/*! 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 87 ssM A car weighing \(10.7 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

ssM A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m}\). (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is \(0.350\), is the attempt at taking the curve successful?

Short Answer

Expert verified
The required frictional force is 3194.75 N, and yes, the attempt is successful.

Step by step solution

01

Determine the Centripetal Force Requirement

To keep a car moving in a circular path, it needs a centripetal force. The formula for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the car, \( v \) is its velocity, and \( r \) is the radius of the curve. First, convert the car's weight from kilonewtons to mass in kilograms:\[m = \frac{10.7 \times 10^3}{9.8} \approx 1091.84 \text{ kg}\]Now, calculate the required centripetal force:\[F_c = \frac{1091.84 \times (13.4)^2}{61.0} \approx 3194.75 \text{ N}\]
02

Calculate the Maximum Static Friction Force

The maximum static friction force, which can act as the centripetal force, is calculated using the coefficient of static friction \( \mu_s \) and the normal force. The formula is \( F_{s, \text{max}} = \mu_s \cdot N \), where \( N \) is the normal force, equivalent to the weight of the car.\[F_{s, \text{max}} = 0.350 \times 10.7 \times 10^3 \approx 3745 \text{ N}\]
03

Compare Frictional Force Requirements

To determine if the car can safely negotiate the curve, compare the required centripetal force with the maximum static friction force. If the maximum static friction is greater than or equal to the required centripetal force, the car can safely make the turn.Since the maximum static friction force (\( 3745 \text{ N} \)) is greater than the required centripetal force (\( 3194.75 \text{ N} \)), the car can safely negotiate the curve.

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.

Centripetal Force
When an object travels in a circular path, a force is required to continuously change its direction. This force is known as centripetal force. It acts towards the center of the circular path and is responsible for keeping the object moving along the curve without veering off.
For a car rounding a curve, the centripetal force is determined by the car's mass, the speed at which it is traveling, and the radius of the curve. The formula for calculating this force is given by:
  • \( F_c = \frac{mv^2}{r} \)
Here, \( m \) represents the mass of the car, \( v \) is the velocity, and \( r \) is the radius of the curve. In the given problem, the centripetal force needed to keep the car on its path was calculated to be approximately 3194.75 N, using the car's weight and speed.
Understanding the role of centripetal force is crucial in physics problem solving, especially in scenarios involving circular motion.
Static Friction
Friction is the resistance that one surface encounters when moving over another. When it comes to circular motion, static friction plays a key role. Static friction is the force that keeps an object at rest until a certain threshold force is applied.
For a car on a curve, the static friction between the tires and the road provides the centripetal force needed for circular motion. This force is calculated using the coefficient of static friction (\( \mu_s \)) and the normal force (\( N \)):
  • \( F_{s, \text{max}} = \mu_s \cdot N \)
In the problem, the maximum static friction force is 3745 N, which is enough to meet the required centripetal force needed to keep the car on the road. This means that the car can safely travel around the curve without slipping.
Static friction is essential in preventing sliding and ensuring safe handling of vehicles during turns.
Circular Motion
In physics, circular motion refers to the movement of an object along the circumference of a circle or a circular path. This type of motion can be either uniform or non-uniform, depending on whether the speed is constant or varies.
Circular motion requires a constant change in direction, which is facilitated by the centripetal force that acts inward towards the center of the circular path. For a car traveling around a bend, maintaining circular motion ensures that all parts of the vehicle remain on the intended path.
  • Uniform circular motion: Speed is constant as the car moves around the curve.
  • Non-uniform circular motion: Speed changes as the car goes around the bend.
The calculation of the necessary forces for maintaining circular motion is essential for designing safe curves on roads and tracks. In this problem, determining the right balance of forces made it possible for the car to move smoothly around the curve, maintaining a safe path.
Normal Force
The normal force is a support force exerted perpendicular to the surface an object rests on. It is one of the crucial forces at play in this physics problem, especially when calculating other forces like static friction.
In a situation with a car and a curve, the normal force is equal to the weight of the car, assuming the road is flat. This force acts upward, balancing the downward gravitational force. The normal force can be calculated as:
  • \( N = mg \)
Where \( m \) is the mass of the vehicle and \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²).
In the context of this exercise, the normal force directly influences the maximum static friction available to keep the car from slipping. By effectively utilizing the normal force, engineers can calculate safe speeds and curve radii for vehicles. Understanding normal force helps demystify the interactions between objects and surfaces in everyday physics problems.

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

\( \Rightarrow E\) In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. (a) Suppose the slope angle is \(\theta=40.0^{\circ}\), the snow is dry snow with a coefficient of kinetic friction \(\mu_{k}=0.0400\), the mass of the skier and equipment is \(m=85.0 \mathrm{~kg}\), the cross-sectional area of the (tucked) skier is \(A=1.30 \mathrm{~m}^{2}\), the drag coefficient is \(C=0.150\), and the air density is \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the terminal speed? (b) If a skier can vary \(C\) by a slight amount \(d C\) by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed?

A \(100 \mathrm{~N}\) force, directed at an angle \(\theta\) above a horizontal floor, is applied to a \(25.0 \mathrm{~kg}\) chair sitting on the floor. If \(\theta=0^{\circ}\), what are (a) the horizontal component \(F_{h}\) of the applied force and (b) the magnitude \(F_{N}\) of the normal force of the floor on the chair? If \(\theta=30.0^{\circ}\), what are (c) \(F_{h}\) and \((\) d \() F_{N}\) ? If \(\theta=60.0^{\circ}\), what are (e) \(F_{h}\) and (f) \(F_{N}\) ? Now assume that the coefficient of static friction between chair and floor is \(0.420\). Does the chair slide or remain at rest if \(\theta\) is \((\mathrm{g}) 0^{\circ},(\mathrm{h}) 30.0^{\circ}\), and (i) \(60.0^{\circ} ?\)

Calculate the magnitude of the drag force on a missile \(53 \mathrm{~cm}\) in diameter cruising at \(250 \mathrm{~m} / \mathrm{s}\) at low altitude, where the density of air is \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(C=0.75\).

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

A locomotive accelerates a 25 -car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v\), where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is \(30 \mathrm{~km} / \mathrm{h}\), the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h}\) ?

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.