/*! 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 61 A square, \(0.40 \mathrm{~m}\) o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A square, \(0.40 \mathrm{~m}\) on a side, is mounted so that it can rotate about an axis that passes through the center of the square. The axis is perpendicular to the plane of the square. A force of \(15 \mathrm{~N}\) lies in this plane and is applied to the square. What is the magnitude of the maximum torque that such a force could produce?

Short Answer

Expert verified
The maximum torque is 4.24 N·m.

Step by step solution

01

Understand Torque

Torque \((\tau)\) is the measure of the force that can cause an object to rotate about an axis. It depends on the force \(F\), the distance \(r\) from the axis to the point of force application (lever arm), and the angle \(\theta\) between the force and the lever arm. The formula for torque is \(\tau = r \cdot F \cdot \sin(\theta)\).
02

Identify Maximum Torque Conditions

The maximum torque is obtained when the force \(F\) is applied perpendicular to the lever arm, making the angle \(\theta = 90^\circ\) or \(\sin(\theta) = 1\). Hence, the formula simplifies to \(\tau = r \cdot F\).
03

Calculate Lever Arm

The problem specifies a square with sides of \(0.40 \) m. The maximum distance \(r\) from the center of the square to the point where the force can be applied (lever arm) is half the diagonal of the square. The diagonal \(d\) of a square with side length \(a\) is given by \(d = a\sqrt{2}\). So, \(r = \frac{0.40\sqrt{2}}{2} = 0.20\sqrt{2} \) m.
04

Calculate Maximum Torque

Substitute the values of \(F\) and \(r\) into the simplified torque equation \(\tau = r \cdot F\). Here, \(F = 15\) N and \(r = 0.20\sqrt{2} \) m. Thus, \(\tau = 0.20\sqrt{2} \times 15 = 3\sqrt{2} = 4.24\) N·m.

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!

Key Concepts

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

Rotational Motion
When we talk about rotational motion, we're diving into the world where objects move in circles. Consider a spinning wheel or a merry-go-round. This motion is about everything relating to how things rotate around a center or axis.
In the exercise, the square rotates around an axis. This axis goes through its center and is perpendicular to its plane. This means it spins as if it were lying flat on a table and turning like a pinwheel.
Rotational motion plays a key role in how forces affect an object. It's not just about linear movement from one point to another. Instead, it's about how forces can change the rotational state of an object. Keeping in mind that when a force is applied away from the axis, it can create what's called torque.
Moment of Force
Moment of force, also commonly referred to as torque, is how we quantify the effectiveness of a force in causing rotation. Think of it as the twist that makes things turn!
The formula for torque is crucial: \( \tau = r \cdot F \cdot \sin(\theta) \). Here, \( \tau \) is the torque, \( r \) is the lever arm length (distance from the axis to where the force is applied), and \( \theta \) is the angle between the force and lever arm direction.
In our exercise, we are looking for the maximum torque possible. This happens when the force is applied perpendicularly, which maximizes its twisting effect. The concept of torque is essential in designing and understanding systems involving rotational motion, such as engines and gears.
Lever Arm
The lever arm is the distance from the axis of rotation to where the force is applied. This distance significantly impacts how much torque a force can produce.
In the case of the exercise, the square's side is 0.40 m. To find the maximum possible lever arm, we need half of its diagonal length. For any square, the diagonal \( d \) is found using the formula \( d = a\sqrt{2} \), where \( a \) is the side length.
Therefore, the lever arm \( r \) becomes \( 0.20\sqrt{2} \) m. A longer lever arm means greater torque, assuming the force applied remains the same. This principle is why tools like wrenches have longer handles — to increase the leverage and thus, the torque.
Perpendicular Force Application
Applying a force perpendicularly maximizes its effect on producing rotation. This is because, when \( \theta \) is 90 degrees, \( \sin(\theta) = 1 \).
In simpler terms, the force is being used in the most effective way possible to twist or rotate the object. In the context of the given problem, achieving maximum torque requires that the force acts at a right angle to the lever arm.
This idea is similar to using a door handle; the force applied perpendicularly to the door results in the easiest and most efficient door opening. Applying the force in any other direction would require more effort for the same rotation. This principle is fundamental not only in physics but also in everyday practical applications.

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

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel?

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\underline{\theta}\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650\). Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.

Two thin rectangular sheets \((0.20 \mathrm{~m} \times 0.40 \mathrm{~m})\) are identical. In the first sheet the axis of rotation lies along the \(0.20-\mathrm{m}\) side, and in the second it lies along the \(0.40-\mathrm{m}\) side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in \(8.0 \mathrm{~s}\). How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

A small 0.500 -kg object moves on a frictionless horizontal table in a circular path of radius \(1.00 \mathrm{~m} .\) The angular speed is \(6.28 \mathrm{rad} / \mathrm{s} .\) The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than \(105 \mathrm{~N}\), what is the radius of the smallest possible circle on which the object can move?

A flat uniform circular disk (radius \(=2.00 \mathrm{~m}\), mass \(=1.00 \times 10^{2} \mathrm{~kg}\) ) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A \(40.0-\mathrm{kg}\) person, standing \(1.25 \mathrm{~m}\) from the axis, begins to run on the disk in a circular path and has a tangential speed of \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. Find the resulting angular speed of the disk (in \(\mathrm{rad} / \mathrm{s}\) ) and describe the direction of the rotation.

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.