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Chapter 9: Problem 6

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 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 approximately \( 4.24 \mathrm{N} \cdot \mathrm{m} \).

Step by step solution

01

Understand the problem

We need to find the maximum torque produced by a force applied to a square. Torque depends on the force applied, the distance from the axis of rotation, and the angle of application.
02

Identify variables and formula

The side of the square is given as 0.40 m, and the force is 15 N. The formula for torque is \( \tau = r \cdot F \cdot \sin(\theta) \), where \( \tau \) is the torque, \( r \) is the distance from the axis, \( F \) is the force, and \( \theta \) is the angle between the force and the position vector.
03

Determine distance to apply force

To maximize torque, the force should be applied at the farthest point from the axis of rotation. The distance \( r \) is half the diagonal of the square, since the square rotates about its center.
04

Calculate diagonal of the square

The diagonal \( d \) of the square can be calculated using the formula for the diagonal of a square: \( d = \sqrt{2} \times \text{side} = \sqrt{2} \times 0.40 \). This gives \( d = 0.40\sqrt{2} \).
05

Calculate the effective radius

The effective distance from the axis is half the diagonal: \( r = \frac{0.40\sqrt{2}}{2} = 0.20\sqrt{2} \).
06

Calculate maximum torque

Since torque is maximized when \( \sin(\theta) = 1 \) (i.e., \( \theta = 90^\circ \)), the maximum torque is: \( \tau = 0.20\sqrt{2} \cdot 15 \cdot 1 = 3\sqrt{2} \). Calculating gives \( \tau \approx 4.24 \mathrm{N} \cdot \mathrm{m} \).

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Key Concepts

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

Force
In physics, force is any interaction that, when unopposed, changes the motion of an object. Forces can cause objects to start moving, stop moving, or change their velocity. In the context of this exercise, we have a force of 15 N applied to a square that can rotate around an axis. This force lies in the plane of the square, playing a crucial role in creating torque.
  • A force is defined as a vector, which means it has both magnitude and direction.
  • The unit of force is the Newton (N), named after Sir Isaac Newton.
  • The force here is what causes the square to rotate around its center.
When discussing rotational motion, it's essential to understand how forces affect objects. The magnitude of this force, combined with its point of application, influences how much torque, or rotational force, is generated. The greater the magnitude of the force, when appropriately applied, the greater the torque created.
Axis of Rotation
The axis of rotation is an imaginary line around which an object rotates. In this problem, the square rotates around an axis that passes through its center and is perpendicular to its plane.
  • This axis is crucial because it determines the leverage a force has to produce torque.
  • The distance from the axis to the point where the force is applied is called the "lever arm" or "radius." In this scenario, it is half the diagonal of the square.
  • The orientation and position of the axis influence the calculation of torque.
When a force is applied to the square, the axis of rotation doesn't change, but the torque depends on how far and in what direction the force is applied relative to this axis. In geometric figures, like squares or rectangles, finding the axis of rotation helps you understand the mechanics of the entire system.
Square Geometry
Understanding the geometry of a square is vital in solving problems related to torque and rotational motion. A square is a regular quadrilateral, which means all four sides are of equal length, and all angles are right angles.
  • The side length of the square in this problem is 0.40 meters.
  • To calculate torque, understanding the diagonal of the square is key. The diagonal divides the square into two equal right triangles.
  • You can find the diagonal of a square using the formula: the square root of the sum of the squares of its side lengths, or simply: \( d = s\sqrt{2} \).
Since the square rotates about its center, the effective lever arm is half of this diagonal. For this problem, the distance from the center to any corner, which is half the diagonal, becomes the pivotal factor for maximum torque calculation.

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Most popular questions from this chapter

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s}\). Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{kg} \cdot \mathrm{m}^{2} .\) From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{kg}\). After all the sand is in place, what is the angular velocity of the disk?

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{kg})\) is rotating at an angular velocity of 13.1 rad/s (see the drawing). The front brake is then applied for \(3.0 \mathrm{s},\) and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{m}\). The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

In outer space two identical space modules are joined together by a massless cable. These modules are rotating about their center of mass, which is at the center of the cable because the modules are identical (see the drawing). In each module, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is \(v_{0}=17 \mathrm{m} / \mathrm{s} .\) Then they pull together until the distance between them is reduced by a factor of two. Each module has a final tangential speed of \(v_{\mathrm{r}}\). Find the value of \(v_{\mathrm{f}}\)

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.08 above the horizontal. One child, with a mass of 44.0 kg, is hanging 1.30 m along the branch from the tree trunk. The other child, with a mass of 35.0 kg, is hanging 2.10 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.
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