/*! 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

Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice skater is gliding horizontally across the ice with an initial velocity of \(+6.3 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the ice and the skate blades is \(0.081,\) and air resistance is negligible. How much time elapses before her velocity is reduced to \(+2.8 \mathrm{~m} /\) s?

A large rock and a small pebble are held at the same height above the ground. (a) Is the gravitational force exerted on the rock greater than, less than, or equal to that exerted on the pebble? Justify your answer. (b) When the rock and the pebble are released, is the downward acceleration of the rock greater than, less than, or equal to that of the pebble? Why? Problem A \(5.0-\mathrm{kg}\) rock and a \(3.0 \times 10^{-4} \mathrm{~kg}\) pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the acceleration of each object when released. Verify that your answers are consistent with your answers to the Concept Questions.

A 15 -g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3} \mathrm{~s} \mathrm{~s}\) for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{~m} / \mathrm{s}\). Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

The drawing shows box 1 resting on a table, with box 2 resting on top of box \(1 .\) A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2 and the other end is connected to box \(3 .\) The weights of the three boxes are \(W_{1}=55 \mathrm{~N}, W_{2}=35 \mathrm{~N},\) and \(W_{3}=28 \mathrm{~N}\). Determine the magnitude of the normal force that the table exerts on box \(1 .\)

Refer to Concept Simulation \(4.4\) at for background relating to this problem. The drawing shows a large cube (mass \(=25 \mathrm{~kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\vec{P}\). A small cube \((\mathrm{mass}=4.0 \mathrm{~kg})\) is in contact with the front surface of the large cube and will slide downward unless \(\vec{P}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71\). What is the smallest magnitude that \(\overrightarrow{\mathrm{P}}\) can have in order to keep the small cube from sliding downward?

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.