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Chapter 4: Problem 120

A flatbed truck is carrying a crate up a hill of angle of inclination \(\theta=10.0^{\circ},\) as the figure illustrates. The coefficient of static friction between the truck bed and the crate is \(\mu_{\mathrm{s}}=0.350 .\) Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

Short Answer

Expert verified
The maximum acceleration is approximately 2.11 m/s虏.

Step by step solution

01

Analyze the forces involved

When the truck accelerates up the hill, several forces act on the crate. These include the gravitational force, normal force, and frictional force. The gravitational force can be split into components parallel and perpendicular to the inclined surface.
02

Calculate the forces

The component of gravitational force parallel to the incline is given by \( mg \sin \theta \), where \( m \) is the mass of the crate and \( g \) is the acceleration due to gravity. The component perpendicular to the incline is \( mg \cos \theta \).
03

Static Friction Force Calculation

The maximum static friction force, which prevents slipping, is calculated as \( f_s = \mu_s N \), where \( N \) is the normal force on the crate. The normal force here is equal to the perpendicular component of gravity, \( N = mg \cos \theta \).
04

Set up the equation for maximum frictional force

Since the static friction force prevents slipping, set its maximum value to the force that would cause slipping: \( f_s = mg \sin \theta + ma \). Substitute \( f_s = \mu_s mg \cos \theta \) into this equation.
05

Solve for acceleration \( a \)

Rearrange the equation \( \mu_s mg \cos \theta = mg \sin \theta + ma \) to solve for \( a \). By dividing through by \( m \), and rearranging, you get \( a = \mu_s g \cos \theta - g \sin \theta \).
06

Calculate the numerical value

Substitute the given values into the equation: \( a = 0.350 \times 9.81 \times \cos(10.0^{\circ}) - 9.81 \times \sin(10.0^{\circ}) \). Calculate \( \cos(10.0^{\circ}) \) and \( \sin(10.0^{\circ}) \), then plug them into the equation.

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Key Concepts

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

Static Friction
Static friction is a type of friction that prevents surfaces from sliding past each other. It acts parallel to the surfaces in contact and comes into play when there is an attempt of relative motion between the two surfaces. In the case of the crate on the inclined truck, static friction prevents the crate from slipping backward as the truck accelerates.

Key points about static friction are:
  • It has a maximum value before sliding begins, calculated as \( f_s = \mu_s N \).
  • The coefficient of static friction, \( \mu_s \), is a measure of how much force can be exerted without movement. Here, it is 0.350.
  • The normal force \( N \) is the force perpendicular to the surfaces in contact.
Understanding static friction is crucial in physics, as it explains how objects remain stationary under various forces until a threshold is surpassed.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, similar to the truck's bed carrying the crate. When analyzing situations on an inclined plane, it's important to consider the forces acting in both parallel and perpendicular directions relative to the plane.

Here are the forces that come into play:
  • The gravitational force component parallel to the incline is calculated using \( mg \sin \theta \), pulling the object down the plane.
  • The perpendicular component, \( mg \cos \theta \), contributes to the normal force.
The angle of the incline, \( \theta \), plays a significant role. A steeper angle increases the parallel gravitational component, requiring more static friction to prevent slipping. Understanding these components helps in designing systems like ramps or studying motion on slopes.
Newton's Laws
Newton's laws are fundamental in analyzing motion and exerted forces. In this crate problem, Newton鈥檚 second law, \( F = ma \), is particularly relevant as it relates acceleration to force.

Here's how Newton's laws apply:
  • The gravitational force and static friction maintain equilibrium until a maximum force is applied.
  • Net force on the crate is calculated by considering different force components acting on it.
  • The equation \( \mu_s mg \cos \theta = mg \sin \theta + ma \) represents the balance between the frictional force and other forces acting on the crate.
By rearranging this equation, you can solve for maximal acceleration, keeping in mind that force, mass, and acceleration are interconnected.
Trigonometry in Physics
Trigonometry is essential when resolving forces on inclined planes. It helps break down forces into perpendicular and parallel components using angles. In this exercise, angles are used to determine the components of gravitational force.

Applications of trigonometry include:
  • Using \( \sin \theta \) for calculating the parallel component of gravity.
  • Applying \( \cos \theta \) for finding the normal force component.
  • Helping in rearranging equations to compute the external factors, like acceleration in this case.
These trigonometric relationships aid in translating real-world problems into solvable mathematics, making the analysis of forces much simpler and more intuitive.

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Most popular questions from this chapter

\(\mathrm{mmh}\) A boat has a mass of \(6800 \mathrm{kg}\). Its engines generate a drive force of \(4100 \mathrm{N}\) due west, while the wind exerts a force of \(800 \mathrm{N}\) due east and the water exerts a resistive force of \(1200 \mathrm{N}\) due east. What are the magnitude and direction of the boat's acceleration?

At a distance \(H\) above the surface of a planet, the true weight of a remote probe is one percent less than its true weight on the surface. The radius of the planet is \(R\). Find the ratio \(H / R\).

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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?

A person with a black belt in karate has a fist that has a mass of \(0.70 \mathrm{kg} .\) Starting from rest, this fist attains a velocity of \(8.0 \mathrm{m} / \mathrm{s}\) in \(0.15 \mathrm{s}\) What is the magnitude of the average net force applied to the fist to achieve this level of performance?
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