/*! 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 4 A force of \(100 \mathrm{~N}\) m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures, whereas \(30 \mathrm{~N}\) has only two.) Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{X}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ}\). Then, using the \(\cos \theta\), $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

Short Answer

Expert verified
The angle \( \theta \) is \( 17^\circ \), and the x-component is \( 95 \text{ N} \).

Step by step solution

01

Understanding the Problem

We are given a force vector of magnitude \( 100 \text{ N} \) which makes an angle \( \theta \) with the x-axis. The scalar y-component of this force is given as \( 30 \text{ N} \). We need to determine the x-component \( F_{x} \) and the angle \( \theta \).
02

Drawing a Force Diagram

To visualize the problem, draw a right triangle where the force vector forms the hypotenuse. The x-axis and y-axis form legs of the triangle. The angle \( \theta \) is between the x-axis and the force vector. The y-component is given by the opposite side and the x-component is given by the adjacent side.
03

Calculate \( \theta \) Using Sine Function

The sine of angle \( \theta \) is the ratio of the opposite side (y-component) to the hypotenuse (magnitude of force). Use the formula:\[\sin \theta = \frac{30 \text{ N}}{100 \text{ N}} = 0.30\]Calculate \( \theta \) by taking the inverse sine:\[\theta = \arcsin(0.30) \approx 17.46^\circ\]Since \( 30 \text{ N} \) has two significant figures, round \( \theta \) to \( 17^\circ \).
04

Calculate the x-component Using Cosine Function

To find the x-component \( F_{x} \), use the cosine function, which relates the adjacent side (x-component) to the hypotenuse:\[\cos \theta = \frac{F_{x}}{100 \text{ N}}\]Calculate \( F_{x} \) using:\[F_{x} = 100 \text{ N} \times \cos(17.46^\circ) \approx 95 \text{ N}\]
05

Verify Significant Figures

Make sure the results respect the number of significant figures: \( \theta \) was rounded to two significant figures as \( 17^\circ \), and \( F_x = 95 \text{ N} \) keeps the three significant figures from the original force magnitude.

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.

Force vectors
A force vector is a quantity that has both magnitude and direction. In physics, forces are represented as vectors to show not just how strong a force is, but also the direction in which the force is applied. Vectors can be visually represented using arrows where the length indicates the magnitude and the arrowhead points in the direction of the force.

Understanding force vectors is essential when analyzing situations where multiple forces act at once or when decomposing a force into its components. By breaking a force vector into two perpendicular components, namely horizontal (x-axis) and vertical (y-axis) components, we can calculate various parameters, such as the angle between the force and a reference axis.
  • Magnitude: Refers to the size or strength of the force, which in this case, was 100 N.
  • Direction: Shown by the angle \(\theta\) with reference to the x-axis.

This understanding helps in resolving forces for practical applications, like determining how much of a force is acting in a certain direction.
Angle of force
The angle of a force is defined as the angle that the force vector makes with a given reference direction, typically the x-axis in a Cartesian coordinate system. It is a crucial factor as it influences how the force is split between its components. A force acting at an angle can be decomposed into horizontal and vertical components using trigonometric functions like sine and cosine.

  • The sine of an angle \(\theta\) is used to determine the vertical (y-axis) component, calculated as \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\).

In this exercise, the angle \(\theta\) was calculated using the y-component and the overall force magnitude. It was found by solving the equation \(\sin \theta = 0.30\). This value was then transformed using the inverse sine function to find that \(\theta \approx 17.46^\circ\). For practical purposes, it was rounded to \(17^\circ\) respecting significant figures.
The influence of the angle of force cannot be overstated, as different angles can result in drastically different force impacts on an object.
Components of force
The components of a force refer to the projections of the force vector along the coordinate axes, specifically the x and y axes in a 2D plane. These components are extremely useful in solving physics problems, especially when analyzing forces in different directions. In this exercise, you are often given the components and asked to find one or both using trigonometric relations.

To resolve a force into its components:
  • The horizontal component \(F_{x}\) is calculated using the cosine function: \[F_{x} = F \times \cos \theta\]. \(F_{x}\) can be interpreted as the effective force acting along the horizontal direction.
  • The vertical component \(F_{y}\) is calculated using the sine function: \[F_{y} = F \times \sin \theta\]. In this problem, \(F_{y}\) was given as 30 N.

Each component describes how much of the overall force acts in a particular direction and provides valuable information for analyzing the effect of the force on an object. By understanding and calculating these components, students can predict how forces will interact in practical scenarios, such as when guiding the direction of movement or applying a force at an angle.

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 20 -kg wagon is pulled along the level ground by a rope inclined at \(30^{\circ}\) above the horizontal. A friction force of \(30 \mathrm{~N}\) opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and \((b)\) an acceleration of \(0.40 \mathrm{~m} / \mathrm{s}^{2}\) ?

An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon? (c) What will be its acceleration on the Moon under the action of a \(0.500-\mathrm{N}\) resultant force?

A child pulls on a rope attached to a sled with a force of \(60 \mathrm{~N}\). The rope makes an angle of \(40^{\circ}\) to the ground. (a) Compute the effective value of the pull tending to move the sled along the ground. (b) Compute the force tending to lift the sled vertically. As depicted in Fig. \(3-5\), the components of the \(60 \mathrm{~N}\) force are \(39 \mathrm{~N}\) and \(46 \mathrm{~N}\). ( \(a\) ) The pull along the ground is the horizontal component, 46 N. (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

Two forces act on a point object as follows: \(100 \mathrm{~N}\) at \(170.0^{\circ}\) and \(100 \mathrm{~N}\) at \(50.0^{\circ}\). Find their resultant.

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.