/*! 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

The Trendelenburg Position. After emergencies with major blood loss, a patient is placed in the Trendelenburg position, in which the foot of the bed is raised to get maximum blood flow to the brain. If the coefficient of static friction between a typical patient and the bed sheets is \(1.20,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

A large crate with mass \(m\) rests on a horizontal floor. The coefficients of friction between the crate and the floor are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}} .\) A woman pushes downward with a force \(\overrightarrow{\boldsymbol{F}}\) on the crate at an angle \(\theta\) below the horizontal. (a) What magnitude of force \(\vec{F}\) is required to keep the crate moving at constant velocity? (b) If \(\mu_{\mathrm{s}}\) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \(\mu_{\mathrm{s}}\)

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}\) without causing the box to slide? Draw a free-body diagram of the toolbox.

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) 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?

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