/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Aparticle moves through an \(x y... [FREE SOLUTION] | 91Ó°ÊÓ

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Aparticle moves through an \(x y z\) coordinate system while a force acts on the particle. When the particle has the position vector \(\vec{r}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}-(3.00 \mathrm{~m}) \hat{\mathrm{j}}+(2.00 \mathrm{~m}) \hat{\mathrm{k}}\), the force is given by \(\vec{F}=F_{x} \hat{\mathrm{i}}+(7.00 \mathrm{~N}) \hat{\mathrm{j}}-(6.00 \mathrm{~N}) \hat{\mathrm{k}}\) and the corresponding torque about the origin is \(\vec{\tau}=(4.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{i}}+(2.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{j}}-(1.00 \mathrm{~N} \cdot \mathrm{m}) \hat{\mathrm{k}}\). Determine \(F_{x^{*}}\)

Short Answer

Expert verified
The value of \( F_x \) is 5 N.

Step by step solution

01

Understanding Torque

Torque, denoted as \( \vec{\tau} \), is a measure of the rotational force around an axis. It is calculated as the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \). In this case, \( \vec{\tau} = \vec{r} \times \vec{F} \).
02

Express Vectors in Unit Notation

Since \( \vec{r} = (2.00 \, \mathrm{m}) \hat{\mathrm{i}} - (3.00 \, \mathrm{m}) \hat{\mathrm{j}} + (2.00 \, \mathrm{m}) \hat{\mathrm{k}} \) and \( \vec{F} = F_x \hat{\mathrm{i}} + (7.00 \, \mathrm{N}) \hat{\mathrm{j}} - (6.00 \, \mathrm{N}) \hat{\mathrm{k}} \), write these vectors clearly to set up the cross product.
03

Set Up Cross Product for Torque

The cross product \( \vec{r} \times \vec{F} \) is given by the determinant of the matrix: \[\begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \2 & -3 & 2 \F_x & 7 & -6 \end{vmatrix}\]This results in the vector components for torque along each axis.
04

Calculate Determinants for Each Component

Compute each component using the determinant: - For \( \hat{\mathrm{i}} \): \([(-3)(-6) - (2)(7)]\)- For \( \hat{\mathrm{j}} \): \( -[(2)(-6) - (2)(F_x)] \)- For \( \hat{\mathrm{k}} \): \([2(7) - (-3)(F_x)]\)
05

Simplify and Solve for \( F_x \)

Compute the separate expressions:- \( \hat{\mathrm{i}} \): \(18 - 14 = 4 \)- \( \hat{\mathrm{j}} \): \(-12 + 2F_x = -2 \)- \( \hat{\mathrm{k}} \): \(14 + 3F_x = -1 \)Using \( -12 + 2F_x = -2\), solve for \(F_x\):\( 2F_x = 10 \rightarrow F_x = 5 \, \mathrm{N} \)

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Key Concepts

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

Cross Product
The cross product is a mathematical operation used to find the vector that is perpendicular, or orthogonal, to two given vectors in three-dimensional space. It is especially useful in physics when dealing with rotational forces and torque.
The result of a cross product is a vector, and its magnitude is found by multiplying the magnitudes of the two vectors and the sine of the angle between them. Imagine the vectors as directions in space. Then, the cross product vector points in a direction that is perpendicular to both.
  • The cross product is denoted by the symbol \( \times \), such that for vectors \( \vec{a} \) and \( \vec{b} \), the cross product is \( \vec{a} \times \vec{b} \).
  • The formula is \( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin{\theta} \hat{n} \), where \( \hat{n} \) is the unit vector perpendicular to \( \vec{a} \) and \( \vec{b} \).
In our exercise, the cross product is used to compute the torque, which represents the rotational effect of a force applied at a point a distance from an axis.
Vector Components
In physics and mathematics, a vector is a quantity with both magnitude and direction. When dealing with vectors in three-dimensional space, it is common to break them into components since this simplifies the calculation process, like finding torque.
Each vector can be decomposed into three mutually perpendicular components along the x, y, and z axes, often represented using the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).
  • A vector \( \vec{V} \) can be written as \( V_x \hat{i} + V_y \hat{j} + V_z \hat{k} \).
  • Here, \( V_x, V_y, \) and \( V_z \) are the magnitudes of the components along the respective axes.
In the exercise, the position and force vectors are expressed in component form, allowing easy setup of the cross product to determine the resulting torque.
Determinants
Determinants play a crucial role in calculating the cross product of two vectors, especially when dealing with torque. A determinant is a special number extracted from a matrix, and it aids in solving systems of linear equations. For our purposes, the determinant helps in finding the component of the torque along each axis.
To compute the cross product using determinants, we set up a 3x3 matrix where:
  • The first row consists of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).
  • The second row contains the components of the first vector (position vector in our example).
  • The third row contains the components of the second vector (force vector in our example).
The resulting determinant gives the formula needed to calculate each component of the torque vector. In our exercise, this method is used to simplify and find the correct value for the unknown force component.
Rotational Force
Rotational force, often referred to as torque, is a fundamental concept in physics explaining how forces cause objects to rotate. Torque arises when a force is applied at a distance from a pivot point or axis and is dependent on how much force is applied and where it is applied, relative to the axis.
This is represented mathematically as:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
where \( \vec{r} \) is the position vector and \( \vec{F} \) is the force vector. The product (cross product in this context) gives us the torque vector, which points along the axis of rotation.
  • The magnitude of torque can be calculated as \( \tau = r F \sin{\theta} \), where \( \theta \) is the angle between the force vector and the position vector.
  • In practical terms, higher torque implies a greater ability to spin or rotate an object.
In our exercise, understanding rotational force is key to determining the effect of the applied forces on the particle at a specific point in its path.

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Most popular questions from this chapter

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

A top spins at 30 rev/s about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.0 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

A \(1000 \mathrm{~kg}\) car has four \(10 \mathrm{~kg}\) wheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the wheels about their axles? Assume that the wheels are uniform disks of the same mass and size. Why do you not need to know the radius of the wheels?

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m}\). Just then, in unitvector notation, what is its angular momentum relative to (a) the origin and (b) the point located at \((-2.0,-2.0) \mathrm{m} ?\)

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