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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 5, Problem 48a

A flat (unbanked) curve on a highway has a radius of 170.0170.0 m. A car rounds the curve at a speed of 25.025.0 m/s. What is the minimum coefficient of static friction that will prevent sliding?

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Identify the forces acting on the car: The car is moving in a circular path, so the centripetal force is required to keep it on the curve. This force is provided by the static friction between the tires and the road. The forces involved are the gravitational force, the normal force, and the static frictional force.
Write the equation for centripetal force: The centripetal force is given by \( F_c = \frac{m v^2}{r} \), where \( m \) is the mass of the car, \( v \) is the speed of the car, and \( r \) is the radius of the curve. Here, \( v = 25.0 \ \text{m/s} \) and \( r = 170.0 \ \text{m} \).
Relate the static frictional force to the centripetal force: The static frictional force \( F_f \) must equal the centripetal force to prevent the car from sliding. Therefore, \( F_f = \frac{m v^2}{r} \). The static frictional force is also given by \( F_f = \mu_s F_N \), where \( \mu_s \) is the coefficient of static friction and \( F_N \) is the normal force.
Determine the normal force: On a flat (unbanked) curve, the normal force \( F_N \) is equal to the gravitational force \( F_g \), which is \( F_N = m g \), where \( g = 9.8 \ \text{m/s}^2 \). Substitute this into the frictional force equation: \( \mu_s m g = \frac{m v^2}{r} \).
Solve for the coefficient of static friction \( \mu_s \): Cancel out \( m \) from both sides of the equation (since it appears in every term), and rearrange to find \( \mu_s = \frac{v^2}{r g} \). Substitute the known values of \( v \), \( r \), and \( g \) to calculate \( \mu_s \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. For a car rounding a curve, this force is provided by the friction between the tires and the road. The formula for centripetal force is F_c = (mv^2)/r, where m is the mass of the car, v is its speed, and r is the radius of the curve.
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Friction and Coefficient of Friction

Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of static friction (μ_s) quantifies the frictional force before sliding occurs. It is defined as the ratio of the maximum static frictional force to the normal force acting on the object. To prevent sliding, the frictional force must equal or exceed the required centripetal force.
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Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). In the context of a car rounding a curve, this law helps to relate the forces acting on the car, including the centripetal force needed for circular motion and the frictional force that provides this centripetal acceleration.
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Related Practice
Textbook Question

The 'Giant Swing' at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.005.00 m long, and the upper end of the cable is fastened to the arm at a point 3.003.00 m from the central shaft (Fig. E5.505.50). Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30.0°30.0° with the vertical.

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Textbook Question

A 11251125-kg car and a 22502250-kg pickup truck approach a curve on a highway that has a radius of 225 225 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 65.065.0 mi/h can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car?

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Textbook Question

A small remote-controlled car with mass 1.601.60 kg moves at a constant speed of v=12.0v = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.005.00 m (Fig. E5.455.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point AA (bottom of the track)?

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Textbook Question

In another version of the 'Giant Swing' (see Exercise 5.505.50), the seat is connected to two cables, one of which is horizontal (Fig. E5.515.51). The seat swings in a horizontal circle at a rate of 28.028.0 rpm (rev/min). If the seat weighs 255255 N and an 825825-N person is sitting in it, find the tension in each cable.

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Textbook Question

A small remote-controlled car with mass 1.601.60 kg moves at a constant speed of v=12.0v = 12.0 m/s in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of 5.005.00 m (Fig. E5.455.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point BB (top of the track)?

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Textbook Question

A small car with mass 0.8000.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.005.00 m (Fig. E5.455.45). If the normal force exerted by the track on the car when it is at the top of the track (point BB) is 6.006.00 N, what is the normal force on the car when it is at the bottom of the track (point AA)?

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