | \vec{A} | \) is thus given by:
  • \[ | \vec{A} | = \sqrt{(A_x)^2 + (A_y)^2} \]
This elegant equation allows you to compute how long the vector actually is, providing a single number that represents its total impact in the space.
Trigonometric Functions
Trigonometric functions play a vital role in vector calculations, especially when you know the angle a vector makes with one of the coordinate axes.
These functions, such as sine, cosine, and tangent, relate angles in right triangles to the ratios of lengths of their sides.

For instance, if a vector \( \vec{A} \) makes an angle of \(32.0^{\circ} \) with the positive y-axis, you can use trigonometric functions to find its components.
  • For the y-component, you might use cosine if the angle is measured from the x-axis, but here, from the y-axis, cosine gives the x-component.
  • For the x-component, since it makes an angle of \(32.0^{\circ} \) with the y-axis, use sine: \(A_x = A_y \times \sin(32.0^{\circ}) \)
Understanding these functions makes it easier to break down vectors into their respective components and reveals their full directional impact.

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Most popular questions from this chapter

Vectors \(\vec{A}\) and \(\vec{B}\) are in the \(x y\) -plane. Vector \(\vec{A}\) is in the \(+x\) - direction, and the direction of vector \(\overrightarrow{\boldsymbol{B}}\) is at an angle \(\theta\) from the \(+x\) -axis measured toward the \(+y\) -axis. (a) If \(\theta\) is in the range \(0^{\circ} \leq \theta \leq 180^{\circ}\), for what two values of \(\theta\) does the scalar product \(\vec{A} \cdot \vec{B}\) have its maximum magnitude? For each of these values of \(\theta,\) what is the magnitude of the vector product \(\vec{A} \times \vec{B} ?(b)\) If \(\theta\) is in the range \(0^{\circ} \leq \theta \leq 180^{\circ}\) for what value of \(\theta\) does the vector product \(\vec{A} \times \vec{B}\) have its maximum value? For this value of \(\theta,\) what is the magnitude of the scalar product \(\vec{A} \cdot \vec{B} ?(\mathrm{c})\) What is the angle \(\theta\) in the range \(0^{\circ} \leq \theta \leq 180^{\circ}\) for which \(\vec{A} \cdot \vec{B}\) is twice \(|\vec{A} \times \vec{B}| ?\)

White Dwarfs and Neutron Stars. Recall that density is mass divided by volume, and consult Appendix \(\mathrm{B}\) as needed. (a) Calculate the average density of the earth in \(\mathrm{g} / \mathrm{cm}^{3}\), assuming our planet is a perfect sphere. (b) In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf that has about the same mass as it does now but is reduced to about \(15,000 \mathrm{~km}\) in diameter. What will be its density at that stage? (c) A neutron star is the remnant of certain supernovae (explosions of giant stars). Typically, neutron stars are about \(20 \mathrm{~km}\) in diameter and have about the same mass as our sun. What is a typical neutron star density in \(\mathrm{g} / \mathrm{cm}^{3}\) ?

You are given two vectors \(\vec{A}=-3.00 \hat{\imath}+6.00 \hat{\jmath} \quad\) and \(\vec{B}=7.00 \hat{\imath}+2.00 \hat{\jmath} . \quad\) Let \(\quad\) coun- terclockwise angles be positive. (a) What angle does \(\vec{A}\) make with the \(+x\) -axis? (b) What angle does \(\boldsymbol{B}\) make with the \(+x\) -axis? (c) Vector \(\boldsymbol{C}\) is the sum of \(\vec{A}\) and \(\vec{B},\) so \(\vec{C}=\vec{A}+\vec{B} .\) What angle does \(\vec{C}\) make with the \(+x\) -axis?

A rectangular piece of aluminum is \(7.60 \pm 0.01 \mathrm{~cm}\) long and \(1.90 \pm 0.01 \mathrm{~cm}\) wide. (a) Find the area of the rectangle and the uncertainty in the area. (b) Verify that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width. (This is a general result.)

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 \(21.0 \mathrm{~m}\) from yours, in the direction \(23.0^{\circ}\) south of east. Karl's tent is \(32.0 \mathrm{~m}\) 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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