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Chapter 4: Problem 1

Two forces have the same magnitude \(F .\) What is the angle between the two vectors if their sum has a magnitude of (a) 2\(F ?\) (b) \(\sqrt{2} F ?\) (c) zero? Sketch the three vectors in each case.

Short Answer

Expert verified
(a) 0掳, (b) 90掳, (c) 180掳.

Step by step solution

01

Understanding Force Vector Addition

To find the angle between two force vectors with the same magnitude when their sum has a specific magnitude, we will use the formula for the magnitude of the sum of two vectors: \[ |\vec{F}_1 + \vec{F}_2| = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos(\theta)} \]Since both vectors have the same magnitude, let \( F_1 = F_2 = F \). Thus, we simplify the formula to: \[ |\vec{F}_1 + \vec{F}_2| = \sqrt{2F^2 + 2F^2 \cos(\theta)} \]
02

Condition (a): Finding Angle for Sum = 2F

Substitute \(|\vec{F}_1 + \vec{F}_2| = 2F \) in the equation: \[ 2F = \sqrt{2F^2 + 2F^2 \cos(\theta)} \]Squaring both sides gives:\[ 4F^2 = 2F^2 + 2F^2 \cos(\theta) \]Solving for \( \cos(\theta) \): \[ 4F^2 - 2F^2 = 2F^2 \cos(\theta) \] \\[ 2F^2 = 2F^2 \cos(\theta) \] \\[ \cos(\theta) = 1 \] \The angle \( \theta \) is \( 0^\circ \), indicating the vectors are in the same direction.
03

Condition (b): Finding Angle for Sum = \sqrt{2}F

Set \(|\vec{F}_1 + \vec{F}_2| = \sqrt{2}F \) and substitute in the equation:\[ \sqrt{2}F = \sqrt{2F^2 + 2F^2 \cos(\theta)} \]Squaring both sides, we have:\[ 2F^2 = 2F^2 + 2F^2 \cos(\theta) \]Solving for \( \cos(\theta) \):\[ 0 = 2F^2 \cos(\theta) \]\[ \cos(\theta) = 0 \]This indicates that \( \theta = 90^\circ \); the vectors are perpendicular.
04

Condition (c): Finding Angle for Sum = 0

When \(|\vec{F}_1 + \vec{F}_2| = 0 \), the vectors must cancel each other out. Therefore, we apply the equation:\[ 0 = \sqrt{2F^2 + 2F^2 \cos(\theta)} \]Squaring both sides gives:\[ 0 = 2F^2 + 2F^2 \cos(\theta) \]Solving for \( \cos(\theta) \):\[ -2F^2 = 2F^2 \cos(\theta) \]\[ \cos(\theta) = -1 \]Thus, \( \theta = 180^\circ \); the vectors are in opposite directions.

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Key Concepts

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

Understanding Force Vectors
In physics, force vectors are used to describe directional forces experienced by an object. They carry both magnitude and direction, essential for understanding how objects interact. When adding multiple force vectors, their combined effect can vary depending on their respective directions. Each vector is represented as an arrow where:
  • The length of the arrow indicates the force magnitude.
  • The direction of the arrow indicates the force direction.
The addition of force vectors often employs vector geometry principles, creating a resultant vector that represents the total force acting upon the object. Understanding how vector addition works is crucial, especially when determining the magnitude and direction of forces in equilibrium or motion scenarios.
Calculating the Magnitude of Vectors
The magnitude of a vector indicates its size or strength, essentially how much force it carries. For any vector \( \vec{F} \) with components \( (x, y) \), the magnitude \( |\vec{F}| \) is calculated using the formula:
  • \( |\vec{F}| = \sqrt{x^2 + y^2} \)
This formula stems from the Pythagorean theorem, considering vectors as sides of a right triangle. For vectors with equal magnitudes, adjustments through trigonometric relationships, such as finding angles between vectors, can help derive results like in our example, where specific conditions lead to unique outcomes:
  • If the sum of two equal vectors is twice their magnitude, they align perfectly, reinforcing each other.
  • If the sum is zero, the vectors fully oppose one another, canceling each other out.
  • A sum of \( \sqrt{2} \) times the magnitude implies a perpendicular relationship, creating a right angle.
Determining the Angle Between Vectors
The angle between two vectors is a critical parameter determining how they interact. Calculating this requires using the cosine rule, where the angle \( \theta \) between two vectors \( \vec{F}_1 \) and \( \vec{F}_2 \) with equal magnitudes \( F \) follows:
  • \( |\vec{F}_1 + \vec{F}_2|^2 = 2F^2 + 2F^2 \cos(\theta) \)
Solving for the cosine of the angle gives insights into their relationship:
  • If \( \cos(\theta) = 1 \), then \( \theta = 0^\circ \). The vectors point in the same direction, maximizing their combined effect.
  • If \( \cos(\theta) = 0 \), then \( \theta = 90^\circ \). The vectors are perpendicular, resulting in the smallest resultant magnitude for non-opposing forces.
  • If \( \cos(\theta) = -1 \), then \( \theta = 180^\circ \). They oppose each other completely, nullifying the resultant force.
Exploring Vector Geometry
Vector geometry offers a visual and mathematical framework for understanding how vectors behave in space. It involves not only magnitudes and directions but also the relationships between multiple vectors when combined. Through graphical representation, vectors can be added tip-to-tail:
  • Place the tail of the second vector at the tip of the first.
  • The resultant vector spans from the tail of the first to the tip of the second.
In exercises like finding the resultant of two equal force vectors under different conditions, vector geometry aids in visualizing outcomes. Sketching vectors:
  • Helps understand alignment for \( \theta = 0^\circ \) with maximum addition.
  • Showcases the perpendicular arrangement at \( \theta = 90^\circ \) for minimal non-zero resultant.
  • Clearly contrasts opposite vectors at \( \theta = 180^\circ \) when they cancel each other.
Visual aids and geometric interpretations solidify the understanding of vector addition, critical when examining forces and movements in physics.

