/*! 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 56 A banked circular highway curve ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A banked circular highway curve is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). The radius of the curve is \(200 \mathrm{~m}\). Traffic is moving along the highway at \(40 \mathrm{~km} / \mathrm{h}\) on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Short Answer

Expert verified
The minimum coefficient of friction needed is approximately 0.103.

Step by step solution

01

Convert Speeds to Meters per Second

First, we convert the speeds from kilometers per hour to meters per second. The design speed is given as \(60 \, \text{km/h}\).\[60 \, \text{km/h} = \frac{60 \times 1000}{3600} \, \text{m/s} = 16.67 \, \text{m/s}\]The actual speed on a rainy day is \(40 \, \text{km/h}\).\[40 \, \text{km/h} = \frac{40 \times 1000}{3600} \, \text{m/s} = 11.11 \, \text{m/s}\]
02

Understand the Forces at Play

When the car takes the curve, several forces are at play: the gravitational force acting downwards, the normal force perpendicular to the road surface, and the friction force. The centripetal force required to keep the car moving in a circle is provided by the horizontal component of the normal force and the friction.
03

Calculate the Banking Angle with Frictionless Design Speed

Using the design speed, calculate the banking angle \(\theta\) without considering friction. The centripetal force formula for a banked curve without friction is:\[\tan(\theta) = \frac{v^2}{rg}\]Substitute the values:\[v = 16.67 \, \text{m/s}, \, r = 200 \, \text{m}, \, g = 9.8 \, \text{m/s}^2\]\[\tan(\theta) = \frac{(16.67)^2}{200 \times 9.8} = 0.141\]\[\theta = \arctan(0.141) \approx 8.05^\circ\]
04

Calculate Net Horizontal Force Requirement

For the actual speed of \(11.11 \, \text{m/s}\), calculate the horizontal force needed (centripetal force):\[F_{\text{centripetal}} = \frac{mv'^2}{r}\]Where \(m\) is the mass of the car. But we can leave \(m\) as a variable because it will cancel out later.
05

Calculate the Frictional Force Needed

The differences in forces due to speed lowering are compensated by friction. The equation relating them is:\[\frac{mv'^2}{r} = mg(\mu + \tan(\theta)) - mv'g\tan(\theta)\]The friction force \(f = \mu mg\), where \(\mu\) is the minimum coefficient of friction. Rearranging leads to solving for \(\mu\):\[\mu = \frac{v'^2/g + \tan(\theta)(r^{-1}v'^2 - g)}{g + r^{-1}v'^2}\]Substitute values to find \(\mu\):\[\mu = \frac{(11.11)^2/9.8 + 0.141 ((11.11)^2/200 - 9.8)}{9.8 + (11.11)^2/200} \approx 0.103\]
06

Conclusion

The calculated minimum coefficient of friction \(\mu = 0.103\) indicates the road surface needs to be sufficiently frictional to allow safe passage of cars during rain conditions without sliding off the banked 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
Centripetal force is a fascinating concept that keeps objects moving in a circular path. When it comes to driving around a curved path, say a highway turn, the centripetal force is what keeps your car from veering off the road. It's always directed towards the center of the circle or curve you're moving around.

This force doesn't come from just anywhere. When a car negotiates a banked curve, the centripetal force needed is primarily provided by a combination of gravitational forces and frictional forces. In mathematical terms, the centripetal force (\( F_{\text{centripetal}} = \frac{mv^2}{r} \)) requires a proper balance of these components to keep vehicles safely on the inclined road.
  • Mass (\( m \)): Represents the mass of the vehicle; it's crucial in determining the centripetal force but often cancels out when considering force equations related to banking angle and friction.
  • Velocity (\( v \)): This is the speed of the car along the curve. Speed adjustments alter the required centripetal force significantly.
  • Radius of Curve (\( r \)): A longer radius means a gentler curve, requiring less centripetal force to maintain the path.
Frictional Force
Frictional force plays a critical role in vehicle dynamics, especially on curves. It provides the necessary grip between the car tires and the road surface, preventing the car from sliding off. On a banked curve, friction acts in two possible directions: it can enhance stability when the car speed is too low or help counteract the banking effects at higher speeds.

Friction is quantified by the coefficient of friction (\( \mu \)
  • Static Friction: This is the frictional force that prevents sliding without any relative motion. It's vital for starting and stopping movement around a curve.
  • Kinetic Friction: Acts when a vehicle slides; less than static friction and usually not desirable in terms of control and safety.
  • Banking Angle
    The banking angle is the tilt of the road surface from the horizontal. Engineers design banking angles to aid vehicles in maintaining path stability without relying heavily on friction. Think of it as the road helping a car "lean" into a turn, just as cyclists do when they bank around corners.

    The design of a banked curve aims to use gravitational and normal forces for centripetal force, minimizing friction's role. When calculated without friction present, the formula \( \tan(\theta) = \frac{v^2}{rg} \) describes the ideal banking angle \( \theta \) that a vehicle should have at a designated speed for safekeeping.
    • Optimum Speed: The banked angle is reliant on the assumption that vehicles maintain a certain design speed.
    • Combination of Forces: At the correct speed, normal force and gravity naturally combine to provide the necessary centripetal force.
    Coefficient of Friction
    The coefficient of friction (\( \mu \)) is an essential parameter that indicates how much stickiness or grip exists between two surfaces, such as car tires and road pavement. In situations like wet weather, the coefficient of friction becomes a crucial safety factor.

    In a banked curve, you calculate the necessary coefficient of friction when design parameters change, like lower speeds on rainy days. This value ensures vehicles can still negotiate the curve without slipping off due to insufficient grip. Typically, you find \( \mu \) using equations that balance potential forces of motion and friction, such as:\[ \mu = \frac{v'^2/g + \tan(\theta)(r^{-1}v'^2 - g)}{g + r^{-1}v'^2} \]
    • Road Conditions: Weather, such as rain, can drastically reduce the coefficient of friction, requiring adjustments in driving behavior.
    • Safety Margins: A sufficient coefficient of friction is necessary to provide a safety cushion against unexpected speed or direction changes.

    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

    The mysterious sliding stones. Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-18). For years curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard- baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about \(0.80 .\) What horizontal force must act on a \(20 \mathrm{~kg}\) stone (a typical mass) to maintain the stone's motion once a gust has started it moving? (Story continues with Problem \(37 .)\)

    A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

    A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\) (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

    You testify as an expert witness in a case involving an accident in which car \(A\) slid into the rear of car \(B,\) which was stopped at a red light along a road headed down a hill (Fig. 6 -25). You find that the slope of the hill is \(\theta=12.0^{\circ},\) that the cars were separated by distance \(d=24.0 \mathrm{~m}\) when the driver of car \(A\) put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car \(A\) at the onset of braking was \(v_{0}=18.0 \mathrm{~m} / \mathrm{s} .\) With what speed did car \(A\) hit car \(B\) if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)?

    A sling-thrower puts a stone \((0.250 \mathrm{~kg})\) in the sling's pouch \((0.010 \mathrm{~kg})\) and then begins to make the stone and pouch move in a vertical circle of radius \(0.650 \mathrm{~m}\). The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is \(33.0 \mathrm{~N}\) or more. Suppose the slingthrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?

    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.