Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

One force acting on a machine part is \(\overrightarrow{F} = (-5.00 N)\hat{\imath} + (4.00 N)\hat{\jmath}\). The vector from the origin to the point where the force is applied is \(\overrightarrow{r} = (-0.450 m)\hat{\imath} + (0.150 m)\hat{\jmath}\). (a) In a sketch,show \(\overrightarrow{r}, \overrightarrow{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin 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 is \( -1.05\hat{k} \ Nm \), directed into the page.

Step by step solution

01

Understand the Problem

We are given a force vector \( \overrightarrow{F} = (-5.00 \ N)\hat{\imath} + (4.00 \ N)\hat{\jmath} \) and a position vector \( \overrightarrow{r} = (-0.450 \ m)\hat{\imath} + (0.150 \ m)\hat{\jmath} \). We need to find the torque produced by the force about the origin.
02

Sketch the Vectors

Draw a diagram with the origin marked as the reference point. Draw the vector \( \overrightarrow{r} \) originating from the origin and pointing to the point \((-0.450, 0.150)\) on the XY plane. Then, draw the vector \( \overrightarrow{F} \) starting from this point.
03

Determine Torque Direction Using Right-Hand Rule

To determine the torque direction using the right-hand rule, point your fingers in the direction of \( \overrightarrow{r} \) and curl them towards \( \overrightarrow{F} \). Your thumb will point in the direction of the torque, which is out of the page along the positive \( \hat{k} \) direction.
04

Calculate the Torque

The torque \( \overrightarrow{\tau} \) is given by the cross product \( \overrightarrow{r} \times \overrightarrow{F} \). Calculate the determinant:\[\overrightarrow{\tau} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ -0.450 & 0.150 & 0 \ -5.00 & 4.00 & 0 \end{vmatrix} = (0.150 \times 0 - 0 \times 4.00)\hat{\imath} - (0 \times -5.00 - (-0.450) \times 0)\hat{\jmath} + ((-0.450) \times 4.00 - 0.150 \times -5.00)\hat{k}\]This results in:\[\overrightarrow{\tau} = (0)\hat{\imath} - (0)\hat{\jmath} + (-1.80 + 0.75)\hat{k} = -1.05\hat{k} \ Nm\]
05

Verify the Direction of Torque

The calculated torque \( -1.05\hat{k} \ Nm \) points in the negative \( \hat{k} \) direction, indicating it is into the page, opposite to our earlier right-hand rule prediction. This discrepancy shows a sign error in the right-hand rule prediction; whenever the vector calculations differ from initial predictions, recalibrate to ensure the computation is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Vector Cross Product
The vector cross product is a crucial operation when calculating torque. It's mainly used to find a vector that is orthogonal to two given vectors. In mathematical terms, for two vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \), the cross product \( \overrightarrow{A} \times \overrightarrow{B} \) results in a vector that is perpendicular to the plane formed by \( \overrightarrow{A} \) and \( \overrightarrow{B} \).

In terms of physical interpretation, the cross product is vital for determining rotational forces, like torque, in physics. Torque, essentially the rotational equivalent of linear force, can be found by applying the cross product equation \( \overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F} \), where \( \overrightarrow{r} \) is the position vector and \( \overrightarrow{F} \) is the force vector.

The magnitude of the cross product depends on the sine of the angle between the two vectors and the magnitudes of the vectors themselves. This means the cross product is directly influenced by both the size and angle of interaction between the two vectors, making it foundational in torque calculations.
Right-Hand Rule
The right-hand rule is a simple way to determine the direction of the resultant vector in a cross product situation. It's particularly useful for finding the direction of torque, which is a pseudo-vector. For torque, specifically, you apply the rule to the position vector \( \overrightarrow{r} \) and the force vector \( \overrightarrow{F} \).

Here's how it works: point your fingers along the direction of \( \overrightarrow{r} \), and then curl them towards \( \overrightarrow{F} \). Your thumb will naturally point in the direction of the torque vector \( \overrightarrow{\tau} \).

This rule promotes a three-dimensional understanding, since torque often acts along the \( \hat{k} \) axis when \( \overrightarrow{r} \) and \( \overrightarrow{F} \) lie in the \( x-y \) plane. It's key to check the orientation of the angles and the axis for precision when predicting torque directions.
Force Vector
A force vector represents the magnitude and direction of force acting at a point. It's defined in a certain dimension, typically using the standard unit vectors \( \hat{\imath} \), \( \hat{\jmath} \), and \( \hat{k} \) in three-dimensional space. In this exercise, the force vector \( \overrightarrow{F} = (-5.00 \ N)\hat{\imath} + (4.00 \ N)\hat{\jmath} \) lies in the \( x-y \) plane.

This vector notation helps in breaking down the force into its components along each axis, which is essential for understanding its complete orientation and influence on an object. The negative \( \hat{\imath} \) component indicates force acting in the negative \( x \)-direction, while the positive \( \hat{\jmath} \) component means the force acts in the positive \( y \)-direction.

Knowing how to dissect these vectors is fundamental in physical calculations, as it helps in understanding how forces can generate movement or rotation.
Position Vector
The position vector, \( \overrightarrow{r} \), describes a point in space relative to the origin. It provides crucial information needed to calculate torque or understand an object's spatial orientation.

Here, the position vector is the origin point to the force's application point: \( \overrightarrow{r} = (-0.450 \ m)\hat{\imath} + (0.150 \ m)\hat{\jmath} \). This vector indicates that the point is moved to the left and slightly up from the origin in the 2D plane.

This vector is instrumental in detailing the 'lever arm' in torque problems, as torque depends on both length and the angle the force is applied. To compute torque, we use the position vector and the direction of the force vector in combination, through the cross product, to obtain an accurate description of the resulting rotational force.

Understanding position vectors allows for the precise calculations of spatial placement, paving the way for effective manipulation and understanding of forces in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 2.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s? (b) Through what angle has it turned during that time? (c) Use Eq. (10.21) to calculate the work done by the torque. (d) What is the grinding wheel's kinetic energy when it is rotating at 1200 rev/min ? Compare your answer to the result in part (c).

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.

A hollow, thin-walled sphere of mass 12.0 kg and diameter 48.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t) = At^2 + Bt^4\), where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10. (a) What are the units of the constants \(A\) and \(B\)? (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad/s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR\(^2\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 2600 J. Calculate \(h\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free