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Most popular questions from this chapter

\(\bullet$$\bullet$$\bullet\) A hot-air balloon consists of a basket, one passenger, and some cargo. Let the total mass be \(M .\) Even though there is an upward lift force on the balloon, the balloon is initially accelerating downward at a rate of \(g / 3 .\) (a) Draw a free-body diagram for the descending balloon. (b) Find the upward lift force in terms of the initial total weight \(M g .\) (c) The passenger notices that he is heading straight for a waterfall and decides he needs to go up. What fraction of the total weight must he drop overboard so that the balloon accelerates upward at a rate of \(g / 2 ?\) Assume that the upward lift force remains the same.

BIO Human Biomechanics. The fastest served tennis ball, served by "Big Bill" Tilden in \(1931,\) was measured at 73.14 \(\mathrm{m} / \mathrm{s} .\) The mass of a tennis ball is \(57 \mathrm{g},\) and the ball is typically in contact with the tennis racquet for \(30.0 \mathrm{ms},\) with the ball starting from rest. Assuming constant acceleration, (a) what force did Big Bill's racquet exert on the tennis ball if he hit it essentially horizontally? (b) Draw free-body diagrams of the tennis ball during the serve and just after it moved free of the racquet.

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \(\mathrm{m} / \mathrm{s}^{2}\) . How much horizontal force must a 55 -kg sprinter exert on the starting blocks during a start to produce this acceleration? Which body exerts the force that propels the sprinter: the blocks or the sprinter herself?

A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 \(\mathrm{kN}\) from its engines slows it down at a rate of \(1.20 \mathrm{m} / \mathrm{s}^{2},\) but it speeds up at a rate of 0.80 \(\mathrm{m} / \mathrm{s}^{2}\) with an upward thrust of 10.0 \(\mathrm{kN}\) (a) In each case, what is the direction of the acceleration of the spacecraft? (b) Draw a free- body diagram for the spacecraft. In each case, speeding up or slowing down, what is the direction of the net force on the spacecraft? (c) Apply Newton's second law to each case, slowing down or speeding up, and use this to find the spacecraft's weight near the surface of Planet X.

Two horses pull horizontally on ropes attached to a stump. The two forces \(\vec{F}_{1}\) and \(\vec{F}_{2}\) that they apply to the stump are such that the net (resultant) force \(\vec{R}\) has a magnitude equal to that of \(\vec{\boldsymbol{F}}_{1}\) and makes an angle of \(90^{\circ}\) with \(\vec{\boldsymbol{F}}_{1 .}\) Let \(F_{1}=1300 \mathrm{N}\) and \(R=1300 \mathrm{N}\) also. Find the magnitude of \(\vec{\boldsymbol{F}}_{2}\) and its direction \((\mathrm{rela}-\)tive to \(\vec{\boldsymbol{F}}_{1} ).\)
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