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Chapter 6: Problem 87

A car weighing \(10.7 \mathrm{kN}\) and traveling at \(13.4 \mathrm{~m} / \mathrm{s}\) without negative lift attempts to round an unbanked curve with a radius of \(61.0 \mathrm{~m}\). (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is 0.350 , is the attempt at taking the curve successful?

Short Answer

Expert verified
(a) 3186.8 N, (b) Yes, the car can successfully round the curve.

Step by step solution

01

Determine the Required Centripetal Force

To keep the car moving in a circular path, a centripetal force is required. This force is provided by the frictional force between the tires and the road. The formula for centripetal force is given by:\[ F_c = \frac{mv^2}{r} \]Where:- \( m \) is the mass of the car,- \( v \) is the velocity of the car,- \( r \) is the radius of the curve.First, convert the weight into mass using \( F = mg \) => \( m = \frac{10.7\,\text{kN}}{9.81\,\text{m/s}^2} = 1090.72\,\text{kg} \). Now substitute these values into the centripetal force formula:\[ F_c = \frac{1090.72\,\text{kg} \times (13.4\,\text{m/s})^2}{61.0\,\text{m}} \approx 3186.8\,\text{N} \].
02

Calculate Maximum Static Frictional Force

The static frictional force is given by the equation:\[ F_s = \mu_s N \]Where:- \( \mu_s \) is the coefficient of static friction (0.350),- \( N \) is the normal force, which equals the weight of the car since it's on an unbanked curve.Thus,\[ F_s = 0.350 \times 10.7\,\text{kN} = 0.350 \times 10700\,\text{N} = 3745\,\text{N} \].
03

Compare Forces for Curve Success

For the car to successfully navigate the curve, the static frictional force must be equal to or greater than the required centripetal force. Compare the calculated forces from Steps 1 and 2:- Required centripetal force = \( 3186.8\,\text{N} \)- Maximum static frictional force = \( 3745\,\text{N} \)Since the static frictional force is greater than the required centripetal force, the car will successfully navigate the curve.

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Key Concepts

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

Static Friction
Static friction plays a vital role in this physics problem, as it prevents the car from sliding or losing grip while turning. Imagine two surfaces in contact, like a car's tires on the road. Static friction is the force that acts to keep the two surfaces from slipping over each other when they are at rest, or in this case, maintaining a circular motion. Always remember:
  • Static friction comes into action just before any motion starts. That means it has a defining threshold beyond which slipping would occur.
  • It's usually larger than dynamic (or kinetic) friction, which plays a role once movement starts.
  • To determine the static frictional force, we use the formula: \( F_s = \mu_s N \).
This problem finds that the static friction ensures the car stays on its path without skidding off. By comparing it with the centripetal force, we determine if the curve can be navigated safely. Remember, higher static friction indicates better grip and better control over circular paths.
Circular Motion
Circular motion is all about movement along a curved path, like a car taking a bend. When a car moves in a circle, it constantly changes direction, and this change in direction is what we call circular motion.
  • For an object to maintain circular motion, it requires a force directed towards the center of the circle 鈥� known as centripetal force. Without this force, the object would merely travel off in a straight line.
  • The formula \( F_c = \frac{mv^2}{r} \) shows that the required centripetal force depends on the mass \( m \), the square of the velocity \( v \), and inversely on the radius \( r \) of the circle.
  • In our car scenario, the centripetal force comes from the friction between the automobile's tires and the road.
Understanding these principles of circular motion can help explain why a tighter turn, a higher speed, or a heavier vehicle demand greater forces.
Physics Problem-Solving
In tackling physics problems, like calculating whether a car can handle a curve without slipping, a structured approach is crucial. Let's break it down into steps that can be used generally for solving similar physics problems:
  • Understand the problem: Ensure you comprehend what is being asked. For instance, determining the frictional force required for circular motion.
  • Identify knowns and unknowns: List out the values given in the problem, such as the weight or velocity of the car, and what you need to find.
  • Use relevant equations: Select the physical laws or formulas that apply to the problem. Here,
    1. For centripetal force: \( F_c = \frac{mv^2}{r} \).
    2. For static friction: \( F_s = \mu_s N \).
  • Perform calculations: Execute the mathematical operations needed to find your answer.
  • Check the solutions: Compare the results to deduce if the car will successfully navigate the curve, as in checking if \( F_s \geq F_c \).
  • Interpret the results: Reflect on the meaning of the calculated forces to understand their real-world implications.
This method helps simplify complex physics problems and aids in reaching a clear solution.

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Most popular questions from this chapter

A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the net force on the bicycle from the road.

Calculate the ratio of the drag force on a jet flying at \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a propdriven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km} .\) Assume that the airplanes have the same effective cross-sectional area and drag coefficient \(C\).

A 3.5 kg block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters can alter the net force on a passenger. Consider a passenger of mass \(m\) riding around a horizontal circle of radius \(r\) at speed \(v .\) What is the variation \(d F\) in the net force magnitude for (a) a variation \(d r\) in the radius with \(v\) held constant, (b) a variation \(d v\) in the speed with \(r\) held constant, and (c) a variation \(d T\) in the period with \(r\) held constant?

An old streetcar rounds a flat corner of radius \(9.1 \mathrm{~m},\) at \(16 \mathrm{~km} / \mathrm{h} .\) What angle with the vertical will be made by the loosely hanging hand straps?
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