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A plum is located at coordinates \((-2.0 \mathrm{~m}, 0,4.0 \mathrm{~m}) .\) In unit- vector 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) \( \vec{\tau} = -24.0 \hat{j} + 12.0 \hat{k} \), (b) \( \vec{\tau} = 24.0 \hat{j} - 12.0 \hat{k} \), (c) \( \vec{\tau} = 12.0 \hat{j} \), (d) \( \vec{\tau} = -12.0 \hat{j} \).

Step by step solution

01

Understanding Torque

Torque, \( \vec{\tau} \), is the cross product of the position vector, \( \vec{r} \), and the force vector, \( \vec{F} \). In unit-vector notation, it is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). Torque is a vector that is perpendicular to both \( \vec{r} \) and \( \vec{F} \).
02

Identifying the Position Vector

The position vector \( \vec{r} \) of the plum is given as \((-2.0 \hat{i} + 0 \hat{j} + 4.0 \hat{k}) \mathrm{~m}\). This vector is the displacement from the origin to the point where the plum is located.
03

Step 3a: Calculating Torque for \( F_x = 6.0 \mathrm{~N} \)

The force vector is \( \vec{F} = 6.0 \hat{i} \). Using the cross product formula, \( \vec{\tau} = \vec{r} \times \vec{F} \):\[\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2.0 & 0 & 4.0 \ 6.0 & 0 & 0 \end{vmatrix}= (0\cdot0 - 4.0\cdot 0)\hat{i} - (0\cdot0 - 4.0\cdot6.0)\hat{j} + (0\cdot0 - 6.0\cdot(-2.0))\hat{k}= 0\hat{i} - 24.0\hat{j} + 12.0\hat{k}\]Thus, \( \vec{\tau} = -24.0 \hat{j} + 12.0 \hat{k} \).
04

Step 3b: Calculating Torque for \( F_x = -6.0 \mathrm{~N} \)

The force vector is \( \vec{F} = -6.0 \hat{i} \). Using the cross product formula:\[\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2.0 & 0 & 4.0 \ -6.0 & 0 & 0 \end{vmatrix}= (0\cdot0 - 4.0\cdot 0)\hat{i} - (0\cdot0 - 4.0\cdot(-6.0))\hat{j} + (0\cdot0 - (-6.0)\cdot(-2.0))\hat{k}= 0\hat{i} + 24.0\hat{j} - 12.0\hat{k}\]Thus, \( \vec{\tau} = 24.0 \hat{j} - 12.0 \hat{k} \).
05

Step 3c: Calculating Torque for \( F_z = 6.0 \mathrm{~N} \)

The force vector is \( \vec{F} = 6.0 \hat{k} \). Using the cross product formula:\[\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2.0 & 0 & 4.0 \ 0 & 0 & 6.0 \end{vmatrix}= (0\cdot6.0 - 4.0\cdot0)\hat{i} - (-2.0\cdot6.0 - 4.0\cdot0)\hat{j} + (-2.0\cdot0 - 0\cdot0)\hat{k}= 0\hat{i} - (-12.0)\hat{j} + 0\hat{k}= 12.0 \hat{j}\]Thus, \( \vec{\tau} = 12.0 \hat{j} \).
06

Step 3d: Calculating Torque for \( F_z = -6.0 \mathrm{~N} \)

The force vector is \( \vec{F} = -6.0 \hat{k} \). Using the cross product:\[\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2.0 & 0 & 4.0 \ 0 & 0 & -6.0 \end{vmatrix}= (0\cdot(-6.0) - 4.0\cdot0)\hat{i} - (-2.0\cdot(-6.0) - 4.0\cdot0)\hat{j} + (-2.0\cdot0 - 0\cdot0)\hat{k}= 0\hat{i} - 12.0\hat{j} \]Thus, \( \vec{\tau} = -12.0 \hat{j} \).

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Key Concepts

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

Cross Product
To understand how torque is calculated in physics, it's essential to grasp the concept of the cross product. The cross product is a mathematical operation performed on two vectors, resulting in a third vector that is perpendicular to the plane of the two original vectors. This operation is used when dealing with quantities like torque and angular momentum. The formula for the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is:\[\vec{C} = \vec{A} \times \vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}\]Here, the \( \hat{i}, \hat{j}, \hat{k} \) are the unit vectors along the x, y, and z axes respectively. This operation is particularly vital when calculating torque because torque is defined as the cross product of the position vector and the force vector applied on the object, which makes it inherently perpendicular to both.
Force Vector
In physics, a force vector is a vector that represents both the magnitude and the direction of a force applied on an object. It is typically denoted by \( \vec{F} \). In this exercise, we have varying components of the force vector along different axes. This vector will change based on the condition given such as \( F_x=6.0 \text{ N} \), \( F_z=6.0 \text{ N} \), and their respective negative values.Calculating torque requires the specific component of the force vector. For example, if the force only acts along the \(x\)-axis, the force vector will be \( \vec{F} = F_x \hat{i} \). Similarly, when it acts along the \(z\)-axis, the force vector becomes \( \vec{F} = F_z \hat{k} \). Knowing the specific direction of the applied force is crucial because it directly affects how the torque is calculated through the cross product.
Position Vector
The position vector \( \vec{r} \) defines a specific point in space, starting from the origin (0, 0, 0) to the location of an object. In this context, the position vector is given for the location of the plum as \((-2.0 \hat{i} + 0 \hat{j} + 4.0 \hat{k}) \text{ m}\). This vector describes where the plum is relative to the origin in three-dimensional space.The position vector is an essential part of calculating torque using the cross product formula. It conveys information about the distance and orientation from the pivot point or axis of rotation, which in this problem is the origin, to the point where force is applied.
Unit Vector Notation
Unit vector notation is a method of expressing vectors as a combination of magnitudes and directions along the Cartesian axes, using \( \hat{i}, \hat{j}, \hat{k} \). This notation is widely used because it allows clear representation of vectors in three-dimensional space, where \( \hat{i}, \hat{j}, \hat{k} \) correspond to the x, y, and z directions respectively.Each vector such as the position vector \(\vec{r}\) and the force vector \(\vec{F}\) is expressed in terms of these unit vectors. In the exercise, the position vector is \(-2.0 \hat{i} + 0 \hat{j} + 4.0 \hat{k} \). This indicates the specific influences of each axis on the overall vector. By using unit vectors, the representation simplifies the understanding and calculation of operations like cross products needed to find torque.

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Most popular questions from this chapter

At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m}\). It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

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