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A flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of static friction that will prevent sliding? (b) Suppose that the highway is icy and the coefficient of static friction between the tires and pavement is only one- third of what you found in part (a). What should be the maximum speed of the car so that it can round the curve safely?

Short Answer

Expert verified
(a) Minimum coefficient of static friction: 0.375. (b) Maximum speed on ice: 14.5 m/s.

Step by step solution

01

Understanding the Forces

To solve this problem, we need to consider that for a car to round a curve without sliding, the static frictional force must be at least equal to the required centripetal force to keep the car moving in a circle. This condition prevents sliding.
02

Calculate Required Centripetal Force

The centripetal force necessary to keep the car moving in a circle is given by the expression \( f_c = \frac{mv^2}{r} \), where \( m \) is the mass of the car, \( v \) is the velocity (25 m/s), and \( r \) is the radius of the curve (170 m). Although the mass isn’t specified, it will cancel out later.
03

Express Static Friction Force

The static frictional force \( f_s \) can be expressed as \( f_s = \mu_s F_n = \mu_s mg \), where \( \mu_s \) is the coefficient of static friction, \( F_n \) is the normal force, and \( mg \) is the weight of the car.
04

Set Up the Equation for Static Friction

For the car not to slide, the frictional force must equal the centripetal force: \( \mu_s mg = \frac{mv^2}{r} \). Cancelling \( m \) from both sides, we solve for \( \mu_s \): \( \mu_s = \frac{v^2}{rg} \).
05

Calculate Minimum Coefficient of Static Friction

Substitute \( v = 25 \text{ m/s} \), \( r = 170 \text{ m} \), and \( g = 9.8 \text{ m/s}^2 \) into the equation \( \mu_s = \frac{v^2}{rg} \): \( \mu_s = \frac{(25)^2}{170 \times 9.8} \approx 0.375 \).
06

Adjust for Reduced Coefficient

Given that the icy situation provides a coefficient of static friction that is one-third of \( 0.375 \), the effective \( \mu_{ice} = \frac{0.375}{3} = 0.125 \).
07

Calculate Maximum Safe Speed on Ice

To find the maximum speed \( v_{max} \) that the car can travel on ice, we use \( \mu_{ice} mg = \frac{mv_{max}^2}{r} \). Solving for \( v_{max} \), we get \( v_{max} = \sqrt{\mu_{ice} rg} \). Substitute \( \mu_{ice} = 0.125 \): \( v_{max} = \sqrt{0.125 \times 170 \times 9.8} \approx 14.5 \text{ m/s} \).

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Key Concepts

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

Centripetal Force
When a car makes a turn on a circular track, it requires a force to keep it moving along that curved path. This force, known as centripetal force, acts towards the center of the circle. It's responsible for changing the direction of the car's velocity, allowing it to continue in a circular motion.
To compute the centripetal force (\[ f_c \]), we use the formula \[ f_c = \frac{mv^2}{r} \]. In this equation, \( m \) represents the mass of the car, \( v \) is its speed, and \( r \) is the radius of the curve. Although the mass doesn't directly affect the determination of the required coefficient of friction, it's crucial to note that this force essentially compels the car to seek the center of the circle as it moves around the curve.
In the exercise, the centripetal force is counteracted by the static frictional force coming from the contact between the car's tires and the road. If the frictional force weren't present or sufficient, the car would fail to follow the curve and slide away from the circular path.
Coefficient of Friction
The coefficient of friction (\( \mu \)) is a value that quantifies the frictional forces between two surfaces. In this exercise, we are dealing with the static coefficient of friction, which applies when there isn’t relative motion between the contact surfaces yet.
The static frictional force (\( f_s \)) is expressed as \[ f_s = \mu_s F_n \], where \( F_n = mg \) is the normal force, or the perpendicular force that a surface exerts on an object. For the car not to slide off the curve, this static frictional force must be at least equal to the centripetal force. Thus, we have \( \mu_s mg = \frac{mv^2}{r} \), leading to \( \mu_s = \frac{v^2}{rg} \).
In scenarios where surfaces may be less conducive to friction, such as an icy road, the coefficient drops. Consequently, the speed at which the car can handle the turn safely decreases. This concept is crucial, not just for understanding physics problems, but also for real-world driving safety.
Circular Motion
Circular motion refers to the movement of an object along a circular path. In our problem, the car travels around a curved section of a highway. Maintaining this circular motion requires constant acceleration towards the center, known as centripetal acceleration. It simplifies to \( a_c = \frac{v^2}{r} \), resulting in the necessary centripetal force.
For real-world applications, the role of static friction in providing enough grip to allow vehicles to complete such curves cannot be overstated. The static friction must match or exceed the force needed to change the vehicle's straightforward velocity into a circular path.
  • Vehicles must adjust speed, as determined by \( v = \sqrt{\mu r g} \), to safely maneuver around centers with differing coefficients of friction.
  • The concept demonstrates the delicate balance necessary for achieving safe transit during turns.
By understanding the dynamics of circular motion, predictions can be made regarding how alterations in speed or surface condition will affect the path and safety of the transit. Such insights aid in designing safer roadways and vehicles.

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Most popular questions from this chapter

A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 m/s\(^2\)? (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.62 m/s\(^2\). (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

On September 8, 2004, the \(Genesis\) spacecraft crashed in the Utah desert because its parachute did not open. The 210-kg capsule hit the ground at 311 km/h and penetrated the soil to a depth of 81.0 cm. (a) What was its acceleration (in m/s\(^2\) and in g's), assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box's acceleration?

A 25.0-kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp?

A 70-kg person rides in a 30-kg cart moving at 12 m/s at the top of a hill that is in the shape of an arc of a circle with a radius of 40 m. (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart can travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

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