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One force acting on a machine part is \(\overrightarrow{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\boldsymbol{\imath}}+\) \((4.00 \mathrm{N}) \hat{\jmath}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\jmath} .\) (a) In a sketch, show \(\vec{r}, \vec{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

Short Answer

Expert verified
The torque vector is \(0.750 \hat{k} \text{ Nm}\), directed out of the page.

Step by step solution

01

Draw the Vectors

Sketch the vectors \( \vec{r} \) and \( \vec{F} \) on a Cartesian coordinate system. \( \vec{r} = (-0.450 \text{ m}) \hat{\imath} + (0.150 \text{ m}) \hat{\jmath} \) goes from the origin to the point of force application. \( \vec{F} = (-5.00 \text{ N}) \hat{\imath} + (4.00 \text{ N}) \hat{\jmath} \) acts at this point.
02

Determine Torque Direction with Right-Hand Rule

Point your fingers along vector \( \vec{r} \) and curl them towards vector \( \vec{F} \). Your thumb points in the direction of the torque vector. Since \( \vec{r} \) points left and \( \vec{F} \) points upwards, torque will point out of the page (or the positive \( \hat{k} \) direction).
03

Calculate the Torque

Torque \( \vec{\tau} \) is calculated using the cross product: \( \vec{\tau} = \vec{r} \times \vec{F} \). Use the determinant method:\[\vec{\tau} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k} \-0.450 & 0.150 & 0 \-5.00 & 4.00 & 0 \ \end{vmatrix}\]Calculating this determinant, we find:\[ \vec{\tau} = (0)(4.00) - (0.150)(-5.00) \hat{k} = 0.750 \hat{k} \text{ Nm}\]
04

Confirm Torque Direction

The calculated torque vector is \(0.750 \hat{k} \text{ Nm}\), which confirms that the direction is out of the page, consistent with the right-hand rule analysis from Step 2.

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Key Concepts

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

Vector Cross Product
The vector cross product is an essential concept in physics and mathematics used to find a vector that is perpendicular to two given vectors. It is often encountered when calculating torque and rotational effects in mechanics. Given two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), the cross product \( \overrightarrow{A} \times \overrightarrow{B} \) results in a third vector. This resultant vector is orthogonal to both \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

To calculate a vector cross product, one common method is the determinant approach. For vectors with components \( \overrightarrow{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \overrightarrow{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), set up a determinant:
  • Top row: unit vectors \( \hat{i}, \hat{j}, \hat{k} \).
  • Second row: components of the first vector \( (a_1, a_2, a_3) \).
  • Third row: components of the second vector \( (b_1, b_2, b_3) \).
The cross product is calculated by expanding this determinant to find the resulting unit vector components in \( \hat{i}, \hat{j}, \) and \( \hat{k} \). This technique not only provides the magnitude of the torque but also its direction.
Right-Hand Rule
The right-hand rule is a simple, intuitive technique used to determine the direction of the result of a cross product operation, like torque in physics. It is aptly named because it requires you to use your right hand. Let's see how it works.

First, position your right hand so that your fingers point in the direction of the first vector, \( \vec{r} \). Then, curl your fingers toward the second vector, \( \vec{F} \). The direction your thumb points indicates the direction of the torque vector, \( \vec{\tau} \). For the problem at hand, as you curl your fingers from \( \vec{r} \) pointing left to \( \vec{F} \) pointing up, your thumb points out of the page; hence, the torque direction is in the positive \( \hat{k} \) direction.

Remember, the right-hand rule is a handy tool (pun intended) when dealing with cross products in three-dimensional space. It provides a practical way to visualize vector directionality crucial to solving physics problems effectively.
Cartesian Coordinate System
The Cartesian coordinate system is foundational in physics and mathematics for representing vectors. It uses three axes - commonly labeled \( x \), \( y \), and \( z \) - each perpendicular to the others, creating a three-dimensional space. Vectors, such as force or position, are resolved into these axes using unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \). This decomposition allows for complex spatial problems to be tackled using simple algebraic methods.

In this coordinate system:
  • The \( \hat{i} \) vector indicates direction along the \( x \)-axis.
  • The \( \hat{j} \) vector indicates direction along the \( y \)-axis.
  • The \( \hat{k} \) vector indicates direction along the \( z \)-axis.
By decomposing vectors into these components, you can easily perform operations like the addition of vectors, determine the magnitude, or perform cross products to calculate resultant vectors like torque. Visualizing vector components in a Cartesian plane simplifies understanding and solving problems involving spatial interactions.
Physics Problem Solving
Physics problem solving often involves breaking down complex scenarios into a series of logical steps. Using a structured approach ensures clarity and accuracy. Consider this methodology when solving problems like calculating torque:

1. **Understand the Problem:** - Carefully read the problem. - Identify what physical quantities you know and what is being asked.

2. **Visual Representation:** - Drawing vectors and coordinate systems helps in visualizing the physical situation.

3. **Mathematical Formulation:** - Translate the physical problem into mathematical language through equations.

4. **Solve Step-by-Step:** - Perform calculations methodically (e.g., use the cross product to find torque). - Keep track of units and vector directions.

5. **Verification:** - Reassess the results through alternate methods or checks like the right-hand rule for direction.

Approaching physics exercises with this structured framework can enhance your analytical skills and confidence in tackling a variety of problems.

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Most popular questions from this chapter

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{x} .\) If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established, A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{k}\) , (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z}\) . Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A high-wheel antique bicycle has a large front wheel with the foot-powered crank mounted on its axle and a small rear wheel turning independently of the front wheel; there is no chain connecting the wheels. The radius of the front wheel is \(65.5 \mathrm{cm},\) and the radius of the rear wheel is 22.0 \(\mathrm{cm} .\) Your modern bike has awheel diameter of 66.0 \(\mathrm{cm}(26 \text { inches) and front and rear sprockets }\) with radii of 11.0 \(\mathrm{cm}\) and \(5.5 \mathrm{cm},\) respectively. The rear sprocket is rigidly attached to the axle of the rear wheel. You ride your modern bike and turn the front sprocket at 1.00 rev \(/ \mathrm{s}\) . The wheels of both bikes roll along the ground without slipping. (a) What is your linear speed when you ride your modern bike? (b) At what rate must you turn the crank of the antique bike in order to travel at the same speed as in part (a)? (c) What then is the angular speed (in rev/s) of the small rear wheel of the antique bike?

The carbide tips of the cutting teeth of a circular saw are 8.6 \(\mathrm{cm}\) from the axis of rotation. (a) The no-toad speed of the saw, when it is not cutting anything, is 4800 rev/min. Why is no-load power output negligible? (b) While the saw is cutting lumber, its angular speed slows to 2400 rev/min and the power outputis 1.9 \(\mathrm{hp}\) . What is the tangential force that the wood exerts on the carbidetips?

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axe. If a constant net torque of 5.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; \((b)\) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . (a) If the disk rolls up a \(30.0^{\circ}\) ramp, how far along the ramp will it move before it stops? (b) Explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

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