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The steering wheel of a car has a radius of 0.19 \(\mathrm{m},\) and the steering wheel of a truck has a radius of \(0.25 \mathrm{m}\). The same force is applied in the same direction to each steering wheel. What is the ratio of the torque produced by this force in the truck to the torque produced in the car?

Short Answer

Expert verified
The ratio of the torque in the truck to the torque in the car is approximately 1.316.

Step by step solution

01

Identify the Given Data

We know the radius of the car's steering wheel is \(0.19 \, \mathrm{m}\), and the radius of the truck's steering wheel is \(0.25 \, \mathrm{m}\). The same force, denoted as \(F\), is applied to both steering wheels.
02

Recall the Torque Formula

Torque (\(\tau\)) is calculated using the formula \(\tau = F \times r\), where \(F\) is the force applied and \(r\) is the radius of the steering wheel.
03

Calculate Torque for the Car

For the car's steering wheel, the torque \(\tau_\text{car}\) is \(F \times 0.19\).
04

Calculate Torque for the Truck

For the truck's steering wheel, the torque \(\tau_\text{truck}\) is \(F \times 0.25\).
05

Calculate the Ratio of Torques

The ratio of the torque produced on the truck to the torque produced on the car is:\[\text{Ratio} = \frac{\tau_\text{truck}}{\tau_\text{car}} = \frac{F \times 0.25}{F \times 0.19} = \frac{0.25}{0.19}.\]Here, the force \(F\) cancels out because it is the same in both cases.
06

Simplify the Ratio

Calculate \(\frac{0.25}{0.19}\) using division to find the numerical ratio. The ratio is approximately \(1.316\).

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Key Concepts

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

Understanding Steering Wheel Radius
The radius of a steering wheel is the distance from the center of the wheel to its outer edge. In the context of vehicles, different types of steering wheels have varied radii due to the different needs of the vehicles.
For instance, a car might have a steering wheel with a smaller radius compared to a truck, as seen in the example where a car's steering wheel has a radius of 0.19 meters, while a truck's has a larger radius of 0.25 meters.
The steering wheel radius plays a crucial role in determining the torque produced when a force is applied. Generally, the larger the radius, the greater the torque for the same amount of force. This relationship explains why trucks, which may require more turning force, are equipped with larger steering wheels than most cars.
The Notion of Force Application
Force application involves how and where a force is exerted onto an object. When applying a force to a steering wheel, it is crucial to understand both the magnitude and the direction of the force. In physics problems like this one, we assume the force is applied perpendicular to the radius to maximize effectiveness.
This perpendicular application is essential because it makes the calculation of torque straightforward, as torque is a product of the radius and the perpendicular force. Understanding how force is applied helps in analyzing the mechanical advantage - or output effectiveness - when turning steering wheels with different radii under the same force conditions.
Exploring Torque Ratio
Knowing about torque ratio is essential when comparing the rotational effectiveness of two systems. In simple terms, torque measures the turning force on an object, calculated as the product of the radius and the force applied.
When the exercise asks for the ratio of the torque produced by a truck's steering wheel to that of a car, it refers to comparing these products for each vehicle:
  • Truck torque: \( au_{\text{truck}} = F \times 0.25\)
  • Car torque: \(\tau_{\text{car}} = F \times 0.19\)
By simplifying the division of these two torques, the force cancels out, leaving the ratio of their radii alone, \(\frac{0.25}{0.19}\). This approach highlights how differences in steering wheel size can affect the torque and thus the ease of steering.
Physics Problem Solving Techniques
Solving physics problems systematically involves several critical steps that help in analyzing even the most complex scenarios. Here’s a straightforward approach you can use:
  • Begin by identifying all given information and writing it down for clarity.
  • Recall relevant formulas, like the one for torque \(\tau = F \times r\), ensuring you understand how each component interacts.
  • Break down the problem into steps, calculate separately for each, and ensure your units are consistent throughout.
  • Finally, compare, contrast, and simplify where necessary, as shown using the torque ratio where division simplifies the comparison to a simple ratio of radii.
By following these steps, complex exercises become more manageable, leading to logical solutions and a deeper understanding of physical interactions.

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Most popular questions from this chapter

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

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?

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius of \(4.0 \mathrm{kg}\) and \(0.35 \mathrm{m}\), respectively. One has the shape of a hoop and the other the shape of a solid disk. The wheels start from rest and have a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each turns through an angle of 13 rad in 8.0 s. Find the net external torque that acts on each wheel.

The drawing shows a rectangular piece of wood. The forces applied to corners \(\mathrm{B}\) and \(\mathrm{D}\) have the same magnitude of \(12 \mathrm{N}\) and are directed parallel to the long and short sides of the rectangle. The long side of the rectangle is twice as long as the short side. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. A third force (not shown in the drawing) is applied to corner A, directed along the short side of the rectangle (either toward \(\mathrm{B}\) or away from \(\mathrm{B}\) ), such that the piece of wood is at equilibrium. Find the magnitude and direction of the force applied to corner A.

The wheel of a car has a radius of \(0.350 \mathrm{m}\). The engine of the car applies a torque of \(295 \mathrm{N} \cdot \mathrm{m}\) 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 countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

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