/*! 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 1 The wheel of a car has a radius ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 1

The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 295 N ? m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a counter- torque. Moreover, the car has a constant velocity, so this counter torque balances the applied torque. What is the magnitude of the static frictional force?

Short Answer

Expert verified
The static frictional force is approximately 842.86 N.

Step by step solution

01

Understand the Relationship Between Torque and Force

Torque \( (\tau) \) is the rotational equivalent of force. It is given by the formula \( \tau = F \, r \), where \( F \) is the force applied, and \( r \) is the radius of rotation.
02

Set Up the Equation for Torque

Since the car travels at constant velocity, the torques are balanced. This means the applied torque \( 295 \, \text{N} \cdot \text{m} \) is equal to the counter-torque due to static friction. \[ \tau_{\text{applied}} = \tau_{\text{friction}} \] \[ 295 = F_{\text{friction}} \, \times \, 0.350 \]
03

Solve for the Static Frictional Force

Rearrange the equation to solve for \( F_{\text{friction}} \): \[ F_{\text{friction}} = \frac{295}{0.350} \]
04

Calculate the Exact Value

Perform the calculation: \[ F_{\text{friction}} = \frac{295}{0.350} \approx 842.86 \, \text{N} \] This is the magnitude of the static frictional force.

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 91影视!

Key Concepts

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

Static Friction
Static friction is the force that prevents two surfaces from sliding past each other. In our exercise, this is the force between the car tire and the road. Think about how your shoes grip the ground. That's static friction at play, stopping you from slipping.
When a wheel does not slip, the static friction is what keeps it rolling smoothly. For the car's wheel to move without skidding, the road must apply an equal and opposite force to the direction of the torque. This counteracts the wheel's tendency to rotate due to the engine's torque.
In our scenario, we calculated this static frictional force to have a magnitude of approximately 842.86 Newtons. It's crucial because it ensures the car moves without slipping and maintains control. Remember, static friction is always acting when there's no relative motion between surfaces.
Constant Velocity
Constant velocity means that an object moves at a steady speed and in a straight line without changing. In physics, this implies two things: no acceleration and balanced forces.
The car wheel in the problem has constant velocity. This means the forces acting on it, such as torques, are in perfect balance. The torque from the engine is matched by the static friction torque from the road, resulting in no net torque. Thus, the car does not speed up or slow down.
This situation is vital for understanding motion. When objects move at a constant velocity, it's because the driving force (engine torque) and the opposing forces (static friction torque) cancel each other out. Simple, yet powerful!
Rotational Force
Rotational force, or torque, is what causes an object to rotate around an axis. It's similar to how you might open a door or turn a wrench.
In our exercise, the engine exerts a rotational force (torque) on the car's wheel. This is calculated by multiplying the force by the wheel's radius. For our car, this torque is 295 Newton-meters, applied to help the wheel rotate properly.
Torque is essential in this context because it helps us determine how the wheel moves rotationally. It's balanced by the static friction torque to maintain the car's constant velocity. Torque plays a pivotal role in many systems, including in engines, motors, and even in simple tools like levers.

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 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 tolcrate 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 thin, rigid, uniform rod has a mass of \(2.00 \mathrm{kg}\) and a length of \(2.00 \mathrm{m}\). (a) Find the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end. (b) Suppose all the mass of the rod were located at a single point. Determine the perpendicular distance of this point from the axis in part (a), such that this point particle has the same moment of inertia as the rod does. This distance is called the radius of gyration of the rod.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh \(438 \mathrm{N}\) and have a center of gravity that is \(1.28 \mathrm{m}\) above the floor. His upper legs weigh \(144 \mathrm{N}\) and have a center of gravity that is \(0.760 \mathrm{m}\) above the floor. Finally, his lower legs and feet together weigh \(87 \mathrm{N}\) and have a center of gravity that is \(0.250 \mathrm{m}\) above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Conceptual Example 13 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2}\). When one person is standing on the carousel at a distance of \(1.50 \mathrm{m}\) from the center, the carousel has an angular velocity of 0.600 rad/s. However, as this person moves inward to a point located \(0.750 \mathrm{m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s} .\) What is the person's mass?

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a $$30.0^{\circ}$$ angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

Read Explanation

Quantum Physics

Read Explanation

Work Energy and Power

Read Explanation

Kinematics Physics

Read Explanation
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.