Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 60

Find an expression for the minimum frictional coefficient needed to keep a car with speed \(v\) on a banked turn of radius \(R\) designed for speed \(v_{0}\)

Short Answer

Expert verified
The minimum friction coefficient required to keep the car on the turn is given by the following expression: \(\mu = \frac{v^2}{v_0^2} - \cos(\theta)\)

Step by step solution

01

Identify Forces

Initially, three forces act on the car: weight due to gravity acts downwards, the normal force from the surface of the banked road acts perpendicular to the surface, and friction acting parallel to the surface. Here, friction acts towards the center, helping the car take the turn.
02

Define Forces

The force due to gravity can be represented as \(mg\), where \(m\) is the mass of the car and \(g\) is the acceleration due to gravity. The force due to friction can be stated as \(f = \mu N\), where \(N\) is the normal force and \(\mu\) is the friction coefficient. The normal force \(N\) can be solved using these forces.
03

Balance of Forces

For the car to successfully navigate the turn without skidding, forces parallel and perpendicular to the incline must be balanced. This gives us two equations: \(-f + N \sin(\theta) = \frac{mv^2}{R}\) for the forces in the horizontal (parallel to the surface of the turn), and \(N \cos(\theta) = mg + f \cos(\theta)\) for forces in the vertical direction (perpendicular).
04

Substitute Equations

Substitute the equations from Step 2 into the equations from Step 3 and solve for the friction coefficient \(\mu\). We will observe that the mass \(m\) can be cancelled out, which leads to the conclusion that the minimum friction coefficient required does not depend on the mass of the car.
05

Finalize the Equation

\(\mu\) equals to \(\frac{v^2}{v_0^2}-\cos(\theta)\), where \(\theta\) is banking angle, \(v_0\) is speed this turn was designed for and \(v\) is actual speed.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 block is launched with speed \(v_{0}\) up a slope making an angle \(\theta\) with the horizontal; the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\) (a) Find an expression for the distance \(d\) the block travels along the slope. (b) Use calculus to determine the angle that minimizes \(d\).

A block is projected up an incline at angle \(\theta\). It returns to its initial position with half its initial speed. Show that the coefficient of kinetic friction is \(\mu_{\mathrm{k}}=\frac{3}{5} \tan \theta\)

Moving through a liquid, an object of mass \(m\) experiences a resistive drag force proportional to its velocity, \(F_{\text {drag }}=-b v,\) where \(b\) is a constant. (a) Find an expression for the object's speed as a function of time, when it starts from rest and falls vertically through the liquid. (b) Show that it reaches a terminal velocity \(m g / b\)

In the process of mitosis (cell division), two motor proteins pull on a spindle pole, each with a 7.3 -pN force. The two force vectors make a \(65^{\circ}\) angle. What's the magnitude of the force the two motor proteins exert on the spindle pole?

Show that the force needed to keep a mass \(m\) in a circular path of radius \(r\) with period \(T\) is \(4 \pi^{2} \mathrm{mr} / T^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Measurements

Read Explanation

Famous Physicists

Read Explanation

Force

Read Explanation

Magnetism

Read Explanation

Electricity and Magnetism

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.

Sign-up for free