/*! 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 84 A \(1250-\mathrm{kg}\) car round... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1250-\mathrm{kg}\) car rounds a curve of radius \(72 \mathrm{~m}\) banked at an angle of \(14^{\circ} .\) If the car is traveling at \(85 \mathrm{~km} / \mathrm{h},\) will a friction force be required? If so, how much and in what direction?

Short Answer

Expert verified
Yes, a friction force of approx. 7284.43 N is required towards the center of the curve.

Step by step solution

01

Convert Speed Units

First, convert the car's speed from kilometers per hour to meters per second. The formula for this conversion is:\[ \text{Speed (m/s)} = \text{Speed (km/h)} \times \frac{1000 \text{ m}}{3600 \text{ s}} \]For a speed of \(85 \; km/h\):\[ \text{Speed (m/s)} = 85 \times \frac{1000}{3600} \approx 23.61 \; m/s \]
02

Write Down Given Data

List all the given information:- Mass of the car \(m = 1250 \, \text{kg}\)- Radius of the curve \(r = 72 \, \text{m}\)- Banking angle \(\theta = 14^\circ\)- Speed of the car \(v = 23.61 \, \text{m/s}\)
03

Calculate the Required Centripetal Force

The centripetal force required to keep the car moving in a circular path is given by:\[ F_c = \frac{mv^2}{r} \]Substitute the given values:\[ F_c = \frac{1250 \times (23.61)^2}{72} \approx 10320.8 \, N \]
04

Calculate the Components Contributing to Centripetal Force

The horizontal component of the normal force provides the required centripetal force. It is given by:\[ F_{n,h} = mg\tan\theta = 1250 \times 9.81 \times \tan(14^\circ) \approx 3036.37 \, N \]
05

Determine if Additional Friction is Needed

Compare the horizontal component of the normal force to the required centripetal force:- Required centripetal force: \( 10320.8 \, N \)- Available from banking: \( 3036.37 \, N \)Since the available centripetal force from banking is less than the required force, an additional frictional force is required.
06

Calculate the Friction Force

Calculate the additional frictional force needed:\[ F_{friction} = F_c - F_{n,h} = 10320.8 - 3036.37 = 7284.43 \, N \]This force is in the horizontal direction, pointing towards the center of 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
In the world of physics, centripetal force keeps an object moving in a circular path. It acts towards the center of the circle, maintaining the object's circular motion.
When you drive a car around a curved path, especially at a high speed, you are experiencing centripetal force. This force is crucial because, without it, your car would simply travel in a straight line off the road.
The formula for calculating the centripetal force, \( F_c \), is \( F_c = \frac{mv^2}{r} \), where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity, and
  • \( r \) is the radius of the circular path.
This equation helps us understand that as the speed of the car increases or as the radius of the turn decreases, the needed centripetal force becomes larger.
In our scenario, the centripetal force must be strong enough to keep the car on the curved path without sliding off.
Banked curves
When roads or tracks have banked curves, they utilize the angle to assist in managing centripetal force naturally. The banking angle, \( \theta \), helps vehicles make turns more smoothly and safely.
Banking is a clever engineering solution to reduce reliance on friction. The idea is simple: by inclining the road or track, the normal force—acting perpendicular to the surface—partially contributes to the centripetal force.
With banked curves, you have two components of force working together:
  • Vertical component which balances the gravitational force,
  • Horizontal component which helps provide the required centripetal force.
By engineering this balance, drivers can navigate curves more safely at higher speeds.
In our specific problem, the normal force's horizontal component was calculated as \( F_{n,h} = mg\tan\theta \). This component alone, however, was insufficient to keep the car safely on the curve, indicating that banked curves don't completely eliminate the need for friction.
Frictional force
Frictional force is that extra friend holding you on the road when banking alone cannot suffice. It acts between the tires and the road surface, playing a critical role in preventing skidding.
In curves where the banking angle doesn’t provide enough centripetal force, friction steps in to bridge the gap. It acts in the direction that helps keep the vehicle moving in its circular path.
For our example, even with a 14-degree banking angle, the horizontal force from banking was not sufficient. Therefore, the frictional force was calculated by subtracting the available banking force from the required centripetal force: \( F_{friction} = F_c - F_{n,h} \).
This resulted in an additional \( 7284.43 \; N \) force needed from friction, working to keep the car safely on track.
Friction, while often unnoticed, silently works to enhance safety in such driving scenarios, proving vital where engineered forces fall short.

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

(II) Police investigators, examining the scene of an accident involving two cars, measure 72 -m-long skid marks of one of the cars, which nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pavement is about \(0.80 .\) Estimate the initial speed of that car assuming a level road.

(III) A bicyclist can coast down a \(7.0^{\circ}\) hill at a steady \(9.5 \mathrm{~km} / \mathrm{h} .\) If the drag force is proportional to the square of the speed \(v,\) so that \(F_{\mathrm{D}}=-c v^{2},\) calculate \((a)\) the value of the constant \(c\) and \((b)\) the average force that must be applied in order to descend the hill at \(25 \mathrm{~km} / \mathrm{h}\). The mass of the cyclist plus bicycle is \(80.0 \mathrm{~kg} .\) Ignore other types of friction.

(II) An object moving vertically has \(\overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{v}}_{0}\) at \(t=0\). Determine a formula for its velocity as a function of time assuming a resistive force \(F=-b v\) as well as gravity for two cases: \((a) \overrightarrow{\mathbf{v}}_{0}\) is downward and \((b) \overrightarrow{\mathbf{v}}_{0}\) is upward.

(II) A bucket of mass \(2.00 \mathrm{~kg}\) is whirled in a vertical circle of radius \(1.10 \mathrm{~m} .\) At the lowest point of its motion the tension in the rope supporting the bucket is \(25.0 \mathrm{~N}\). \((a)\) Find the speed of the bucket. \((b)\) How fast must the bucket move at the top of the circle so that the rope does not go slack?

(III) Assume a net force \(F=-m g-k v^{2}\) acts during the upward vertical motion of a \(250-\mathrm{kg}\) rocket, starting at the moment \((t=0)\) when the fuel has burned out and the rocket has an upward speed of \(120 \mathrm{~m} / \mathrm{s}\). Let \(k=0.65 \mathrm{~kg} / \mathrm{m}\). Estimate \(v\) and \(y\) at 1.0 -s intervals for the upward motion only, and estimate the maximum height reached. Compare to free-flight conditions without air resistance \((k=0)\).

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.