/*! 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 34 A force vector has a magnitude o... [FREE SOLUTION] | 91Ó°ÊÓ

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A force vector has a magnitude of 575 newtons and points at an angle of \(36.0^{\circ}\) below the positive \(x\) axis. What are (a) the \(x\) scalar component and (b) the \(y\) scalar component of the vector?

Short Answer

Expert verified
(a) 465.175 N, (b) -337.485 N.

Step by step solution

01

Understanding the Problem

We have a force vector with a magnitude of 575 N that points 36.0° below the positive x-axis. We need to find the scalar components of this vector along the x-axis and the y-axis.
02

Determine the Angle with the Positive X-axis

Since the vector points below the positive x-axis, the angle the vector makes with the positive x-axis is still 36.0°.
03

Calculate the X-component

The x-component of a vector is calculated using the formula \( F_x = F \cdot \cos(\theta) \), where \( \theta \) is the angle with the positive x-axis. Here, \( F = 575 \ \text{N} \) and \( \theta = 36.0^{\circ} \). Therefore, \( F_x = 575 \ \text{N} \cdot \cos(36.0^{\circ}) \).
04

Compute Cosine of Angle

Calculate \( \cos(36.0^{\circ}) \), which is approximately 0.8090.
05

Calculating the X-component

Use the cosine value to find the x-component: \( F_x = 575 \ \text{N} \times 0.8090 \approx 465.175 \ \text{N} \).
06

Calculate the Y-component

The y-component of a vector can be found using \( F_y = F \cdot \sin(\theta) \). Since the vector points below the x-axis, \( F_y \) will be negative. Thus, \( F_y = -575 \ \text{N} \cdot \sin(36.0^{\circ}) \).
07

Compute Sine of Angle

Calculate \( \sin(36.0^{\circ}) \), which is approximately 0.5878.
08

Calculating the Y-component

Use the sine value to find the y-component: \( F_y = -575 \ \text{N} \times 0.5878 \approx -337.485 \ \text{N} \).

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Key Concepts

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

Force Magnitude
Every vector in physics, including force vectors, has a magnitude. A magnitude simply refers to the size or the amount of the vector, without considering its direction. In this exercise, the force magnitude is given as 575 newtons. When solving problems involving force vectors, it's important to start with the magnitude since it helps define how strong the force is.

Force is measured in newtons (N) and represents how much effect a vector can exert. To fully understand a vector, consider both its magnitude and direction. The magnitude provides the necessary information for calculating vector components using trigonometric functions.
Trigonometric Functions in Physics
Trigonometric functions, specifically sine and cosine, are crucial when it comes to analyzing vectors in physics. They help break down a vector into its components. This is especially useful in solving problems where you need to determine the influence of a vector along the axes of a coordinate system.

  • **Cosine (\(\cos\(\theta\)\))**: Used to find the component of a vector along the horizontal axis (x-axis). When you need to know how much of the vector's magnitude affects the x-axis, cosine comes into play.
  • **Sine (\(\sin\(\theta\)\))**: Used for the vertical axis (y-axis). Sine shows how much of the force acts or is projected along the y-axis.
Cosine and sine functions use the angle made with the x-axis to calculate the respective components. In this exercise, \(\theta = 36.0^{\circ} \), which you use with these trigonometric functions to determine how the force spreads along the axes.

For a vector component problem, mastering trigonometric functions will help you effectively determine components, ensuring accurate calculations. You often encounter these functions in physics exercises dealing with forces.
Angle of Vector
Understanding the angle of a vector is critical. This angle, measured with respect to the positive x-axis, helps in resolving a vector into its components. In the given problem, the vector makes an angle of \(36.0^{\circ}\) below the positive x-axis.

This angle is significant because:
  • It determines the direction of the vector. A vector angled below the x-axis suggests a negative y-component.
  • It influences the calculations of both x and y components using cosine and sine functions.
When working with vectors, always carefully analyze which direction the angle suggests:
  • An angle above the x-axis generally yields a positive y-component.
  • An angle below signals a negative y-component, as seen in this exercise.
Establishing the angle correctly ensures that vectors are properly decomposed into their respective axes, leading to precise component identification and application.

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Most popular questions from this chapter

A force vector points at an angle of \(52^{\circ}\) above the \(+x\) axis. It has a \(y\) component of +290 newtons. Find (a) the magnitude and (b) the \(x\) component of the force vector.

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?

The route followed by a hiker consists of three displacement vectors \(\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}},\) and \(\overrightarrow{\mathrm{C}} .\) Vector \(\overrightarrow{\mathbf{A}}\) is along a measured trail and is \(1550 \mathrm{m}\) in a direction \(25.0^{\circ}\) north of east. Vector \(\overrightarrow{\mathbf{B}}\) is not along a measured trail, but the hiker uses a compass and knows that the direction is \(41.0^{\circ}\) east of south. Similarly, the direction of vector \(\overrightarrow{\mathbf{C}}\) is \(35.0^{\circ}\) north of west. The hiker ends up back where she started. Therefore, it follows that the resultant displacement is zero, or \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=0 .\) Find the magnitudes of (a) vector \(\overrightarrow{\mathbf{B}}\) and \((\mathrm{b})\) vector \(\overrightarrow{\mathbf{C}}\).

A monkey is chained to a stake in the ground. The stake is \(3.00 \mathrm{m}\) from a vertical pole, and the chain is \(3.40 \mathrm{m}\) long. How high can the monkey climb up the pole?

Three deer, \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) are grazing in a field. Deer \(\mathrm{B}\) is located \(62 \mathrm{m}\) from deer \(\mathrm{A}\) at an angle of \(51^{\circ}\) north of west. Deer \(\mathrm{C}\) is located \(77^{\circ}\) north of east relative to deer A. The distance between deer \(\mathrm{B}\) and \(\mathrm{C}\) is \(95 \mathrm{m}\). What is the distance between deer A and C? (Hint: Consider the law of cosines given in Appendix E.

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