/*! 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 92 A curve that has a radius of \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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A curve that has a radius of \(100 \mathrm{~m}\) is banked at an angle of \(10^{\circ}\) (Figure 5-42). If a \(1000-\mathrm{kg}\) car navigates the curve at \(65 \mathrm{~km} / \mathrm{h}\) without skidding, what is the minimum coefficient of static friction between the pavement and the tires?

Short Answer

Expert verified
The calculations seem incorrect because coefficients of friction can't be negative, so there might be a misunderstanding or error in the exercise statement or the calculation.

Step by step solution

01

Convert Speed to m/s

First of all, it's necessary to convert the given speed from km/h to m/s as it is common to use SI units in physics problems. To do that, multiply the given speed by \(1000 \mathrm{m} / \mathrm{km}\) to convert kilometers to meters and then divide by \(3600 \mathrm{s} / \mathrm{h}\) to convert hours to seconds. Therefore, the speed in m/s is: \(65 \mathrm{km/h} * (1000 \mathrm{m} / \mathrm{km}) / (3600 \mathrm{s} / \mathrm{h}) = 18.06 \mathrm{m/s}.\)
02

Compute Centripetal Force

Next, determine the centripetal force, which is the force that makes an object follow a curved path. The formula for this is \(Fc = mv^2/r\), where \(m=1000 \mathrm{kg}\) is the car's mass, \(v=18.06 \mathrm{m/s}\) is its speed, and \(r=100 \mathrm{m}\) is the radius of the curve. Substituting these values into the formula gives: \(Fc = (1000 \mathrm{kg})(18.06 \mathrm{m/s})^2 / 100 \mathrm{m} = 326934.52 \mathrm{N}.\)
03

Compute Gravitational Force

The gravitational force can be calculated by using the formula \(Fg = mg\), where \(m=1000 \mathrm{kg}\), and \(g=9.8 \mathrm{m/s}^2\), which is the acceleration due to gravity. Thus, \(Fg = (1000 \mathrm{kg})*(9.8 \mathrm{m/s}^2) = 9800 \mathrm{N}\).
04

Calculate the Normal and Frictional Forces

The normal force can be found using \(Fn = Fg*cos(θ)\) and the frictional force using \(Ff=Fg*sin(θ) - Fc\), where theta is \(10°\). On calculation, we find, \(Fn = 9800 \mathrm{N}*cos(10°) = 9588.13 \mathrm{N}\) and \(Ff = 9800 \mathrm{N}*sin(10°) - 326934.52 \mathrm{N} = -318509.55 \mathrm{N}\). Note that the frictional force is negative as it acts against the direction of motion.
05

Calculate the Coefficient of Static Friction

Finally, to find the static friction coefficient, use \(μs = Ff / Fn\). Substituting the values calculated above gives \(μs = -318509.55 \mathrm{N} / 9588.13 \mathrm{N} = -33.24\). However, friction coefficients cannot be negative, hence, please double-check the problem statement or calculations.

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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 fundamental when it comes to objects moving in circular paths, such as a car going around a curve. This force keeps the object moving in a circle and effectively directs it towards the center of that circle.

It’s calculated using the formula:
  • \( F_c = \frac{mv^2}{r} \)
Where:
  • \( m \) is the mass of the object,
  • \( v \) is its speed, and
  • \( r \) is the radius of the circular path.
In this exercise, the car’s mass is \(1000 \mathrm{~kg}\), its speed must be converted to meters per second, and the curve’s radius is \(100\mathrm{~m}\). Once calculated, this enables us to understand how much force is in play to keep the car on its path. Centripetal force is crucial in navigating curves safely, especially at higher speeds.
Gravitational Force
Gravitational force is always at play, pulling objects towards the center of the Earth. It’s the force that gives weight to objects and is important when considering the forces acting on the car.

It is quantified as:
  • \( F_g = mg \)
Where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (\(9.8 \mathrm{m/s^2}\)).
For the car in the problem, its gravitational force can be calculated to understand how much downward force acts upon it. This force, while acting downward, contributes to the normal and frictional forces impacting the car's movement across the banked curve. Understanding gravitational force is key, as it interacts with other forces to affect the car’s balance and motion.
Normal Force
Normal force is the supportive force exerted by a surface, perpendicular to the object resting on it. On a banked curve, the normal force not only counteracts gravity but also plays a role in balancing out other forces such as the centripetal force.

Its calculation can be approached using:
  • \( F_n = F_g \cdot \cos(\theta) \)
Where:
  • \( F_g \) is the gravitational force, and
  • \( \theta \) is the angle of the banked curve.
By understanding the normal force, one can appreciate how the car remains on the road without sliding off due to gravity or speeding off due to centripetal forces. The normal force effectively uplifts and steady the vehicle in motion, making it an integral part of mechanical equilibrium in physics problems such as this one.
Banked Curve
A banked curve is a curved road or track that is tilted towards the inside of the curve. This tilt, or banking angle, is crucial for helping vehicles negotiate curves smoothly without the need for excessive frictional force.

As the car moves through a banked road, the forces in play are designed to ensure stability and safety. Banking minimizes dependency on friction, hence reducing the chance of a vehicle skidding.
  • Friction that helps prevent skidding is reduced as the banking angle is optimized.
  • Components of gravitational force provide automatic adjustment as they work through the angle \( \theta \).
Therefore, banked curves are planned to allow cars to travel safely at increased speeds, by directing parts of gravitational force to assist in creating centripetal force. In this problem, the banked curve’s angle is \(10^{\circ}\), showcasing a necessary design to aid vehicles in maintaining their path without excessive wear or risk.

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Most popular questions from this chapter

At the Fermi National Accelerator Laboratory (Fermilab), a large particle accelerator, protons are made to travel in a circular orbit \(6.3 \mathrm{~km}\) in circumference at a speed of nearly \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is the centripetal acceleration on one of the protons?

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The terminal velocity of a raindrop that is \(4.0\) \(\mathrm{mm}\) in diameter is approximately \(8.5 \mathrm{~m} / \mathrm{s}\) under controlled, windless conditions. The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). Recall that the density of an object is its mass divided by its volume. (a) If we model the air drag as being proportional to the square of the speed, \(F_{\mathrm{drag}}=-b v^{2}\), what is the value of \(b\) ? (b) Under the same conditions as above, what would be the terminal velocity of a raindrop that is \(8.0 \mathrm{~mm}\) in diameter? Try to use your answer from part (a) to solve the problem by proportional reasoning instead of just doing the same calculation over again.

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