/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Explain how to find the torque p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Explain how to find the torque produced by a force using cross products.

Short Answer

Expert verified
**Question:** Explain how to find the torque produced by a force using cross products. Describe the steps involved and the interpretation of the resulting torque vector. **Answer:** To find the torque produced by a force using cross products, follow these steps: 1. Understand the concept of torque: Torque is a measure of the rotational force acting on an object, and it is a vector quantity with both magnitude and direction. 2. Recall the cross product: The cross product of two vectors is a vector that is perpendicular to the plane formed by the two vectors and has a magnitude equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. 3. Identify the position vector and force vector: The position vector (r) connects the pivot point to the point where the force is applied, while the force vector (F) represents the applied force. 4. Calculate the cross product of position vector and force vector: Compute the determinant of a matrix formed by the coordinate unit vectors (i, j, k) and the coordinates of the position and force vectors. The cross product r×F will give you the torque vector (τ). 5. Result and interpretation: The resulting torque vector (τ) represents the rotational force and its direction. The magnitude of τ can be found using the formula |τ| = |r||F|sin(θ), where θ is the angle between the position and force vectors. The direction of the torque vector indicates the direction of rotation, either clockwise or counterclockwise.

Step by step solution

01

Understand the concept of torque

Torque (Ï„) is a measure of the rotational force that causes an object to rotate around a specified axis or pivot point. Torque is a vector quantity, meaning it has both magnitude and direction. In the context of this exercise, we will find the torque produced by a force using cross products.
02

Recall the cross product

The cross product of two vectors A and B, denoted as A×B, is a vector that is perpendicular to the plane formed by A and B and follows the right-hand rule. The magnitude of the cross product is equal to the product of the magnitudes of the two vectors multiplied by the sine of the angle between them. Mathematically, this can be expressed as |A×B| = |A||B|sin(θ), where θ is the angle between vectors A and B.
03

Identify the position vector and force vector

To find the torque produced by a force, we need two vectors: the position vector (r) and the force vector (F). The position vector is the vector that connects the pivot point to the point where the force is applied. The force vector represents the applied force.
04

Calculate the cross product of position vector and force vector

To find the torque, we need to calculate the cross product of the position vector and the force vector. The cross product can be computed using the determinant of a matrix formed by the coordinate unit vectors (i, j, k) and the coordinates of the position and force vectors. Let the position vector r = and the force vector F = , the cross product r×F can be found using the following determinant: | i j k | | r1 r2 r3 | | F1 F2 F3 | Compute the cross product by calculating the determinant, which gives the torque vector (τ) as: τ = <(r2*F3 - r3*F2), (r3*F1 - r1*F3), (r1*F2 - r2*F1)>
05

Result and interpretation

The resulting vector τ is the torque produced by the force. The magnitude of τ represents the rotational force, and the direction of τ is perpendicular to the plane formed by the position and force vectors, following the right-hand rule. To find the magnitude of the torque, you can use the formula |τ| = |r||F|sin(θ), where θ is the angle between the position and force vectors. Note that the direction of the torque vector denotes the direction of rotation, either clockwise or counterclockwise.

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 91Ó°ÊÓ!

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 pair of nonzero vectors in the plane is linearly dependent if one vector is a scalar multiple of the other. Otherwise, the pair is linearly independent. a. Which pairs of the following vectors are linearly dependent and which are linearly independent: \(\mathbf{u}=\langle 2,-3\rangle\) \(\mathbf{v}=\langle-12,18\rangle,\) and \(\mathbf{w}=\langle 4,6\rangle ?\) b. Geometrically, what does it mean for a pair of nonzero vectors in the plane to be linearly dependent? Linearly independent? c. Prove that if a pair of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is linearly independent, then given any vector \(w\), there are constants \(c_{1}\) and \(c_{2}\) such that \(\mathbf{w}=c_{1} \mathbf{u}+c_{2} \mathbf{v}\)

Proof of Product Rule By expressing \(\mathbf{u}\) in terms of its components, prove that $$\frac{d}{d t}(f(t) \mathbf{u}(t))=f^{\prime}(t) \mathbf{u}(t)+f(t) \mathbf{u}^{\prime}(t)$$

An ant walks due east at a constant speed of \(2 \mathrm{mi} / \mathrm{hr}\) on a sheet of paper that rests on a table. Suddenly, the sheet of paper starts moving southeast at \(\sqrt{2} \mathrm{mi} / \mathrm{hr} .\) Describe the motion of the ant relative to the table.

Evaluate the following definite integrals. $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t$$

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\), the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C,\) the outwardpointing unit vector normal to \(C\) is also \(\mathbf{n}=\langle\cos \theta, \sin \theta\rangle\) c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\), find the component of \(\mathbf{v}\) normal to \(C\). Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

See all solutions

Recommended explanations on Math 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.