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Chapter 4: Problem 41

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 \((\mathbf{b}) \theta=35^{\circ}\).

Short Answer

Expert verified
The ratios are approximately 0.9659 for \( \theta = 15^{\circ} \) and 0.8192 for \( \theta = 35^{\circ} \).

Step by step solution

01

Identify Components of Forces

The weight of the car acts downward with a magnitude equal to its weight \( mg \), where \( m \) is the mass of the car and \( g \) is acceleration due to gravity. The normal force acts perpendicular to the surface of the hill. Due to the incline, part of the weight acts to push the car into the hill and part acts along the hill.
02

Resolve Weight into Components

Divide the weight into two components: one perpendicular to the hill and one parallel to it. The component of the weight perpendicular to the hill (which the normal force balances) is \( mg \cos \theta \). The component parallel to the hill is \( mg \sin \theta \).
03

Calculate the Normal Force

The normal force \( N \) is equal and opposite to the component of the gravitational force perpendicular to the hill, so \( N = mg \cos \theta \).
04

Determine the Ratio of Normal Force to Weight

The ratio of the normal force to the weight is given by \( \frac{N}{mg} = \cos \theta \).
05

Substitute Given Angles

(a) For \( \theta = 15^{\circ} \), substitute into the ratio to get \( \frac{N}{mg} = \cos 15^{\circ} \). (b) For \( \theta = 35^{\circ} \), substitute to get \( \frac{N}{mg} = \cos 35^{\circ} \).
06

Calculate Specific Ratios

Use a calculator to find \( \cos 15^{\circ} \approx 0.9659 \) and \( \cos 35^{\circ} \approx 0.8192 \). Thus, \( \frac{N}{mg} \) is approximately 0.9659 for \( \theta = 15^{\circ} \) and approximately 0.8192 for \( \theta = 35^{\circ} \).

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Key Concepts

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

Inclined Plane Basics
An inclined plane is a flat surface that is tilted at an angle, often used to make it easier to lift or lower objects. This angle, known as the angle of inclination, affects how forces act on an object positioned on the slope. When a car travels up or down such a plane, the direction and magnitude of forces like gravity and friction change. The key force we deal with is the weight of the car, which acts downward but feels different effects due to the incline. Understanding how an inclined plane alters the components of forces is essential in solving many physics problems, especially those involving motion and equilibrium.
Understanding Trigonometric Components
In physics, dealing with inclined planes requires understanding trigonometric components because forces operate in different directions. When forces like gravity act on an inclined plane, they can be broken down into components: one perpendicular and the other parallel to the plane.

  • The perpendicular component helps us find the normal force, which acts against gravity.
  • The parallel component influences motion along the plane, such as sliding down due to gravity.
To determine these components, trigonometry is used. For example, if the total force is the weight of the car, and the plane is inclined at an angle \( \theta \), the component of the force perpendicular to the plane is \( mg \cos \theta \), and the component parallel is \( mg \sin \theta \). Understanding these components helps us determine the forces acting on the car and solve the problem accurately.
Resolving Forces for Normal Force Calculation
The process of force resolution allows us to understand how different forces act on an object on an inclined plane. Since gravity pulls the car directly downward, resolving this force into two orthogonal components simplifies calculations.

  • The normal force is directly influenced by the component perpendicular to the plane, which is given by \( mg \cos \theta \).
  • This component counteracts the weight of the car, maintaining equilibrium and preventing the car from sinking into the plane.
  • The calculation becomes straightforward once the forces are resolved along the axes defined by the plane.
Resolving forces in this way simplifies understanding force interactions, aiding in solving various physics problems involving inclined planes.
Tackling Physics Problem Solving
Physics problem solving involves methodically dissecting a problem to find solutions. When facing a challenge like calculating normal force on an inclined plane, the process can be streamlined by following consistent steps.

  • Identify all forces at play and determine how they're influencing the object on the plane.
  • Breakdown complex forces into components using trigonometry. This helps visualize and calculate what each force contributes.
  • Apply known physics equations to solve for unknowns, such as using \( \cos \theta \) to find the ratio of the normal force to weight.
Using structured steps enhances understanding and accuracy, making seemingly complex physics problems manageable and solvable.

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Most popular questions from this chapter

A block is pressed against a vertical wall by a force \(\overrightarrow{\mathbf{P}},\) as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.250 . The weight of the block is \(39.0 \mathrm{N},\) and the directional angle for the force \(\overrightarrow{\mathbf{P}}\) is \(\theta=30.0^{\circ} .\) Determine the magnitude of \(\overrightarrow{\mathbf{P}}\) when the block slides \((\mathbf{a})\) up the wall and (b) down the wall.

ssm mmh A 1580-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m?

ssm A \(325-\mathrm{kg}\) boat is sailing \(15.0^{\circ}\) north of east at a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) . Thirty seconds later, it is sailing \(35.0^{\circ}\) north of east at at a speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . During this time, three forces act on the boat: a \(31.0 \mathrm{N}\) force directed \(15.0^{\circ}\) north of east (due to an auxiliary engine), a \(23.0-\mathrm{N}\) force directed \(15.0^{\circ}\) south of west (resistance due to the water), and \(\overrightarrow{\mathbf{F}}_{\mathrm{w}}\) (duc to the wind). Find the magnitude and direction of the force \(\overrightarrow{\mathbf{F}}_{\mathrm{W}}\) . Express the direction as an angle with respect to due east.

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 \(\mathrm{N}\) and the drag force has a magnitude of 1027 \(\mathrm{N}\) . The mass of the sky diver is 93.4 \(\mathrm{kg}\) . What are the magnitude and direction of his acceleration?

ssm A person is trying to judge whether a picture (mass 1.10 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.660. What is the minimum amount of pressing force that must be used?
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