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Chapter 6: Problem 4

Given \(\mathbf{A}=\mathbf{i}+\mathbf{j}-2 \mathbf{k}, \mathbf{B}=2 \mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{C}=\mathbf{j}-5 \mathbf{k}\): Let \(O\) be the tail of \(\mathbf{B}\) and let \(\mathbf{A}\) be a force acting at the head of \(\mathbf{B}\). Find the torque of \(\mathbf{A}\) about \(O ;\) about a line through \(O\) perpendicular to the plane of \(\mathbf{A}\) and \(\mathbf{B}\); about a line through \(O\) parallel to \(\mathbf{C}\).

Short Answer

Expert verified
The torque \(\boldsymbol{\tau} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\). The torque about the perpendicular line is \sqrt{59}\ units and about the line parallel to \(\mathbf{C}\) is \frac{-8}{\sqrt{26}}\ units.

Step by step solution

01

- Find the torque \(\boldsymbol{\tau}\) of \(\mathbf{A}\) about \(O\)

The torque \(\boldsymbol{\tau}\) is given by the cross product \(\mathbf{r} \times \mathbf{F}\) where \(\mathbf{r}\) is the position vector from \(O\) to the point where the force \(\mathbf{A}\) is applied and \(\mathbf{F} = \mathbf{A}\). Since \(\mathbf{r}\) is \(\mathbf{B}\), the torque becomes \[\boldsymbol{\tau} = \mathbf{B} \times \mathbf{A}\].
02

- Calculate the cross product \[\mathbf{B} \times \mathbf{A}\]

First, write \(\mathbf{B} = 2 \mathbf{i} - \mathbf{j} + 3 \mathbf{k}\) and \(\mathbf{A} = \mathbf{i} + \mathbf{j} - 2 \mathbf{k}\). The cross product is calculated using the determinant form: \[ \mathbf{B} \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & 3 \ 1 & 1 & -2 \end{vmatrix} \].
03

- Solve the determinant

Expanding the determinant, we get: \[ \mathbf{B} \times \mathbf{A} = \mathbf{i}((-1)(-2) - 3(1)) - \mathbf{j}((2)(-2) - 3(1)) + \mathbf{k}((2)(1) - (-1)(1)) \].This simplifies to: \[ \mathbf{B} \times \mathbf{A} = \mathbf{i}(2 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(2 + 1) \]\[ \mathbf{B} \times \mathbf{A} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k} \].
04

- Determine the torque about the line through \(O\) perpendicular to the plane of \((\mathbf{A}, \mathbf{B})\)

A line through \(O\) perpendicular to the plane of \(\mathbf{A}\) and \(\mathbf{B}\) has a direction vector that is parallel to \(\mathbf{B} \times \mathbf{A} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\). The magnitude of this cross product vector represents the torque about the line.\[\text{Magnitude} = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59} \]. Hence, the torque is \sqrt{59}\ units.
05

- Determine the torque about the line through \(O\) parallel to \(\mathbf{C}\)

First, calculate \(\mathbf{u}_{\mathbf{C}}\), the unit vector of \(\mathbf{C} = \mathbf{j} - 5 \mathbf{k}\), as: \[ \mathbf{u}_{\mathbf{C}} = \frac{\mathbf{C}}{\| \mathbf{C} \|} = \frac{\mathbf{j} - 5 \mathbf{k}}{\sqrt{1 + 25}} = \frac{\mathbf{j} - 5 \mathbf{k}}{\sqrt{26}} \]. The component of \(\boldsymbol{\tau} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\) along \(\mathbf{u}_{\mathbf{C}}\) is given by the dot product: \(\boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}}\).\[ \boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}} = (-1 \mathbf{i} + 7 \mathbf{j} + 3 \mathbf{k}) \cdot \left(\frac{0 \mathbf{i} + 1 \mathbf{j} - 5 \mathbf{k}}{\sqrt{26}}\right)\]\[ \boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}} = \frac{(0)(-1) + (7)(1) + (3)(-5)}{\sqrt{26}} = \frac{7 - 15}{\sqrt{26}} = \frac{-8}{\sqrt{26}} \]. Therefore, the torque about a line parallel to \(\mathbf{C}\) is \(\frac{-8}{\sqrt{26}}\) units.

