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Chapter 11: Problem 22

$$ \vec{\mathbf{A}} \text { and } \vec{\mathbf{B}} $$ $$ \vec{\mathbf{A}} \times \vec{\mathbf{B}} $$ $$\vec{\mathbf{B}} \text { points south. }(b) \overline{\mathbf{A}} \text { points cast, } \vec{\mathbf{B}} $$ straight down. (c) \(\overline{\mathbf{A}}\) points straight up, \(\vec{\mathbf{B}}\) points north. (d) \(\vec{\mathbf{A}}\) points straight up, \(\vec{\mathbf{B}}\) points straight down.

Short Answer

Expert verified
(a) Up, (b) South, (c) West, (d) Zero Vector

Step by step solution

01

Understand the Vector Cross Product

The cross product of two vectors, \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} \), results in a vector that is perpendicular to both \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \) following the right-hand rule. Additionally, its magnitude is given by \( |\vec{\mathbf{A}}| \times |\vec{\mathbf{B}}| \times \sin(\theta) \), where \( \theta \) is the angle between \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \).
02

Evaluate Part (a)

\( \vec{\mathbf{B}} \) points south and \( \vec{\mathbf{A}} \) points east. Applying the right-hand rule, point your index finger in the direction of \( \vec{\mathbf{A}} \) (east) and your middle finger in the direction of \( \vec{\mathbf{B}} \) (south). Your thumb will point upwards, indicating the direction of \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} \).
03

Evaluate Part (b)

\( \vec{\mathbf{A}} \) points east and \( \vec{\mathbf{B}} \) points straight down. Applying the right-hand rule, point your index finger in the direction of \( \vec{\mathbf{A}} \) (east) and your middle finger in the direction of \( \vec{\mathbf{B}} \) (down). Your thumb will point south, indicating the direction of \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} \).
04

Evaluate Part (c)

\( \vec{\mathbf{A}} \) points straight up and \( \vec{\mathbf{B}} \) points north. Applying the right-hand rule, point your index finger in the direction of \( \vec{\mathbf{A}} \) (up) and your middle finger in the direction of \( \vec{\mathbf{B}} \) (north). Your thumb will point west, indicating the direction of \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} \).
05

Evaluate Part (d)

\( \vec{\mathbf{A}} \) points straight up and \( \vec{\mathbf{B}} \) points straight down. In this case, the angle \( \theta \) between \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \) is 180 degrees, making \( \sin(\theta) = 0 \). Therefore, \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} = 0 \).

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Key Concepts

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

Right-hand Rule
When dealing with vector cross products, the right-hand rule helps determine the direction of the resulting vector. It's a simple but powerful tool. To use it, follow these steps:
  • Point your index finger in the direction of the first vector, say \( \vec{\mathbf{A}} \).
  • Extend your middle finger perpendicular to your index finger, in the direction of the second vector, \( \vec{\mathbf{B}} \).
  • Your thumb, sticking out in another perpendicular direction, will give the direction of the cross product, \( \vec{\mathbf{A}} \times \vec{\mathbf{B}} \).
So, remember: Index finger for the first vector, middle finger for the second, thumb for the resultant. This method works consistently when dealing with physic situations involving forces or rotation.
Angle Between Vectors
The angle \( \theta \) between two vectors is crucial for calculating the cross product. It affects both the direction and magnitude of the result. The magnitude is determined by the formula:

\[ |\vec{\mathbf{A}} \times \vec{\mathbf{B}}| = |\vec{\mathbf{A}}| \times |\vec{\mathbf{B}}| \times \sin(\theta) \]

The angle ranges from 0 to 180 degrees:
  • If \( \theta = 0 \) degrees, the vectors are parallel; the cross product is zero since \( \sin(0) = 0 \).
  • If \( \theta = 90 \) degrees, the vectors are perpendicular; the cross product is maximized because \( \sin(90) = 1 \).
  • If \( \theta = 180 \) degrees, the vectors are anti-parallel; again, the cross product is zero as \( \sin(180) = 0 \).
The sine function in the formula highlights how only the perpendicular component contributes to the cross product.
Magnitude of a Vector
Magnitude signifies the length or size of a vector in space. In mathematical terms, it can be described as:

\[ |\vec{\mathbf{A}}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]

Here, \( A_x \), \( A_y \), and \( A_z \) are the components of vector \( \vec{\mathbf{A}} \) in three-dimensional space. For real-world applications:
  • An object's velocity has a magnitude, or speed, irrespective of its direction.
  • In force vectors, magnitude determines the strength of the force being applied.
Magnitude is essential in physics and engineering to quantify vectors' size, separate from their direction, enabling accurate calculations of physical phenomena.
Direction of Vectors
A vector's direction is as vital as its magnitude, indicating where the vector is pointed. Understanding direction involves recognizing:
  • The orientation in three-dimensional space based on vector components \( (A_x, A_y, A_z) \).
  • The use of angles like pitch, roll, and yaw in navigation and aerodynamics.
In the cross product, the resulting vector is orthogonal (perpendicular) to the plane formed by the original two vectors. This is particularly important in operations like:
  • Describing rotational forces, where the direction is pivotal; e.g., rotational motion around an axis.
  • Electrical currents within magnetic fields, relating to electromagnetic force calculations.
Direction, alongside magnitude, gives a complete description of a vector, making it indispensable in fields ranging from physics to computer graphics.

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Most popular questions from this chapter

(III) Suppose a 65 -kg person stands at the edge of a 6.5 -m dia- meter merry-go-round turntable that is mounted on frictionless bearings and has a moment of incrtia of 1850 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . The tumtable is at rest initially, but when the person begins rumning at a speed of 3.8 \(\mathrm{m} / \mathrm{s}\) (with respect to the tumtablc) around its edge, the turntable begins to rotate in the opposite direction. Cakulate the angular velocity of the tumtable.

(III) An ant crawls with constant speed outward along a radial spoke of a wheel rotating at constant angular velocity \(\omega\) about a vertical axis. Write a vector equation for all the forces (including inertial forces) acting on the ant. Take the \(x\) axis along the spoke, \(y\) perpendicular to the spoke pointing to the ant's left, and the \(z\) axis vertically upward. The wheel rotates counterclockwise as seen from above.

(II) A diver (such as the one shown in Fig. 2\()\) can reduce her moment of inertia by a factor of about 35 when changing from the straight position to the tuck position. If she makes 20 rotations in 1.5 \(\mathrm{s}\) when in the tuck position, what is her angular spced (rev/s) when in the straight position?

A spherical asteroid with radius \(r=123 \mathrm{m}\) and mass \(M=2.25 \times 10^{10} \mathrm{kg}\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the pole (as defined by the axis of rotation) and fires its engine, applying a force \(F \quad\) tangentially to the asteroid's surface as shown in Fig. \(44 .\) If \(F=265 \mathrm{N},\) how long will it take the tug to rotate the asteroid's axis of rotation through an angle of \(10.0^{a}\) by this method?

For what directions of velocity would the Coriolis effect on an object moving at the Earth's equator be zero?
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