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Chapter 6: Problem 6

Find the torque about the point (1,-2,1) due to the force \(\mathbf{F}=2 \mathbf{i}-\mathbf{j}+3 \mathbf{k}\) acting at the point (1,1,-3)

Short Answer

Expert verified
The torque is \(5\mathbf{i} + 8\mathbf{j} - 6\mathbf{k}\).

Step by step solution

01

Determine the Position Vector

Find the position vector \(\textbf{r}\) from the point of rotation (1, -2, 1) to the point where the force is applied (1, 1, -3). The position vector can be calculated by subtracting the coordinates of the point of rotation from the coordinates of the point where the force acts. So \(\textbf{r} = (1 - 1)\mathbf{i} + (1 - (-2))\mathbf{j} + (-3 - 1)\mathbf{k} = 0\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\).
02

Write Down the Given Force Vector

Recall the given force vector \(\textbf{F} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\).
03

Use the Torque Formula

Torque \(\textbf{\tau}\) is given by \(\textbf{\tau} = \textbf{r} \times \textbf{F}\). Use the cross product formula for vectors: \(\textbf{r} \times \textbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 3 & -4 \ 2 & -1 & 3 \ \end{vmatrix}\).
04

Compute the Determinant

Compute the cross product by calculating the determinant: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 3 & -4 \ 2 & -1 & 3 \ \end{vmatrix} = \mathbf{i} (3 \cdot 3 - (-4) \cdot (-1)) - \mathbf{j} (0 \cdot 3 - (-4) \- 2) + \mathbf{k} (0 \cdot (-1) - 3 \cdot 2) \). This simplifies to \(\textbf{\tau} = \mathbf{i} (9 - 4) - \mathbf{j} (-8) + \mathbf{k} (0 - 6) = 5\mathbf{i} + 8\mathbf{j} - 6\mathbf{k}\).

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Key Concepts

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

Torque Calculation
Torque is a measure of the rotational force an object experiences. It is calculated as the cross product of the position vector and the force vector. The formula is given by \(\tau = \mathbf{r} \times \mathbf{F}\). To compute torque, we follow these steps:
  • Identify the position vector (\textbf{r}) and the force vector (\textbf{F}).
  • Use the cross product formula to find the torque.

In this particular exercise, we have a given force and a point of rotation. By finding the position vector from the rotation point to where the force is applied, you can easily calculate the torque.
Position Vector
The position vector (\textbf{r}) connects the point of rotation to the point where the force is applied. It is crucial for calculating torque. To find it, you subtract the coordinates of the initial point from the final point coordinates.
Consider points (1, -2, 1) (rotation) and (1, 1, -3) (force application). The position vector is:
  • Step 1: Subtract x-coordinates: \[1 - 1 = 0\mathbf{i}\]
  • Step 2: Subtract y-coordinates: \[1 - (-2) = 3\mathbf{j}\]
  • Step 3: Subtract z-coordinates: \[-3 - 1 = -4\mathbf{k}\]

So, \textbf{r} = \(0\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\)
Cross Product
The cross product is a mathematical operation that returns a vector perpendicular to the plane formed by two independent vectors. In the context of torque, the cross product of the position and force vectors gives the torque vector.
The cross product formula is:
\(\mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 3 & -4 \ 2 & -1 & 3 \end{vmatrix}\)
Using the determinant method, we can solve this matrix to obtain the torque:
\(\mathbf{i} (3 \cdot 3 - (-4) \cdot (-1)) - \mathbf{j} (0 \cdot 3 - (-4) \cdot 2) + \mathbf{k} (0 \cdot (-1) - 3 \cdot 2)\)
So the torque vector is:\(\tau = 5\mathbf{i} + 8\mathbf{j} - 6\mathbf{k}\).

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Most popular questions from this chapter

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.

The force \(\mathbf{F}=2 \mathbf{i}-5 \mathbf{k}\) acts at the point \((3,-1,0) .\) Find the torque of \(\mathbf{F}\) about each of the following lines. (a) \(\quad \mathbf{r}=(2 \mathbf{i}-\mathbf{k})+(3 \mathbf{j}-4 \mathbf{k}) t\). (b) \(\quad \mathbf{r}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t\).

If \(\mathbf{A}\) and \(\mathbf{B}\) are unit vectors with an angle \(\theta\) between them, and \(\mathbf{C}\) is a unit vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B},\) evaluate \([(\mathbf{A} \times \mathbf{B}) \times(\mathbf{B} \times \mathbf{C})] \times(\mathbf{C} \times \mathbf{A})\).

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of the volume in the first octant bounded by \(x^{2}+y^{2}+z^{2}=16\) and the coordinate planes, where $$\mathbf{V}=\left(x+x^{2}-y^{2}\right) \mathbf{i}+(2 x y z-2 x y) \mathbf{j}-x z^{2} \mathbf{k}$$

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. $$\iiint \nabla \cdot \mathbf{V} d \tau \text { over the volume } x^{2}+y^{2} \leq 4,0 \leq z \leq 5, \mathbf{V}=(\sqrt{x^{2}+y^{2}})(\mathbf{i} x+\mathbf{j} y)$$.
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