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Key Concepts

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

Cross Product
In vector calculus, the cross product of two vectors produces a third vector that is perpendicular to the plane formed by the initial two vectors. It's often used in physics to determine the direction and magnitude of quantities like torque and magnetic force. The cross product of two vectors \(\textbf{A}\) and \(\textbf{B}\), denoted as \(\textbf{A} \times \textbf{B}\), can be calculated using a determinant matrix:
\[\textbf{A} \times \textbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \ \end{vmatrix} \]
The resulting vector's components are derived by solving this determinant.
Torque
Torque, also known as the moment of force, is a measure of how much a force acting on an object causes that object to rotate. The magnitude of the torque \(\boldsymbol{\tau}\) depends on three factors: the magnitude of the force \(\textbf{F}\), the distance \(\textbf{r}\) from the point where the force is applied to the axis of rotation, and the angle between the force vector and the position vector. Mathematically, it is given by the cross product:
\[\boldsymbol{\tau} = \textbf{r} \times \textbf{F} \]
The direction of torque is perpendicular to the plane formed by \(\textbf{r}\) and \(\textbf{F}\), which complies with the right-hand rule.
Unit Vector
A unit vector is a vector with a magnitude of 1 and points in a specific direction. Unit vectors are used to specify directions and are particularly helpful in physics and engineering. To find the unit vector \(\textbf{u}_{\textbf{A}}\) of any vector \(\textbf{A}\), you divide the vector by its magnitude \(\textbf{|A|}\):
\[\textbf{u}_{\textbf{A}} = \frac{\textbf{A}}{|\textbf{A}|}\]
For example, if vector \(\textbf{A} = a\textbf{i} + b\textbf{j} + c\textbf{k}\), its magnitude \(|\textbf{A}|\) is calculated as:
\[\textbf{|A|} = \sqrt{a^2 + b^2 + c^2} \]
The unit vector is then:
\[\textbf{u}_{\textbf{A}} = \frac{a\textbf{i} + b\textbf{j} + c\textbf{k}}{\textbf{|A|}} \]
Dot Product
The dot product (or scalar product) of two vectors results in a scalar value and is a measure of how much one vector goes in the direction of another. The dot product of vectors \(\textbf{A}\) and \(\textbf{B}\), denoted as \(\textbf{A} \bullet \textbf{B}\), is calculated using their components:
\[\textbf{A} \bullet \textbf{B} = A_xB_x + A_yB_y + A_zB_z \]
Alternatively, it can be related to the magnitudes of the vectors and the cosine of the angle \(\theta\) between them:
\[\textbf{A} \bullet \textbf{B} = |\textbf{A}| |\textbf{B}| \cos \theta \]
This product provides insights into the alignment of the vectors: if \(\theta\) is 0 degrees (vectors are parallel), the dot product is maximized. If \(\theta\) is 90 degrees (vectors are perpendicular), the dot product is zero.

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Most popular questions from this chapter

Use Green's theorem [formula }(9.7)] to evaluate the given integrals. \(\int_{C} e^{x} \cos y d x-e^{x} \sin y d y,\) where \(C\) is the broken line from \(\bar{A}=(\ln 2,0)\) to \(D=(0,1)\) and then from \(D\) to \(B=\) \((-\ln 2,0) .\) Hint: Apply Green's theorem to the integral around the closed curve \(A D B A\)

Consider a uniform distribution of total mass \(m^{\prime}\) over a spherical shell of radius The potential energy \(\phi\) of a mass \(m\) in the gravitational field of the spherical shell is. $$\phi=\left\\{\begin{array}{ll}\text { const. } & \text { if } m \text { is inside the spherical shell, } \\ -\frac{C m^{\prime}}{r} & \text { if } m \text { is outside the spherical shell, where } r \text { is the distance } \\\ & \text { from the center of the sphere to } m, \text { and } C \text { is a constant. }\end{array}\right.$$ Assuming that the earth is a spherical ball of radius \(R\) and constant density, find the potential and the force on a mass \(m\) outside and inside the earth. Evaluate the constants in terms of the acceleration of gravity \(g,\) to get \(\mathbf{F}=-\frac{m g R^{2}}{r^{2}} \mathbf{e}_{r}, \quad\) and \(\quad \phi=-\frac{m g R^{2}}{r}\) \(m\) outside the earth; \(\mathbf{F}=-\frac{m g r}{R} \mathbf{e}_{r}, \quad\) and \(\quad \phi=\frac{m g}{2 R}\left(r^{2}-3 R^{2}\right)\) \(m\) inside the earth.

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint_{\text {surface } \sigma} \operatorname{curl}\left(x^{2} \mathbf{i}+z^{2} \mathbf{j}-y^{2} \mathbf{k}\right) \cdot \mathbf{n} d \sigma,\) where \(\sigma\) is the part of the surface \(z=4-x^{2}-y^{2}\) above the \((x, y)\) plane.

Verify that each of the following force fields is conservative. Then find, for each, a scalar potential \(\phi\) such that \(\mathbf{F}=-\nabla \phi\). $$\mathbf{F}=z^{2} \sinh y \mathbf{j}+2 z \cosh y \mathbf{k}$$

If \(\mathbf{F}=x \mathbf{i}+y \mathbf{j},\) calculate \(\iint \mathbf{F} \cdot \mathbf{n} d \sigma\) over the part of the surface \(z=4-x^{2}-y^{2}\) that is above the \((x, y)\) plane, by applying the divergence theorem to the volume bounded by the surface and the piece that it cuts out of the \((x, y)\) plane. Hint: What is \(\mathbf{F} \cdot \mathbf{n}\) on the \((x, y)\) plane?
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