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

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m} .\) (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

Short Answer

Expert verified
(a) \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \). (b) Magnitude: 37.00 Nā‹…m, Direction: into the plane.

Step by step solution

01

Find the Position Vector

The position vector \( \vec{r} \) is given by the coordinates of the point where the force is applied, \( x = 3.00 \, \text{m} \) and \( y = 4.00 \, \text{m} \). Therefore, the position vector in terms of the unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \) is \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \).
02

Calculate the Torque

Torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \). The cross product in two dimensions is calculated as follows: \( \vec{\tau} = \vec{r} \times \vec{F} = |\begin{array}{cc} 3.00 \, \text{m} & 4.00 \, \text{m} \ 7.00 \, \text{N} & -3.00 \, \text{N} \end{array}| \). This results in \( \vec{\tau} = (3.00)(-3.00) - (4.00)(7.00) \).
03

Evaluate the Determinant

Calculate the determinant to find the torque: \( \vec{\tau} = (3.00)(-3.00) - (4.00)(7.00) = -9.00 - 28.00 = -37.00 \, \text{N}\cdot\text{m} \).
04

Determine Direction of Torque

In the 2-dimensional problem, the direction of torque can be inferred as perpendicular to the plane, along the \( \hat{k} \) direction. Since the torque \( \vec{\tau} \) is negative, this indicates the direction is into the plane (negative \( \hat{k} \)).

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.

Cross Product
The cross product is an essential tool in physics and engineering, especially for calculating torque. It involves determining the product of two vectors, resulting in a third vector that is perpendicular to the plane containing the initial vectors.
In this exercise, we calculate the cross product of a position vector \( \vec{r} \) and a force vector \( \vec{F} \) to find the torque \( \vec{\tau} \). The cross product is typically defined in a three-dimensional space, but it simplifies in two dimensions where results are along the \( \hat{k} \) direction, indicating the torque's perpendicular nature to the xy-plane.
The cross product in two dimensions is represented by the determinant \( | \begin{array}{cc} x & y \ f_x & f_y \end{array} | \), which simplifies to express a scalar value indicating the magnitude of the torque.
Position Vector
The position vector is crucial when calculating torque, as it defines the point of application of force relative to the origin or another reference point.
In our problem, the position vector \( \vec{r} \) represents the coordinates of the force's point of application in the xy-plane, given by \( \vec{r} = (3.00 \, \text{m}) \hat{\imath} + (4.00 \, \text{m}) \hat{\jmath} \). This vector quantifies the position of the force relative to the origin (0,0), providing a way to determine how effectively a force can induce rotational motion.
The position vector plays a critical role when we calculate torque, as it determines the lever arm which directly influences the torque's magnitude.
Force Vector
The force vector shows the direction and magnitude of the given force applied to an object. In this exercise, the force vector \( \vec{F} \) is \( (7.00 \, \text{N}) \hat{\imath} + (-3.00 \, \text{N}) \hat{\jmath} \).
This vector indicates that:
  • \(7.00 \, \text{N}\) of force is acting in the positive x-direction.
  • \(-3.00 \, \text{N}\) points in the negative y-direction.
Understanding the components of a force vector is imperative for calculating resulting effects like torque, where different vector components interact to produce rotation.
By manipulating force vectors, we can predict and analyze the impact they will have on an object or a system.
Determinants
Determinants provide us with a mathematical function to extract useful scalar information from a matrix, often used to solve systems of linear equations and calculate vector products like the cross product.
In torque calculations, the determinant simplifies finding the cross product in two dimensions. By organizing the components of our position and force vectors into a 2x2 matrix, we can quickly evaluate the determinant which corresponds to the magnitude of the torque.
In our exercise, the cross product and hence the determinant \( | \begin{array}{cc} 3.00 \, \text{m} & 4.00 \, \text{m} \ 7.00 \, \text{N} & -3.00 \, \text{N} \end{array} | \) is calculated as \( -37.00 \, \text{N}\cdot\text{m} \), indicating both the torque's magnitude and its sense of rotation. Understanding how to evaluate determinants is key in applications like these, ensuring accurate calculations and analyses.

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

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [\(Hint:\) Integrating Eq. (10.29) yields \(\Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z)dt = (\Sigma\tau_z)_av \Delta t\). The quantity \(\int_{t1}^{t2} (\Sigma\tau_z)dt\) is called the angular impulse.]

We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 kg and diameter 1.0 m. If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a \(neutron star\). The density of a neutron star is roughly 10\(^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0 \(\times 10^5\) km (comparable to our sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star.

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass \(M\), for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0\(^\circ\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 kg?

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