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A plum is located at coordinates \((-2.0 \mathrm{~m}, 0,4.0 \mathrm{~m})\). In unitvector notation, what is the torque about the origin on the plum if that torque is due to a force \(\vec{F}\) whose only component is (a) \(F_{x}=\) \(6.0 \mathrm{~N},(\mathrm{~b}) F_{x}=-6.0 \mathrm{~N},(\mathrm{c}) F_{z}=6.0 \mathrm{~N}\), and \((\mathrm{d}) F_{z}=-6.0 \mathrm{~N} ?\)

Short Answer

Expert verified
(a) \( \vec{\tau} = -24.0 \, \hat{j} \) Nm, (b) \( \vec{\tau} = 24.0 \, \hat{j} \) Nm, (c) \( \vec{\tau} = 12.0 \, \hat{j} \) Nm, (d) \( \vec{\tau} = -12.0 \, \hat{j} \) Nm.

Step by step solution

01

Understand Torque Formula

Torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \): \( \vec{\tau} = \vec{r} \times \vec{F} \). Here, \( \vec{r} = (-2.0 \,\hat{i} + 0 \,\hat{j} + 4.0 \,\hat{k}) \, \text{m} \).
02

Calculate Torque for (a) \( F_{x} = 6.0 \, \text{N} \)

For this situation, \( \vec{F} = 6.0 \, \hat{i} \, \text{N} \). The cross product \( \vec{\tau} = \vec{r} \times \vec{F} \) is calculated as:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \-2.0 & 0 & 4.0 \6.0 & 0 & 0 \end{vmatrix} = 0 \, \hat{i} - 24.0 \, \hat{j} + 0 \, \hat{k} \]So, \( \vec{\tau} = -24.0 \, \hat{j} \, \text{Nm} \).
03

Calculate Torque for (b) \( F_{x} = -6.0 \, \text{N} \)

Here, \( \vec{F} = -6.0 \, \hat{i} \, \text{N} \). The cross product is:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \-2.0 & 0 & 4.0 \-6.0 & 0 & 0 \end{vmatrix} = 0 \, \hat{i} + 24.0 \, \hat{j} + 0 \, \hat{k} \]So, \( \vec{\tau} = 24.0 \, \hat{j} \, \text{Nm} \).
04

Calculate Torque for (c) \( F_{z} = 6.0 \, \text{N} \)

For \( \vec{F} = 6.0 \, \hat{k} \, \text{N} \), the cross product is:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \-2.0 & 0 & 4.0 \0 & 0 & 6.0 \end{vmatrix} = 0 \, \hat{i} + 12.0 \, \hat{j} + 0 \, \hat{k} \]So, \( \vec{\tau} = 12.0 \, \hat{j} \, \text{Nm} \).
05

Calculate Torque for (d) \( F_{z} = -6.0 \, \text{N} \)

With \( \vec{F} = -6.0 \, \hat{k} \, \text{N} \), the cross product is:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \-2.0 & 0 & 4.0 \0 & 0 & -6.0 \end{vmatrix} = 0 \, \hat{i} - 12.0 \, \hat{j} + 0 \, \hat{k} \]So, \( \vec{\tau} = -12.0 \, \hat{j} \, \text{Nm} \).

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Key Concepts

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

Cross Product
The concept of the cross product is essential in torque calculations. It is a vector operation used predominantly in three dimensions. In physics, the cross product helps determine how two vectors interact in terms of rotational effect, particularly in this exercise. For any two vectors \( \vec{A} \) and \( \vec{B} \), their cross product \( \vec{A} \times \vec{B} \) results in a third vector that is perpendicular to both:
  • The direction of the resultant vector follows the right-hand rule.
  • In mathematical terms, the magnitude equals the area of the parallelogram formed by the two initial vectors: \( |\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}| \sin\theta \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).
Understanding cross products is key to many vector-related calculations in physics, such as forces and torques. This notion is crucial as it directly influences how forces result in rotational actions, exemplified by calculating the torque on the plum.
Vector Notation
Vectors are mathematical entities with both magnitude and direction, represented syntax-wise using unit vectors (\( \hat{i} \), \( \hat{j} \), \( \hat{k} \)) in three dimensions. Unit vectors are:
  • \( \hat{i} \) for the x-direction.
  • \( \hat{j} \) for the y-direction.
  • \( \hat{k} \) for the z-direction.
In the exercise, the vectors are presented in unit vector notation, offering a clear method to express vector positions and vectors forces involved. A position vector \( \vec{r} \) and force vector \( \vec{F} \) are written as \( \vec{r} = (-2.0 \hat{i} + 0 \hat{j} + 4.0 \hat{k}) \) and \( \vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k} \).
This notation simplifies vector operations like addition, subtraction, and the cross product, which are vital for solving torque problems.
Physics Problem Solving
In tackling physics problems like calculating torque, a structured problem-solving approach is vital. The problems typically require:
  • Understanding the relevant physical laws or formulas, such as torque \( \vec{\tau} = \vec{r} \times \vec{F} \).
  • Identifying all given data and required unknowns from the problem statement. For example, knowing the components of \( \vec{F} \).
  • Breaking down vectors into unit vector components, thus simplifying subsequent calculations.
  • Applying mathematical operations like the cross product accurately to find exact solutions.
Problem-solving in physics combines analytical skills with practical knowledge of vector operations. Each step guides you towards the final solution, verifying physical sensibility along the way.
Coordinate Systems
A coordinate system is a framework used to define the positions of points in space. In this problem, we use the Cartesian coordinate system, characterized by the following axes:
  • x-axis: Horizontal direction.
  • y-axis: Vertical direction.
  • z-axis: Depth direction.
The position of the plum \((-2.0, 0.0, 4.0)\) means it is placed on these axes, providing a clear reference for calculating the torque using the torque formula. The force components specified in the problem can leverage this system for more straightforward application of vector operations.
Different coordinate systems can be used depending on the context, but Cartesian is widespread due to its straightforward representation of three-primary perpendicular axes, facilitating vector arithmetic and cross product computations efficiently.

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