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Chapter 1: Problem 40

Your friend has slipped and fallen. To help her up, you pull with a force \(\overrightarrow{\mathbf{F}}\), as the drawing shows. The vertical component of this force is 130 newtons, and the horizontal component is 150 newtons. Find (a) the magnitude of \(\overrightarrow{\mathbf{F}}\) and \((\mathrm{b})\) the angle \(\theta.\)

Short Answer

Expert verified
The magnitude is approximately 198.49 N and the angle is about 41.19掳.

Step by step solution

01

Understand the Components

The problem provides two components of the force \(\overrightarrow{\mathbf{F}}\): a vertical component \(F_v = 130\) newtons and a horizontal component \(F_h = 150\) newtons. We'll use these components to find the magnitude of the force and the angle it makes with the horizontal.
02

Calculate the Magnitude

The magnitude of \(\overrightarrow{\mathbf{F}}\) can be found using the Pythagorean theorem: \( F = \sqrt{F_h^2 + F_v^2} \). Substitute the values: \( F = \sqrt{150^2 + 130^2} \).
03

Perform the Calculation

Calculate \( F = \sqrt{150^2 + 130^2} = \sqrt{22500 + 16900} = \sqrt{39400} \). Therefore, the magnitude of the force is approximately \( 198.49 \) newtons.
04

Determine the Angle

The angle \(\theta\) can be found using the tangent function: \( \tan(\theta) = \frac{F_v}{F_h} = \frac{130}{150} \). Then, \( \theta = \tan^{-1}(\frac{130}{150}) \).
05

Calculate the Angle

Compute \( \theta = \tan^{-1}(\frac{130}{150}) \approx 41.19^\circ \). This is the angle between the force \(\overrightarrow{\mathbf{F}}\) and the horizontal.

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Key Concepts

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

Force Components
When considering a vector, such as a force, it is essential to understand the concept of components. A vector can have both a vertical and a horizontal component, which are projections of the vector along the vertical and horizontal axes, respectively.
  • The vertical component of the force, noted as \( F_v \), acts in the up-and-down direction.
  • The horizontal component, denoted \( F_h \), acts side-to-side against a reference direction.
Breaking down a force into these components helps in analyzing and understanding physical problems better, as you can study each component separately before considering their combined effect.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, often used to relate the sides of a right triangle. In vector analysis, this theorem helps calculate the magnitude of a vector when its components are known. To use this theorem:
  • Imagine a right triangle where the magnitude of the force is the hypotenuse, and the force components are the legs.
  • The theorem is expressed as: \( a^2 + b^2 = c^2 \), where \( a \) and \( b \) are the legs, and \( c \) is the hypotenuse.
In our exercise, the components are 150 and 130 newtons, allowing us to find the vector magnitude using \( F = \sqrt{F_h^2 + F_v^2} \). This application confirms the power of the Pythagorean theorem in solving vector problems.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent help connect angles in a triangle to its sides' ratios, providing a powerful tool in vector analysis. In our exercise, the tangent function is particularly useful. It is defined as:
  • \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \)
Regarding the angle \( \theta \), opposite to it is the vertical component \( F_v = 130 \) newtons, and adjacent is the horizontal component \( F_h = 150 \) newtons. By applying the tangent function, we can derive the angle \( \theta \) through \( \theta = \tan^{-1}\left(\frac{F_v}{F_h}\right) \).
Magnitude of a Vector
The magnitude of a vector represents its total size or strength in a given direction. It is crucial in understanding how powerful a force is when both its components are considered.
  • To find it, combine the effects of the horizontal and vertical components.
  • In vector analysis, the magnitude is typically found using the Pythagorean theorem: \( F = \sqrt{F_h^2 + F_v^2} \).
In practice, calculate the magnitude by substituting the given values of the components, resulting in a single scalar quantity that summarizes the vector's impact, which in this exercise is approximately 198.49 newtons.
Angle Calculation
Calculating the angle a vector makes with a reference direction is often vital in physics and engineering. This angle gives insight into the vector's orientation and can be critical in applications requiring precise directionality. In this context, the angle \( \theta \) can be determined using:
  • Trigonometric identities like tangent, where \( \theta = \tan^{-1}\left(\frac{F_v}{F_h}\right) \).
Given our example, where \( F_v = 130 \) newtons and \( F_h = 150 \) newtons, the calculation simplifies to finding the arctan of the ratio between these components, resulting in an angle of approximately \( 41.19^\circ \), indicating the vector's inclination relative to the horizontal axis.

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Most popular questions from this chapter

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A jogger travels a route that has two parts. The first is a displacement \(\overrightarrow{\mathbf{A}}\) of \(2.50 \mathrm{km}\) due south, and the second involves a displacement \(\overrightarrow{\mathbf{B}}\) that points due east. (a) The resultant displacement \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) has a magnitude of \(3.75 \mathrm{km} .\) What is the magnitude of \(\overrightarrow{\mathbf{B}},\) and what is the direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) relative to due south? (b) Suppose that \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) had a magnitude of \(3.75 \mathrm{km}\). What then would be the magnitude of \(\overrightarrow{\mathbf{B}},\) and what is the direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) relative to due south?
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