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Chapter 6: Problem 92

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h} .\) Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m},\) what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
(a) 10.7掳; (b) 0.188.

Step by step solution

01

Convert Speed to Meters per Second

The speed is given in kilometers per hour. First, convert it to meters per second using the conversion factor: \[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} \]So, \[ 60 \, \text{km/h} = 60 \times \frac{1000}{3600} \, \text{m/s} = 16.67 \, \text{m/s} \]
02

Determine the Banking Angle

For the banking angle \( \theta \) with no friction, use the formula:\[ \tan(\theta) = \frac{v^2}{r g} \]Where \( v = 16.67 \, \text{m/s} \), \( r = 150 \, \text{m} \), and \( g = 9.81 \, \text{m/s}^2 \). Substitute the values:\[ \tan(\theta) = \frac{(16.67)^2}{150 \times 9.81} = 0.188 \]Find \( \theta \) by taking the arctan:\[ \theta = \arctan(0.188) \approx 10.7^\circ \]
03

Calculate the Minimum Coefficient of Friction

If the road is not banked, the frictional force must provide the necessary centripetal force. Use the equation:\[ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} \]Where \( \mu \) is the coefficient of friction. Solving for \( \mu \), we have:\[ \mu = \frac{v^2}{r \cdot g} \]Substitute the values:\[ \mu = \frac{(16.67)^2}{150 \times 9.81} = 0.188 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal force is the force that keeps an object moving in a curved path. It acts towards the center of the circle. Imagine you're driving around a curve. The centripetal force prevents your car from sliding out of the curve. It's provided by the friction between the tires and the road, as well as any other forces like banking in our exercise.

The formula to calculate centripetal force (\( F_c \)) is:
  • \[ F_c = \frac{m \cdot v^2}{r} \]
  • \( m \) is the mass of the vehicle.
  • \( v \) is the velocity (speed).
  • \( r \) is the radius of the path.
As speed increases, more centripetal force is needed. If friction is not enough, the car could skid off the road. That's where banking angles and friction coefficients come into play.
Coefficient of Friction
The coefficient of friction, \( \mu \), is a measure of how much frictional force can be generated before an object begins to slide. In the context of curved roads, it indicates how well tires can grip the road surface.

When a car goes around a curve, the friction between the road and tires acts as the centripetal force. The formula used to find the minimum coefficient of friction necessary is:
  • \[ \mu = \frac{v^2}{r \cdot g} \]
  • Here, \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
  • This shows \( \mu \) depends on speed (\( v \)), radius of the curve (\( r \)), and gravity.
Higher speed and smaller radius need more friction to prevent skidding. This is why careful consideration of \( \mu \) is vital for road safety, especially on curves.
Angle of Banking
The angle of banking is the angle at which the road is inclined to help vehicles stay on the curve without relying solely on friction. Proper banking allows safe navigation at certain speeds. It counteracts the outward force experienced by a vehicle on a curve.

To calculate the correct angle of banking (\( \theta \)), without requiring friction, use the formula:
  • \[ \tan(\theta) = \frac{v^2}{r \cdot g} \]
  • \( v \) is the velocity of the vehicle.
  • \( r \) is the radius of the curve.
  • \( g \) is the acceleration due to gravity.
Banking at an angle helps distribute some of the centripetal force components to prevent skidding, even if the road is wet or icy. The calculated angle for the given exercise is approximately \( 10.7^\circ \), illustrating how subtle inclines can improve vehicle stability on curves.

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Most popular questions from this chapter

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the \(3.5 \mathrm{~kg}\) book is pushed from rest through a distance of \(0.90 \mathrm{~m}\) by the horizontal \(25 \mathrm{~N}\) force from the broom and then has a speed of \(1.60 \mathrm{~m} / \mathrm{s},\) what is the coefficient of kinetic friction between the book and floor?

Brake or turn? Figure 6-44 depicts an overhead view of a car's path as the car travels toward a wall. Assume that the driver begins to brake the car when the distance to the wall is \(d=107 \mathrm{~m},\) and take the car's mass as \(m=1400 \mathrm{~kg},\) its initial speed as \(v_{0}=35 \mathrm{~m} / \mathrm{s},\) and the coefficient of static friction as \(\mu_{s}=0.50 .\) Assume that the car's weight is distributed evenly on the four wheels, even during braking. (a) What magnitude of static friction is needed (between tires and road) to stop the car just as it reaches the wall? (b) What is the maximum possible static friction \(f_{s, \max } ?\) (c) If the coefficient of kinetic friction between the (sliding) tires and the road is \(\mu_{k}=0.40,\) at what speed will the car hit the wall? To avoid the crash, a driver could elect to turn the car so that it just barely misses the wall, as shown in the figure. (d) What magnitude of frictional force would be required to keep the car in a circular path of radius \(d\) and at the given speed \(v_{0}\), so that the car moves in a quarter circle and then parallel to the wall? (e) Is the required force less than \(f_{s, \max }\) so that a circular path is possible?

A bolt is threaded onto one end of a thin horizontal rod, and the rod is then rotated horizontally about its other end. An engineer monitors the motion by flashing a strobe lamp onto the rod and bolt, adjusting the strobe rate until the bolt appears to be in the same eight places during each full rotation of the rod (Fig. \(6-42\) ). The strobe rate is 2000 flashes per second; the bolt has mass \(30 \mathrm{~g}\) and is at radius \(3.5 \mathrm{~cm} .\) What is the magnitude of the force on the bolt from the rod?

In Fig. \(6-39,\) a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is \(0 .\) The driver's mass is \(70.0 \mathrm{~kg}\). What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

A circular-motion addict of mass \(80 \mathrm{~kg}\) rides a Ferris wheel around in a vertical circle of radius \(10 \mathrm{~m}\) at a constant speed of \(6.1 \mathrm{~m} / \mathrm{s}\). (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?
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