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In unit-vector notation, what is the net torque about the origin on a flea located at coordinates \((0,-4.0 \mathrm{~m}, 5.0 \mathrm{~m})\) when forces \(\overrightarrow{\vec{F}}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{j}}\) act on the flea?

Short Answer

Expert verified
The net torque is \(-2.0\hat{i} \mathrm{Nm}\).

Step by step solution

01

Understand the Problem

We need to find the net torque on the flea located at the coordinates \((0, -4.0 \mathrm{~m}, 5.0 \mathrm{~m})\) due to two forces \(\vec{F}_{1}\) and \(\vec{F}_{2}\). Torque \(\vec{\tau}\) is given by the equation \(\vec{\tau} = \vec{r} \times \vec{F}\), where \(\vec{r}\) is the position vector and \(\vec{F}\) is the force vector.
02

Write the Position Vector

The position vector \(\vec{r}\) from the origin to the point where the flea is located is given by \(\vec{r} = 0\hat{i} - 4.0\hat{j} + 5.0\hat{k}\).
03

Calculate Torque Due to \(\vec{F}_1\)

The force \(\vec{F}_1\) is given as \(\vec{F}_1 = 3.0 \mathrm{~N} \hat{k}\). Calculate the torque \(\vec{\tau}_1\) using the cross product formula: \\(\vec{\tau}_1 = \vec{r} \times \vec{F}_1 = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & -4.0 & 5.0 \0 & 0 & 3.0 \end{vmatrix} \).
04

Solve the Cross Product for \(\vec{\tau}_1\)

Calculate the determinant: \\(\vec{\tau}_1= \hat{i}(0*(-3.0) - (5.0)*(0)) - \hat{j}(0*(-3.0) - (5.0)*(0)) + \hat{k}(0*(0) - (-4.0)*(0)) = -12.0\hat{i}\). \So, \(\vec{\tau}_1 = -12.0\hat{i} \mathrm{Nm}\).
05

Calculate Torque Due to \(\vec{F}_2\)

The force \(\vec{F}_2\) is given as \(\vec{F}_2 = -2.0 \mathrm{~N} \hat{j}\). Calculate the torque \(\vec{\tau}_2\) using the cross product formula: \\(\vec{\tau}_2 = \vec{r} \times \vec{F}_2 = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & -4.0 & 5.0 \0 & -2.0 & 0 \end{vmatrix} \).
06

Solve the Cross Product for \(\vec{\tau}_2\)

Calculate the determinant: \\(\vec{\tau}_2 = \hat{i}(-4.0*0 - 5.0*(-2.0)) - \hat{j}(0*0 - 5.0*(0)) + \hat{k}(0*(-2.0) - (0)*(-4.0)) = 10.0\hat{i}\). \So, \(\vec{\tau}_2 = 10.0\hat{i} \mathrm{Nm}\).
07

Sum the Torques to Find the Net Torque

Add the calculated torques \(\vec{\tau}_1\) and \(\vec{\tau}_2\) to find the net torque: \\(\vec{\tau}_{net} = \vec{\tau}_1 + \vec{\tau}_2 = -12.0\hat{i} + 10.0\hat{i} = -2.0\hat{i} \mathrm{Nm}\).

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Key Concepts

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

Cross Product
When working with physics problems involving torque, the cross product is a key mathematical operation. The cross product, also known as the vector product, is used to find the torque vector from given position and force vectors.
The torque \(\vec{\tau}\) is calculated by taking the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\). This is denoted as:\[ \vec{\tau} = \vec{r} \times \vec{F} \]The cross product of two vectors, \(\vec{A}\) and \(\vec{B}\), results in a third vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\). This is highly useful in three-dimensional physics problems.
To solve for the cross product, the components of the vectors are arranged in a determinant, making it essential to understand and calculate determinants for successful torque calculations.
Vector Notation
In physics, vectors are used to represent quantities that have both a magnitude and a direction. Vectors are critical for describing physical phenomena like force, velocity, and torque.
A vector in unit-vector notation is expressed as a combination of its components along the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) unit vectors, which correspond to the x, y, and z axes, respectively.
For example, the position vector \(\vec{r}\) in the problem is written as:\[ \vec{r} = 0\hat{i} - 4.0\hat{j} + 5.0\hat{k} \]This constitutes an organized method to represent complex quantities and perform mathematical operations such as the cross product, facilitating the accurate calculation of quantities like torque in multi-dimensional spaces.
Physics Problem
Physics problems often involve applying theoretical principles to determine unknown quantities. In the case of calculating torque, the challenge typically lies in understanding the relationship between forces, positions, and their effects.
- **Identify forces and their direction:** Examine the given forces and their components, like \(\vec{F}_1 = 3.0 \, \mathrm{N}\, \hat{k}\) and \(\vec{F}_2 = -2.0 \, \mathrm{N}\, \hat{j}\).
- **Position vectors:** Understand the position of the object in three-dimensional space, indicated by coordinates, to properly describe \(\vec{r}\).
- **Calculate step-by-step:** Compute torque for each force separately by plugging vectors into the cross product, then sum to find the net torque.
Physics problems blend mathematical formulations with conceptual physics knowledge, requiring precise application and practice to attain correct solutions.
Net Torque
Net torque is the resultant torque acting on an object, found by summing up individual torques produced by each force applied on that object. It is a determinant of an object's rotational motion.
In our example, the torques due to two different forces were calculated as \(\vec{\tau}_1 = -12.0\hat{i} \, \mathrm{Nm}\) and \(\vec{\tau}_2 = 10.0\hat{i}\, \mathrm{Nm}\).
To find the net torque \(\vec{\tau}_{net}\), simply add the individual torque vectors:\[ \vec{\tau}_{net} = \vec{\tau}_1 + \vec{\tau}_2 \quad = -12.0\hat{i} + 10.0\hat{i} \quad = -2.0\hat{i} \, \mathrm{Nm} \]This shows the combined rotational effect of all forces, guiding conclusions regarding how the object will behave and moving towards solving larger systems in mechanics.

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Most popular questions from this chapter

A particle is acted on by two torques about the origin: \(\vec{\tau}_{1}\) has a magnitude of \(2.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the positive direction of the \(x\) axis, and \(\vec{\tau}_{2}\) has a magnitude of \(4.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the negative direction of the \(y\) axis. In unit-vector notation, find \(d \vec{\ell} / d t\), where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

The angular momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from \(3.00\) to \(0.800 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) in \(1.50 \mathrm{~s}\). (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

A uniform thin rod of length \(0.500 \mathrm{~m}\) and mass \(4.00 \mathrm{~kg}\) can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a \(3.00 \mathrm{~g}\) bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle \(\theta=60.0^{\circ}\) with the rod (Fig. 11-50). If the bullet lodges in the rod and the angular velocity of the rod is \(10 \mathrm{rad} / \mathrm{s}\) immediately after the collision, what is the bullet's speed just before impact?

A ball of mass \(M\) and radius \(R\) rolls smoothly from rest down a ramp and onto a circular loop of radius \(0.48 \mathrm{~m}\). The initial height of the ball is \(h=0.36 \mathrm{~m}\). At the loop bottom, the magnitude of the nor-mal force on the ball is \(2.00 \mathrm{Mg}\). The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central sphere (of a different uniform density). The rotational inertia of the ball can be expressed in the general form \(I=\beta M R^{2}\), but \(\beta\) is not \(0.4\) as it is for a ball of uniform density. Determine \(\beta\).

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of \(0.10 g ?\) (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to \(0.10 g\) ? Why?

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