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Chapter 4: Problem 30

A 650-N force acts in a northwesterly direction. A second 650-N force must be exerted in what direction so that the resultant of the two forces points westward? Illustrate your answer with a vector diagram.

Short Answer

Expert verified
F2 must be 45掳 south of west.

Step by step solution

01

Understanding the Problem

We have two forces - The first force (\( F_1 \)) is 650 N, acting in a northwesterly direction. - The second force (\( F_2 \)) is 650 N, required such that the resultant (\( R \)) of these two forces points directly west.Our goal is to determine the direction in which \( F_2 \) should act.
02

Visualizing the Forces

In a vector diagram, a force held at 45掳 north of west points northwest.Since \( F_1 \) is 650 N northwest, it makes a 45掳 angle above the west axis. We need to find \( F_2 \) such that the resultant of \( F_1 \) and \( F_2 \) points directly west.
03

Using Components to Solve

Decompose \( F_1 \) into its components. \( F_1 \) can be represented as:- The horizontal component (\( F_{1x} \)) = 650 \( \times \) cos(45掳) = 650 \( \times \) \( \frac{\sqrt{2}}{2} \)- The vertical component (\( F_{1y} \)) = 650 \( \times \) sin(45掳) = 650 \( \times \) \( \frac{\sqrt{2}}{2} \)Since the resultant needs to point west (no vertical component), the vertical components of \( F_1 \) and \( F_2 \) should cancel each other out.
04

Calculating Required Direction for F2

Given that \( F_{1y} = 650 \times \frac{\sqrt{2}}{2} \), \( F_2 \) must have an opposite vertical component to cancel it out, meaning:\[ F_{2y} = -650 \times \frac{\sqrt{2}}{2} \]Thus, \( F_2 \) must also be at 45掳 south of west to have the vertical component in the opposite direction.
05

Conclusion and Diagram

In order for the resultant force to be perfectly westward, \( F_2 \) must act 45掳 south of west. Both \( F_1 \) and \( F_2 \) will then contribute their horizontal components to produce a resultant pointing due west.[Insert diagram here illustrating both vectors, \( F_1 \) as northwest and \( F_2 \) as southwest, canceling out vertical components]

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Key Concepts

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

Vector Components
To fully grasp vector addition, it's important to understand vector components. Vectors are represented as arrows that point in a direction. They have both magnitude (how long the arrow is) and direction (where the arrow points).
A vector can be split into parts that show its impact in different directions, usually along the x-axis (horizontal) and y-axis (vertical).
  • For example, if you have a vector pointing northwest, you can break it down into two components.
  • The horizontal component (x) shows how much of the vector is pointing west, and the vertical component (y) shows how much is pointing north.
  • Mathematically, if the vector is at a 45掳 angle, both components will be equal because of the symmetric angle.
Calculating these components involves using trigonometric functions:
  • Horizontal component: \( F_{x} = F \times \cos(\theta) \)
  • Vertical component: \( F_{y} = F \times \sin(\theta) \)
Breaking down vectors into components simplifies calculations, particularly when adding them together to find a resultant vector.
Resultant Vector
The resultant vector is the single vector that has the same effect as applying multiple vectors at once.
Think of it as a summary vector that tells us the total effect of combining other vectors.
In this scenario, you have two forces (vectors) acting on an object. You need to find where these two forces would lead the object, which is shown by the resultant vector.
  • Imagine adding two force vectors. The resultant is a vector that shows the net force.
  • The direction of the resultant depends on both the individual directions and magnitudes of the forces involved.
  • In our exercise, the goal was to make the resultant point directly west.
By calculating how each vector contributes to a specific direction (using components), one can determine the needed direction for another vector to achieve a desired resultant. Here, aligning both the vectors to have their net effect purely in the west direction was key.
Force Decomposition
Force decomposition is the process of breaking a vector, like a force, into perpendicular components. It is essential for simplifying physics problems, where resolving forces into components helps in analyzing different effects.
In this exercise, the first force was split into north and west components. This helps to determine what the second force needs to do.
  • The first force, acting northwest, is decomposed into north (vertical) and west (horizontal) components.
  • By analyzing its vertical component, the needed direction of the second vector emerged.
  • The second force had to completely counter the northward pull of the first, using a south direction.
This approach ensures that the combined vertical components of the forces cancel out, leaving a net horizontal force that leads west. Force decomposition isn't just about solving specific problems. It's a powerful tool for understanding how different forces interact in any system.

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Most popular questions from this chapter

What average force is required to stop a \(950-\mathrm{kg}\) car in \(8.0 \mathrm{~s}\) if the car is traveling at \(95 \mathrm{~km} / \mathrm{h} ?\)

A \(1.5-\mathrm{kg}\) block rests on top of a \(7.5-\mathrm{kg}\) block (Fig. \(4-63)\). The cord and pulley have negligible mass, and there is no significant friction anywhere. ( \(a\) ) What force \(F\) must be applied to the bottom block so the top block accelerates to the right at \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) ? (b) What is the tension in the connecting cord?

(II) At the instant a race began, a \(65-\mathrm{kg}\) sprinter exerted a force of 720 \(\mathrm{N}\) on the starting block at a \(22^{\circ}\) angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? \((b)\) If the force was exerted for \(0.32 \mathrm{s},\) with what speed did the sprinter leave the starting block?

(II) A 27 -kg chandelier hangs from a ceiling on a vertical \(4.0-\mathrm{m}\) -long wire. (a) What horizontal force would be neces- sary to displace its position 0.15 \(\mathrm{m}\) to one side? (b) What will be the tension in the wire?

A \(7650-\mathrm{kg}\) helicopter accelerates upward at 0.80 \(\mathrm{m} / \mathrm{s}^{2}\) while lifting a 1250 -kg frame at a construction site, Fig. \(59 .\) (a) What is the lift force exerted by the air on the helicopter rotors? (b) What is the tension in the cable (ignore its mass) that connects the frame to the helicopter? (c) What force does the cable exert on the helicopter?
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