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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

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. (a) Given \(\phi=x^{2}-y^{2},\) sketch on one graph the curves \(\phi=4, \phi=1, \phi=0\), \(\phi=-1, \phi=-4 .\) If \(\phi\) is the electrostatic potential, the curves \(\phi=\) const. are equipotentials, and the electric field is given by \(\mathbf{E}=-\nabla \phi\). If \(\phi\) is temperature, the curves \(\phi=\) const. are isothermals and \(\nabla \phi\) is the temperature gradient; heat flows in the direction \(-\nabla \phi\). (b) Find and draw on your sketch the vectors \(-\nabla \phi\) at the points \((x, y)=(\pm 1,\pm 1)\), \((0,\pm 2),(\pm 2,0) .\) Then, remembering that \(\nabla \phi\) is perpendicular to \(\phi=\) const., sketch, without computation, several curves along which heat would flow [see(a)].

If the temperature is \(T=x^{2}-x y+z^{2},\) find (a) the direction of heat flow at (2,1,-1) (b) the rate of change of temperature in the direction \(\mathbf{j}-\mathbf{k}\) at (2,1,-1).

Use a computer as needed to make plots of the given surfaces and the isothermal or equipotential curves. Try both 3D graphs and contour plots. If the temperature in the \((x, y)\) plane is given by \(T=x y-x,\) sketch a few isothermal curves, say for \(T=0,1,2,-1,-2 .\) Find the direction in which the temperature changes most rapidly with distance from the point \((1,1),\) and the maximum rate of change. Find the directional derivative of \(T\) at (1,1) in the direction of the vector \(3 \mathbf{i}-4 \mathbf{j} .\) Heat flows in the direction \(-\nabla T\) (perpendicular to the isothermals). Sketch a few curves along which heat would flow.

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

The force \(\mathbf{F}=\mathbf{i}+3 \mathbf{j}+2 \mathbf{k}\) acts at the point (1,1,1). (a) Find the torque of the force about the point (2,-1,5) . Careful! The vector \(\mathbf{r}\) goes from (2,-1,5) to (1,1,1). (b) Find the torque of the force about the line \(\mathbf{r}=2 \mathbf{i}-\mathbf{j}+5 \mathbf{k}+(\mathbf{i}-\mathbf{j}+2 \mathbf{k}) t\). Note that the line goes through the point (2,-1,5).
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