/*! 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 10 A 350 -kg sailboat has an accele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 350 -kg sailboat has an acceleration of \(0.62 \mathrm{~m} / \mathrm{s}^{2}\) at an angle of \(64^{\circ}\) north of east. Finc the magnitude and direction of the net force that acts on the sailboat.

Short Answer

Expert verified
The net force is 217 N at an angle of 64° north of east.

Step by step solution

01

Identify information

First, let's list the given information:- Mass of the sailboat: \(m = 350 \text{ kg}\)- Acceleration: \(a = 0.62 \text{ m/s}^2\)- Direction of acceleration: \(64^{\circ}\) north of east.
02

Apply Newton's Second Law

Newton's Second Law states that the net force \(\mathbf{F}_{\text{net}}\) is given by \(\mathbf{F}_{\text{net}} = m \cdot \mathbf{a}\). The magnitude of the net force can be calculated using the formula:\[ F_{\text{net}} = m \times a \]Substitute the given values:\[ F_{\text{net}} = 350 \times 0.62 = 217 \text{ N}\]
03

Determine the components of the force

The net force has components along the east and north directions. Use trigonometry to find these components.For the eastern component \( F_{E} \):\[ F_{E} = F_{\text{net}} \times \cos(64^{\circ}) \]For the northern component \( F_{N} \):\[ F_{N} = F_{\text{net}} \times \sin(64^{\circ}) \]
04

Calculate the components

Substitute \( F_{\text{net}} = 217 \text{ N} \) into the equations:Eastern component:\[ F_{E} = 217 \times \cos(64^{\circ}) \approx 94.58 \text{ N} \]Northern component:\[ F_{N} = 217 \times \sin(64^{\circ}) \approx 195.08 \text{ N} \]
05

Verify magnitude and direction

The magnitude calculated initially as \(217 \text{ N}\) is accurate since it's computed from the components using Pythagorean theorem:\[ F_{\text{net}} = \sqrt{F_{E}^2 + F_{N}^2} = \sqrt{(94.58)^2 + (195.08)^2} \approx 217 \text{ N}\]The direction is already given as \(64^{\circ}\) north of east.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Net Force Calculation
Newton's Second Law is fundamental for understanding net force. It states: "The net force acting on an object is equal to the product of its mass and acceleration." This principle can be summarized by the equation:
  • Net Force: \( F_{\text{net}} = m \cdot a \)
This means that to find the net force, you multiply the object's mass by its acceleration. In the case of the sailboat, with a mass of 350 kg and an acceleration of 0.62 m/s², the equation becomes:
  • \( F_{\text{net}} = 350 \times 0.62 = 217 \text{ N} \)
The result of 217 N represents the total force required to make the sailboat accelerate at the given rate. This calculation shows how the net force is directly proportional to both the mass and acceleration, which is crucial in understanding how different forces interact to change the motion of an object.
Vector Components
Vectors are quantities that have both magnitude and direction. When analyzing forces in physics, it's essential to break them down into components. For a force acting at an angle, like our sailboat example, these components lie along standard axes—typically east and north for navigation.
  • Eastern Component: \( F_{E} = F_{\text{net}} \times \cos(\theta) \)
  • Northern Component: \( F_{N} = F_{\text{net}} \times \sin(\theta) \)
Here, \( \theta \) is the angle of 64° north of east. Calculating these components involves:
  • \( F_{E} = 217 \times \cos(64^{\circ}) \approx 94.58 \text{ N} \)
  • \( F_{N} = 217 \times \sin(64^{\circ}) \approx 195.08 \text{ N} \)
These calculations help us understand how the force is distributed along different directions. Knowing these vector components is crucial because often, various forces act simultaneously in different directions, affecting an object's movement. By using vector components, we can simplify these complex scenarios into manageable calculations.
Trigonometry in Physics
Trigonometry is an essential tool in physics, especially for resolving forces into their components. It involves using relationships in triangles, primarily right triangles, to find unknown lengths or angles. Angles like the 64° north of east in our problem are common as they describe the direction of forces in a plane.
  • Cosine (\( \cos \)) helps find the adjacent side of a right triangle over the hypotenuse, useful for horizontal components.
  • Sine (\( \sin \)) is used for the opposite side over the hypotenuse, perfect for vertical components.
Using trigonometric functions efficiently is vital to determine the direction and magnitude of forces acting on objects. For instance, the eastern and northern components of the sailboat's net force were found using:
  • \( F_{E} = F_{\text{net}} \times \cos(64^{\circ}) \)
  • \( F_{N} = F_{\text{net}} \times \sin(64^{\circ}) \)
Understanding trigonometric relationships assists in comprehending how forces apply to different dimensions, making complex physics problems simpler and more intuitive. This is why learning trigonometry is indispensable for physics students tackling real-world problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs \(31 \mathrm{~N}\). The coefficient of static friction between his hands and the book is \(0.40 .\) To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

Mars has a mass of \(6.46 \times 10^{23} \mathrm{~kg}\) and a radius of \(3.39 \times 10^{6} \mathrm{~m}\). (a) What is the acceleration due to gravity on Mars? (b) How much would a \(65-\mathrm{kg}\) person weigh on this olanet?

Consult Interactive LearningWare 4.2 at before beginning this problem. A truck is traveling at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is \(0.650 .\) Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

A student is skateboarding down a ram p that is \(6.0 \mathrm{~m}\) long and inclined at \(18^{\circ}\) with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{~m} / \mathrm{s}\). Neglect friction and find the speed at the bottom of the ramp.

A person is attempting to push a refrigerator across a room. He exerts a horizontal force on the refrigerator, but it does not move. (a) What other horizontal force must be acting on the refrigerator? (b) How are the magnitude and direction of this force related to the force that the person exerts? (c) Suppose that the person applies the force such that it is the largest possible force before the refrigerator begins to move. What factors determine the magnitude of this force? Problem Consult Multiple-Concept Example 9 to explore a model for solving this problem. A person pushes on a 57 -kg refrigerator with a horizontal force of \(-267 \mathrm{~N}\); the minus sign indicates that the force is directed along the \(-x\) direction. The coefficient of static friction is \(0.65 .\) (a) If the refrigerator does not move, what is the magnitude and direction of the static frictional force that the floor exerts on the refrigerator? (b) What is the magnitude of the largest pushing force that can be applied to the refrigerator before it just begins to move?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.