/*! 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 102 A racetrack curve has radius \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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A racetrack curve has radius \(120.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.300 .\) A race car with mass \(900 \mathrm{~kg}\) rounds the curve with the minimum speed needed to not slide down the banking. (a) As the car rounds the curve, what is the normal force exerted on it by the road? (b) What is the car's speed?

Short Answer

Expert verified
The normal force exerted on the car by the road is calculated to be approximately 8940 N and the minimum speed of the car to avoid sliding is around 22 m/s.

Step by step solution

01

Calculate the Gravitational Force

The weight of the car can be determined first, as it is the force of gravity acting downwards on the car. It can be calculated using the formula \( F_g = m \cdot g \), where \( m = 900 \mathrm{kg} \) is the mass of the car and \( g = 9.8 \mathrm{m/s^2} \) is the acceleration due to gravity. This results in \( F_g = 900 \mathrm{kg} \cdot 9.8 \mathrm{m/s^2} = 8820 \mathrm{N} \).
02

Break Down the Forces

The forces acting on the car can be resolved in two components - one parallel to the incline (caused by gravity) and the other perpendicular to the incline (the normal force). The force from gravity can be separated into these two components: \( F_{g, parallel} = F_g \cdot sin(\theta) \) and \( F_{g, perpendicular} = F_g \cdot cos(\theta) \), where \( \theta \) is the banking angle, 18.0 degrees in this case.
03

Calculate the Normal Force

The normal force ( \( F_N \) ) is the resultant of the force due to gravity and the frictional force (\( F_f \)). At the minimum speed required to not slide, these forces are in equilibrium. The frictional force is equal to the product of the normal force and the coefficient of static friction (\( \mu_s \)). Based on these, we can write: \( F_N = F_{g, perpendicular} + F_f = F_g \cdot cos(\theta) + \mu_s \cdot F_N \). Solving for \( F_N \) gives: \( F_N = F_g \cdot cos(\theta) / (1 - \mu_s) \).
04

Calculate the Minimum Speed Required

To find the speed of the car, we use the formula for the centripetal force, \( F_c = m \cdot v^2 / r \), which is balanced by the normal force and the parallel component of the gravity force. Therefore, \( m \cdot v^2 / r = F_{g, parallel} + F_f = F_g \cdot sin(\theta) + \mu_s \cdot F_N \). Solving for \( v \) will give us the minimum speed required.

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Key Concepts

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

Centripetal Force
When an object travels along a curved path, the centripetal force is the invisible tether keeping it on its circular trajectory. Think of it as the central force that acts at a right angle to the object's velocity, directing it towards the curve's center.

Imagine a race car zipping around a circular track. The car wants to move straight due to its momentum, yet it continues along the curve. That change of direction is the result of centripetal force, a necessary ingredient for circular motion. In our racetrack problem, the minimum speed is calculated based on the centripetal force, which is the combined effort of static friction and the component of gravitational force parallel to the slope.For an object with mass \(m\) moving at speed \(v\) around a curve of radius \(r\), centripetal force (\(F_c\)) is given by the equation:\[ F_c = \frac{m \cdot v^2}{r} \]In the context of our racetrack scenario, the car's tires grip the road, providing the centripetal force via static friction and, to some extent, the normal force.
Static Friction
Static friction is the force that prevents surfaces from sliding past each other when at rest or moving at a constant speed. It plays the under-appreciated role of a silent guardian, ensuring objects don't slip and slide at the slightest nudge.

In our banked curve example, static friction is the superhero that prevents the race car from drifting downwards on the banked road. This frictional force arises from the contact between the car tires and the road surface. The maximum static frictional force (\(F_{f, \text{max}}\)) that can be exerted before slipping occurs is expressed by the equation:\[ F_{f, \text{max}} = \mu_s \cdot F_N \]where \(\mu_s\) is the coefficient of static friction and \(F_N\) is the normal force. The static friction adjusts according to the needs up to its maximum limit, ensuring the car maintains its circular motion without slipping down.
Normal Force
The normal force is the perpendicular force exerted by a surface on an object in contact with it. Think of it as the Earth's handshake—it's the push you feel under your feet while standing still.

In a banked turn scenario, the normal force acts perpendicularly to the road surface. It's balanced by a portion of the car's weight (the component perpendicular to the incline) and contributes to the centripetal force that keeps the car on track. Through some strategic calculations involving the mass of the car, the angle of the banking, and the gravitational force, we determine the normal force using the formula:\[ F_N = \frac{F_g \cdot \cos(\theta)}{1 - \mu_s} \]Here, \(\theta\) symbolizes the bank angle and \(\mu_s\) is the static friction coefficient. The normal force not only withstands the gravitational pull but, coupled with static friction, aids in creating a stable, non-sliding circular path for the race car.

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Most popular questions from this chapter

If the coefficient of static friction between a table and a uniform, massive rope is \(\mu_{\mathrm{s}},\) what fraction of the rope can hang over the edge of the table without the rope sliding?

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is \(0.400 .\) A force \(\vec{F}\) is applied to the crate in a direction \(30.0^{\circ}\) above the horizontal. The minimum value of \(F\) required to get the crate to start sliding is \(380 \mathrm{~N}\). What is the mass of the crate?

A small rock with mass \(m\) is attached to a light string of length \(L\) and whirled in a vertical circle of radius \(R\). (a) What is the minimum speed \(v\) at the rock's highest point for which it stays in a circular path? (b) If the speed at the rock's lowest point in its circular path is twice the value found in part (a), what is the tension in the string when the rock is at this point?

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