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Chapter 4: Problem 9

Two forces \(\vec{F}_{A}\) and \(\vec{F}_{B}\) are applied to an object whose mass is \(8.0 \mathrm{~kg} .\) The larger force is \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}\). When both forces point due east, the object's acceleration has a magnitude of \(0.50 \mathrm{~m} / \mathrm{s}^{2} .\) However, when \(\overrightarrow{\mathrm{F}}\) s points due east and \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) points due west, the acceleration is \(0.40 \mathrm{~m} / \mathrm{s}^{2},\) due east. Find (a) the magnitude of \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}\) and (b) the magnitude of \(\overrightarrow{\mathrm{F}}\)

Short Answer

Expert verified
(a) \( F_A = 3.6\,\mathrm{N} \), (b) \( F_B = 0.4\,\mathrm{N} \)

Step by step solution

01

Understand the Problem

We have two forces, \(\vec{F}_A\) and \(\vec{F}_B\), acting on an object of mass \(8.0\,\mathrm{kg}\). We are given two scenarios with different acceleration results when these forces act in opposite directions and need to find the magnitudes of \(\vec{F}_A\) and \(\vec{F}_B\).
02

Apply Newton's Second Law for the First Scenario

In the first scenario, both forces \(\vec{F}_A\) and \(\vec{F}_B\) are pointing east, resulting in a net force \(F_1 = F_A + F_B\). The acceleration is given as \(a_1 = 0.50\,\mathrm{m/s^2}\). So, by Newton's second law: \[ F_1 = m \cdot a_1 = 8.0\,\mathrm{kg} \times 0.50\,\mathrm{m/s^2} = 4.0\,\mathrm{N} \]Therefore, \(F_A + F_B = 4.0\,\mathrm{N}\).
03

Apply Newton's Second Law for the Second Scenario

In the second scenario, \(\vec{F}_A\) points east and \(\vec{F}_B\) points west, leading to a net force \(F_2 = F_A - F_B\). The acceleration is \(a_2 = 0.40\,\mathrm{m/s^2}\), with the acceleration also pointing east, so: \[ F_2 = m \cdot a_2 = 8.0\,\mathrm{kg} \times 0.40\,\mathrm{m/s^2} = 3.2\,\mathrm{N} \]Thus, \(F_A - F_B = 3.2\,\mathrm{N}\).
04

Solve the System of Equations

We now have two equations from the scenarios:1. \(F_A + F_B = 4.0\,\mathrm{N}\)2. \(F_A - F_B = 3.2\,\mathrm{N}\)Adding these two equations, we get:\[ 2F_A = 4.0 + 3.2 = 7.2\,\mathrm{N} \]\[ F_A = \frac{7.2}{2} = 3.6\,\mathrm{N} \]Substituting \(F_A = 3.6\,\mathrm{N}\) into the first equation gives:\[ 3.6 + F_B = 4.0 \]\[ F_B = 4.0 - 3.6 = 0.4\,\mathrm{N} \]
05

Conclude with the Magnitudes of Forces

We have found the magnitudes of the forces:- The magnitude of \(\vec{F}_A\) is \(3.6\,\mathrm{N}\).- The magnitude of \(\vec{F}_B\) is \(0.4\,\mathrm{N}\).

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Key Concepts

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

Net Force
To understand the problem and solution better, we start with the concept of net force. In physics, when multiple forces act upon an object, the net force is the vector sum of these forces. It determines the overall movement of the object. In this exercise, the net force is affected by two different configurations of forces applied to the object.
  • In the first scenario, where both forces are in the same direction (east), the net force is simply the sum of both forces: \[ F_1 = F_A + F_B \]
  • In the second scenario, where one force points east and the other west, the net force is the difference between the forces: \[ F_2 = F_A - F_B \]
The direction and magnitude of the net force are critical in determining the resulting acceleration according to Newton's Second Law.
Acceleration
Acceleration is a key concept to grasp when applying Newton's Second Law of motion, as it describes how quickly an object's velocity changes.In this exercise, two scenarios provide different accelerations based on net force influence:
  • Scenario one reports an acceleration of \(0.50\,\mathrm{m/s^2}\) when both forces act together eastward.This scenario's acceleration is a direct result of the net force \(F_1\).
  • Scenario two reports an acceleration of \(0.40\,\mathrm{m/s^2}\) with opposing forces, showing the standalone effect of \(F_A\) plus its conflict with \(F_B\).
These values illustrate that the force configuration changes the accelaration, emphasizing how crucial it is to identify the applied forces precisely when solving problems.
Force Magnitude
Force magnitude refers to the strength of a force without considering its direction. In this exercise, the problem is to find the magnitudes of \(\vec{F}_A\) and \(\vec{F}_B\). We know from the second law of motion, \(F = m \cdot a\), the force acts proportionate to the product of an object's mass and its acceleration. With the mass of the object given as \(8.0\,\mathrm{kg}\), and using the equations derived from each scenario, the magnitudes come up to:
  • \(F_A = 3.6\,\mathrm{N}\)
  • \(F_B = 0.4\,\mathrm{N}\)
Thus, the magnitude of each force tells us how strong each force is acting on the mass to produce the respective accelerations in the given configuration.
System of Equations
To solve the problem and find the magnitudes of the forces, we must form a system of equations from the information given. In the solution process:
  • The first scenario gives us the equation: \[ F_A + F_B = 4.0\,\mathrm{N} \]
  • The second scenario provides: \[ F_A - F_B = 3.2\,\mathrm{N} \]
These two simultaneous equations allow us to solve for two unknowns, \(F_A\) and \(F_B\).By adding and subtracting these equations, we obtain the individual force magnitudes. This method of using a system of equations is an efficient way to handle problems involving multiple forces.

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A \(95.0-\mathrm{kg}\) person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of \(1.80 \mathrm{~m} / \mathrm{s}^{2},\) (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of \(1.30 \mathrm{~m} / \mathrm{s}^{2}\) ?

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