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

Short Answer

Expert verified
(a) \( \overrightarrow{F}_A = 3.6 \text{ N} \), (b) \( \overrightarrow{F}_B = 0.4 \text{ N} \).

Step by step solution

01

Apply Newton's Second Law

When both forces point due east, the net force is the sum of both forces: \( \overrightarrow{F}_{net,1} = \overrightarrow{F}_A + \overrightarrow{F}_B \). According to Newton's Second Law, \( \overrightarrow{F}_{net,1} = m \cdot a_1 \), where \( m = 8.0 \, \text{kg} \) and \( a_1 = 0.50 \, \text{m/s}^2 \). Substitute these values to find \( \overrightarrow{F}_{net,1} = 8.0 \text{kg} \times 0.50 \, \text{m/s}^2 = 4.0 \, \text{N} \). Therefore, \( \overrightarrow{F}_A + \overrightarrow{F}_B = 4.0 \, \text{N} \).
02

Apply Newton's Second Law for Opposite Directions

When \( \overrightarrow{F}_A \) points east and \( \overrightarrow{F}_B \) points west, the net force is the difference: \( \overrightarrow{F}_{net,2} = \overrightarrow{F}_A - \overrightarrow{F}_B \). By Newton's Second Law, \( \overrightarrow{F}_{net,2} = m \cdot a_2 \), where \( a_2 = 0.40 \, \text{m/s}^2 \). Substitute these values to find \( \overrightarrow{F}_{net,2} = 8.0 \text{kg} \times 0.40 \, \text{m/s}^2 = 3.2 \, \text{N} \). Therefore, \( \overrightarrow{F}_A - \overrightarrow{F}_B = 3.2 \, \text{N} \).
03

Solve the System of Equations

We now have two equations:1. \( \overrightarrow{F}_A + \overrightarrow{F}_B = 4.0 \, \text{N} \)2. \( \overrightarrow{F}_A - \overrightarrow{F}_B = 3.2 \, \text{N} \)Add the two equations to eliminate \( \overrightarrow{F}_B \):\[ 2\overrightarrow{F}_A = 7.2 \text{N} \]Thus, \( \overrightarrow{F}_A = 3.6 \, \text{N} \).
04

Find the Magnitude of Force B

Substitute \( \overrightarrow{F}_A = 3.6 \, \text{N} \) back into the first equation:\( 3.6 \, \text{N} + \overrightarrow{F}_B = 4.0 \, \text{N} \).Solve for \( \overrightarrow{F}_B \):\[ \overrightarrow{F}_B = 4.0 \, \text{N} - 3.6 \, \text{N} = 0.4 \, \text{N} \].

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Key Concepts

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

Force and Acceleration
In the realm of physics, force and acceleration are deeply linked through Newton's Second Law of Motion. This fundamental principle tells us that the force applied to an object is equal to the mass of the object multiplied by the acceleration that the force causes: \( F = m \cdot a \).

This equation clearly shows that acceleration is directly proportional to the force when the mass remains constant. Here are a few key points to understand:
  • When you increase the force applied to an object, the acceleration increases if the mass stays the same.
  • A larger mass requires a larger force to achieve the same acceleration as a smaller mass.
  • If there's no force acting on an object, then according to Newton, there will be no acceleration.
In our problem, we observe these dynamics perfectly. With a mass of 8 kg, when the sum of forces acting due east was 4 N, the acceleration was 0.50 m/s² in the same direction. Similarly, when the net force was less, at 3.2 N, the acceleration also decreased to 0.40 m/s². This beautifully illustrates how acceleration alters with changes in the net applied force.
Vector Addition
When dealing with forces, especially in physics problems like this one, understanding vector addition is essential. Forces are vector quantities. This means they have both a magnitude and a direction. Simply put, vector addition involves both adding up the magnitudes of vectors and considering their directions.

In scenarios involving multiple forces, you may need to:
  • Add vectors that are aligned in the same direction, like in step 1 of the solution, where forces pointed eastward, so we add all their magnitudes.
  • Subtract vectors that have opposite directions, like in step 2, where one vector pointed east and another west, and we worked with the difference in their magnitudes.
  • Consider the overall effect of several forces acting on an object, identifying the resultant force which dictates the acceleration.
The concept of vector addition is crucial for accurately calculating the net force and hence the acceleration of an object. In our example, when both forces were directed east, the net force was simply their sum. However, when they opposed each other, the net force needed a subtraction to account for their opposing directions.
System of Equations
Systems of equations are a cornerstone of solving coupled problems in physics, especially concerning different forces at play. Here, we had two separate instances with the same forces but different net outcomes, leading us to develop and solve two equations. These equations need to be solved simultaneously to find individual forces in complex scenarios.

In this exercise:
  • We established two separate equations based on different force alignments.
  • The first equation summarized forces pointing in the same direction.
  • The second equation summarized forces in opposing directions.
To solve these, we use simple addition and subtraction methods to eliminate one of the unknowns, which allows us to find the values of each force specifically. When we added the two equations together, it simplified to solve for
\( \overrightarrow{F}_A \) directly. With that found, it was just a matter of substitution to uncover \( \overrightarrow{F}_B \). This method is efficient and essential for many physics and engineering problems where two or more forces interact.

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Most popular questions from this chapter

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of \(151 \mathrm{N}\) on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{N} .\) Find the magnitude and direction of the tension that the right section of the wire sustains.

A 95.0-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 \(1.30 \mathrm{m} / \mathrm{s}^{2} ?\)

Mars has a mass of \(6.46 \times 10^{23} \mathrm{kg}\) and a radius of \(3.39 \times 10^{6} \mathrm{m}\). (a) What is the acceleration due to gravity on Mars? (b) How much would a \(65-\mathrm{kg}\) person weigh on this planet?

A Mercedes-Benz \(300 \mathrm{SL}\) ( \(m=1700 \mathrm{kg}\) ) is parked on a road that rises \(15^{\circ}\) above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is \(915 \mathrm{N}\) and the drag force has a magnitude of \(1027 \mathrm{N}\). The mass of the sky diver is \(93.4 \mathrm{kg}\). What are the magnitude and direction of his acceleration?

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