/*! 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 The standard test to determine t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200 -ft-diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If this maximum speed is \(35 \mathrm{mi} / \mathrm{hr}\) for a \(3000-\mathrm{lb}\) car, determine its lateral acceleration capability \(a_{n}\) in \(g\) 's and compute the magnitude \(F\) of the total friction force exerted by the pavement on the car tires.

Short Answer

Expert verified
The car's lateral acceleration capability is approximately 0.819 g's, and the total frictional force is 2455.4 lb.

Step by step solution

01

Convert Speed Units

First, we need to convert the speed from miles per hour to feet per second, which is a more suitable unit for calculating acceleration. The conversion factor is 1 mile per hour is approximately 1.467 feet per second.\[v = 35 \text{ mi/hr} \times 1.467 \text{ ft/s per mi/hr} = 51.345 \text{ ft/s}\]
02

Calculate Radius of Circular Path

Given the diameter of the circle is 200 ft, we can calculate the radius of the circular path. The radius is half of the diameter.\[r = \frac{200}{2} = 100 \text{ ft}\]
03

Compute Lateral Acceleration

The formula for lateral (centripetal) acceleration \(a_n\) in terms of speed \(v\) and radius \(r\) is:\[a_n = \frac{v^2}{r}\]Substituting the known values:\[a_n = \frac{(51.345)^2}{100} \approx 26.375 \text{ ft/s}^2\]
04

Convert Acceleration to g's

To express the lateral acceleration in terms of \(g\)'s (where \(g \approx 32.2 \text{ ft/s}^2\)), we divide the acceleration by \(g\).\[a_n \text{ in g's} = \frac{26.375}{32.2} \approx 0.819 \text{ g's}\]
05

Compute Frictional Force

The frictional force \(F\) exerted by the pavement is equal to the centripetal acceleration force required to keep the car moving in a circle. This can be expressed as:\[F = m \cdot a_n\]First, convert the car's weight from pounds to mass. Weight \(W = mg\), so:\[m = \frac{W}{g} = \frac{3000 \text{ lb}}{32.2 \text{ ft/s}^2} \approx 93.17 \text{ slugs}\]Now compute the force:\[F = 93.17 \text{ slugs} \times 26.375 \text{ ft/s}^2 \approx 2455.4 \text{ lb}\]

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.

Circular Motion
Circular motion occurs when an object moves along a curved path or a circle. This means that the direction of the object's velocity is constantly changing, even if its speed remains constant. Circular motion requires a force to keep the object moving in the curved path. In our exercise, when a car goes around a circular track, it undergoes circular motion. The necessary force to keep the car on this path is called the centripetal force. This force acts towards the center of the circle and is crucial in maintaining the motion. Without this force, the car would move off in a straight line due to inertia.
Friction Force
The friction force is a resistive force that acts between surfaces in contact. It prevents two objects from sliding past each other easily. In circular motion, friction from the pavement provides the centripetal force required to keep the car moving along the curved path. The maximum speed the car can maintain without skidding off depends on the friction force between the tires and the pavement. In the context of our car on the circular track, the pavement provides the friction force necessary to keep the car straddling the line while increasing speed until reaching a point where friction is no longer sufficient, and the car starts to skid.
Unit Conversion
Unit conversion is sometimes necessary in physics to apply formulas correctly or to understand measurements in desired units. In our problem, we convert the car's speed from miles per hour to feet per second. This conversion is crucial because the standard unit for calculating acceleration in the context of this exercise is feet per second squared. To make these conversions, we use the conversion factor, where 1 mile per hour is approximately 1.467 feet per second. Proper unit conversion ensures that calculations are accurate and conform to the units required for the physical formulas used, such as calculating centripetal acceleration.
Centripetal Acceleration
Centripetal acceleration is the acceleration directed toward the center of a circular path. It is essential for maintaining an object's circular motion and is determined by the velocity of the object and the radius of the circle. The formula for centripetal acceleration is given by: \[ a_c = \frac{v^2}{r} \]Where \( v \) is the linear velocity and \( r \) is the radius of the circular path. For the car driving around a circle, this acceleration is what allows the car to change direction while moving at a high speed. Expressing centripetal acceleration in terms of \( g \)'s helps relate the force experienced due to gravity, providing an intuitive understanding of how strong the acceleration is compared to gravity's pull. In this exercise, the car's performance around a track is gauged by how much force it can generate laterally relative to gravity, thus calculated in \( g \) units.

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

In a railroad classification yard, a 68 -Mg freight car moving at \(0.5 \mathrm{m} / \mathrm{s}\) at \(A\) encounters a retarder section of track at \(B\) which exerts a retarding force of \(32 \mathrm{kN}\) on the car in the direction opposite to \(\mathrm{mo}\) tion. Over what distance \(x\) should the retarder be activated in order to limit the speed of the car to \(3 \mathrm{m} / \mathrm{s}\) at \(C ?\)

The slider of mass \(m_{1}=0.4 \mathrm{kg}\) moves along the smooth support surface with velocity \(v_{1}=5 \mathrm{m} / \mathrm{s}\) when in the position shown. After negotiating the curved portion, it moves onto the inclined face of an initially stationary block of mass \(m_{2}=2 \mathrm{kg}\) The coefficient of kinetic friction between the slider and the block is \(\mu_{k}=0.30 .\) Determine the velocity \(v^{\prime}\) of the system after the slider has come to rest relative to the block. Neglect friction at the small wheels, and neglect any effects associated with the transition.

The system is at rest with the spring unstretched when \(\theta=0 .\) The 3 -kg particle is then given a slight nudge to the right. \((a)\) If the system comes to momentary rest at \(\theta=40^{\circ},\) determine the spring constant \(k .(b)\) For the value \(k=100 \mathrm{N} / \mathrm{m},\) find the speed of the particle when \(\theta=25^{\circ} .\) Use the value \(b=0.40 \mathrm{m}\) throughout and neglect friction.

The 0.1 -lb projectile \(A\) is subjected to a drag force of magnitude \(k v^{2},\) where the constant \(k=\) \(0.0002 \mathrm{lb}-\mathrm{sec}^{2} / \mathrm{ft}^{2} .\) This drag force always opposes the velocity \(\mathbf{v} .\) At the instant depicted, \(v=100 \mathrm{ft} / \mathrm{sec}\) \(\theta=45^{\circ},\) and \(r=400 \mathrm{ft} .\) Determine the corresponding values of \(\ddot{r}\) and \(\ddot{\theta}\).

An electromagnetic catapult system is being designed to replace a steam-driven system on an aircraft carrier. The requirements include accelerating a 12000 -kg aircraft from rest to a speed of \(70 \mathrm{m} / \mathrm{s}\) over a distance of \(90 \mathrm{m} .\) What constant force \(F\) must the catapult exert on the aircraft?

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.