/*! 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 41 A car is traveling up a hill tha... [FREE SOLUTION] | 91Ó°ÊÓ

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A car is traveling up a hill that is inclined at an angle \(\theta\) above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) \(\theta=15^{\circ}\) and (b) \(\theta=35^{\circ}\).

Short Answer

Expert verified
(a) 0.9659, (b) 0.8192

Step by step solution

01

Identifying Forces

The weight of the car is given by the gravitational force acting downwards, which is \( W = mg \), where \( m \) is the mass of the car and \( g \) is gravitational acceleration. The normal force \( N \) acts perpendicular to the plane of the incline.
02

Determine Components of Forces

The normal force \( N \) balances the component of the gravitational force that is perpendicular to the inclined surface. The component of weight perpendicular to the incline is \( mg \cos \theta \).
03

Calculate the Normal Force

Since the component of the weight perpendicular to the incline must be balanced by the normal force, we find that\[ N = mg \cos \theta \]
04

Derive the Ratio of Normal Force to Weight

The ratio of the magnitude of the normal force to the weight of the car is given by\[ \frac{N}{W} = \frac{mg \cos \theta}{mg} = \cos \theta \]
05

Substitute Values for Part (a)

For part (a), substitute \( \theta = 15^{\circ} \) into the formula:\[ \cos(15^{\circ}) = 0.9659 \]
06

Substitute Values for Part (b)

For part (b), substitute \( \theta = 35^{\circ} \) into the formula:\[ \cos(35^{\circ}) = 0.8192 \]

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Key Concepts

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

Inclined Plane
When we deal with physics problems involving inclined planes, we're looking at situations where an object is situated on a sloped surface. This slope is typically at an angle, denoted as \( \theta \), relative to a horizontal plane. Inclined planes are a great way to study forces because they introduce the concept of components of forces due to the slope. The angle of inclination affects how these forces are divided into components, making it crucial for solving problems involving movement or stability on slopes.

The car traveling up a hill, in this context, is affected by gravity pulling it downward, while the inclined plane's angle partially counteracts this force. The key here is understanding how the forces are distributed, which leads us into the next concept.
Gravitational Force
Gravitational force is what keeps everything anchored to the Earth. It's the force of attraction between the Earth and an object. In this scenario with the car, gravity acts straight down towards the center of the Earth with a force equal to the car's weight, \( W = mg \), where \( m \) is the car's mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \) on Earth).

On an inclined plane, gravity's full force doesn't directly aid or hinder the car's movement. Instead, it splits into components – one perpendicular and the other parallel to the inclined surface. This concept of splitting forces is critical when dealing with slopes.
Trigonometric Functions
Trigonometric functions, such as sine, cosine, and tangent, help us to determine the components of forces on an incline. They are particularly useful in breaking down gravitational force into perpendicular and parallel components relative to the inclined plane. For an inclined angle \( \theta \):

  • \( \cos(\theta) \) helps find the perpendicular component of gravity, which directly influences the normal force.
  • \( \sin(\theta) \) helps in finding the parallel component, which affects any frictional forces and the tendency of the car to slide down the hill.
Knowing the angle, these trigonometric functions allow us to relate the components of gravitational force back to the car's movement or equilibrium on the incline.
Force Components
Force components are essential when analyzing problems on inclined planes. Here, the gravitational force is divided into two parts:

  • Perpendicular Component: This component acts normal (perpendicular) to the inclined plane and is given by \( mg \cos(\theta) \). It's crucial because it balances the normal force exerted by the plane on the car, ensuring the car doesn't sink into or lift off the slope.
  • Parallel Component: This component acts along the plane and is given by \( mg \sin(\theta) \). It represents the force that potentially causes the car to slide down the incline.
Understanding these components helps us figure out the net forces acting on the car, and thus solve for conditions like the normal force, and even predict movement results.

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Most popular questions from this chapter

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