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Chapter 4: Problem 35

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 \(\vec{\boldsymbol{F}}_{1} ).\)

Short Answer

Expert verified
Magnitude of \( \vec{F}_2 \) is 1300 N; direction is 90 degrees relative to \( \vec{F}_1 \).

Step by step solution

01

Identify Given Information

We are given that the magnitudes of forces \( \vec{F}_1 \) and \( \vec{R} \) (the resultant force) are both 1300 N. The angle between \( \vec{F}_1 \) and \( \vec{R} \) is 90 degrees.
02

Understand Resultant Force

The fact that the resultant force \( \vec{R} \) forms a 90 degree angle with \( \vec{F}_1 \) implies that \( \vec{F}_2 \) must be acting perpendicular to \( \vec{F}_1 \) to achieve this equilibrium.
03

Calculate the Magnitude of \( \vec{F}_2 \)

Since \( \vec{R} \) is perpendicular to \( \vec{F}_1 \) and \( R = 1300 \) N, by applying the Pythagorean Theorem, the magnitude of \( \vec{F}_2 \) is equal to \( R \). Therefore, \( F_2 = R = 1300 \) N.
04

Determine \( \vec{F}_2 \)'s Direction

\( \vec{F}_2 \) must be perpendicular to \( \vec{F}_1 \) because \( \vec{R} \) is at a 90-degree angle to \( \vec{F}_1 \). Hence, the direction of \( \vec{F}_2 \) is 90 degrees relative to \( \vec{F}_1 \).

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Key Concepts

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

Vector Addition
When dealing with multiple forces acting on an object, like the forces from two horses pulling on a stump, vector addition enables us to compute the total force acting on it.
  • Vectors have both a magnitude and a direction. Forces are vectors as they not only apply a push or pull but also have a specific direction.
  • To find the total force, also known as the resultant force, we need to add these vectors together. However, this isn't as simple as adding numbers, because direction matters.
  • In two dimensions, we typically use methods like the parallelogram rule or the triangle rule to find the resultant vector.
In this specific example with the two horses, the forces exerted by both horses are vectors that we add together to get a single resultant vector that tells us the net effect of their efforts on the stump. Understanding vector addition is crucial because it allows us to predict how forces combine, making it a foundational concept in physics.
Resultant Force
The resultant force is the single force that represents the combined effect of multiple forces acting on an object. For the stump being pulled by the two horses:
  • The resultant force, denoted as \( \vec{R} \), simplifies the complex interaction of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \).
  • Since the forces are vectors, their directions contribute to the resultant force, not just their magnitudes.
  • In the problem, the resultant force \( \vec{R} \) was given a magnitude equal to \( \vec{F}_1 \), i.e., 1300 N, while forming a 90-degree angle with it.
The significance of the resultant force is in predicting motion: if \( \vec{R} \) were zero, the stump would remain at equilibrium, experiencing balanced forces. In this exercise, the angle between the resultant force and one of the forces reflects how one force modifies the direction of the other.
Perpendicular Forces
Perpendicular forces are forces that act at right angles (90 degrees) to one another. In our example with the two horses, the forces \( \vec{F}_1 \) and \( \vec{F}_2 \) were perpendicular:
  • This relationship is significant because perpendicular vectors have unique properties that can simplify the calculation of the resultant force.
  • When forces are perpendicular, such as \( \vec{F}_2 \) to \( \vec{F}_1 \), the Pythagorean theorem can be used effectively: \( R = \sqrt{F_1^2 + F_2^2} \).
  • However, in this scenario, since both \( R \) and \( F_1 \) are equal at 1300 N, it implies that \( \vec{F}_2 \) is also 1300 N to maintain equilibrium.
The use of perpendicular forces here also helps in determining the direction of \( \vec{F}_2 \), as it needs to be at a right angle to \( \vec{F}_1 \). Such concepts are essential as they allow us to analyze and resolve forces efficiently, especially in mechanical systems.

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Most popular questions from this chapter

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?

At the surface of Jupiter's moon Io, the acceleration due to gravity is \(g=1.81 \mathrm{m} / \mathrm{s}^{2} .\) A watermelon weighs 44.0 \(\mathrm{N}\) at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? (b) What are its mass and weight on the surface of Io?

An astronaut's pack weighs 17.5 \(\mathrm{N}\) when she is on earth but only 3.24 \(\mathrm{N}\) when she is at the surface of an asteroid. (a) What is the acceleration due to gravity on this asteroid? (b) What is the mass of the pack on the asteroid?

\(\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.

You walk into an elevator, step onto a scale, and push the "up" button. You also recall that your normal weight is 625 \(\mathrm{N}\) . Start answering each of the following questions by drawing a free-body diagram. (a) If the elevator has an acceleration of magnitude \(2.50 \mathrm{m} / \mathrm{s}^{2},\) what does the scale read? (b) If you start holding a 3.85 -kg package by a light vertical string, what will be the tension in this string once the elevator begins accelerating?
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