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Chapter 4: Problem 9

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

Short Answer

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

Step by step solution

01

Analyze the scenario with both forces pointing east

When both forces \( \overrightarrow{F}_A \) and \( \overrightarrow{F}_B \) act in the same direction (east), their magnitudes add up. Using Newton's second law \( F = ma \), where \( m = 8.0 \mathrm{~kg} \) and \( a = 0.50 \mathrm{~m/s^2} \), the net force is given by:\[ F_A + F_B = m \cdot a = 8.0 \cdot 0.50 = 4.0 \mathrm{~N} \]
02

Analyze the scenario with opposite forces

Now consider \( \overrightarrow{F}_A \) pointing east and \( \overrightarrow{F}_B \) pointing west. The effective force is the difference \( F_A - F_B \). Using \( a = 0.40 \mathrm{~m/s^2} \):\[ F_A - F_B = m \cdot a = 8.0 \cdot 0.40 = 3.2 \mathrm{~N} \]
03

Solve the system of equations

We now have two equations:1. \( F_A + F_B = 4.0 \mathrm{~N} \)2. \( F_A - F_B = 3.2 \mathrm{~N} \)Add these equations to eliminate \( F_B \):\[ (F_A + F_B) + (F_A - F_B) = 4.0 + 3.2 \]\[ 2F_A = 7.2 \mathrm{~N} \]\[ F_A = 3.6 \mathrm{~N} \]
04

Substitute to find \( F_B \)

Use \( F_A = 3.6 \mathrm{~N} \) in one of the original equations to find \( F_B \). Substituting in equation 1:\[ 3.6 + F_B = 4.0 \mathrm{~N} \]\[ F_B = 4.0 - 3.6 = 0.4 \mathrm{~N} \]

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Key Concepts

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

Net Force
The net force on an object is the combination of all the forces acting upon it. According to Newton's Second Law, the net force is crucial for determining how the object will accelerate. Whenever we know the mass and the acceleration, we can find the net force using the equation:
\[ F_{net} = m \times a \]
  • Eastward Forces: When both forces are pointing east, the forces add up. For the given exercise, the object鈥檚 mass is \(8.0 \text{ kg}\) with an acceleration of \(0.50 \text{ m/s}^2\), leading to a net force:\[ F_A + F_B = 8.0 \times 0.50 = 4.0 \text{ N} \]
  • Opposite Forces: When one force points east and the other west, they work against each other. The larger force subtracts the smaller one. With an acceleration of \(0.40 \text{ m/s}^2\), the net force equation becomes:\[ F_A - F_B = 8.0 \times 0.40 = 3.2 \text{ N} \]
The understanding of net force helps to deduce which forces contribute to motion and in which direction.
Acceleration Calculation
Acceleration is the rate at which an object鈥檚 velocity changes with time. It鈥檚 directly proportional to the net force acting upon an object and inversely proportional to the object鈥檚 mass. The formula \( a = \frac{F_{net}}{m} \) can be used to calculate acceleration.
For instance:
  • When forces were aligned (both eastwards), the object's acceleration was found using the forces鈥� sum which led to: \[ a = \frac{4.0}{8.0} = 0.50 \text{ m/s}^2 \]
  • When the forces opposed each other, the acceleration was computed from the force difference: \[ a = \frac{3.2}{8.0} = 0.40 \text{ m/s}^2 \]
These calculations affirm that understanding acceleration requires knowing both force and mass.
Force Decomposition
Force decomposition involves breaking down a force into its perpendicular components. While in this problem, forces weren't decomposed into perpendicular components, recognizing this concept helps in scenarios involving angles.
  • Forces acting in the same or opposite linear directions, like EAST-WEST, remain in their one-dimensional lines.
  • For forces applied at an angle, you break them into vertical and horizontal components using trigonometry.
In the given case, the decomposition isn't visual because forces are applied linearly, but understanding it sets the foundation for more complex physics problems involving vectors.
System of Equations
A system of equations is used when two or more equations are involved to find a common solution. In this exercise, we handled equations representing the net force in two situations.
The system we tackled comprised two equations:
  • \( F_A + F_B = 4.0 \text{ N} \)
  • \( F_A - F_B = 3.2 \text{ N} \)
To solve these:
  • Add the equations to eliminate \( F_B \), which gave: \[ 2F_A = 7.2 \text{ N} \]
  • Solving for \( F_A \): \[ F_A = 3.6 \text{ N} \]
  • Substitute \( F_A \) into one equation to find \( F_B \): \[ F_B = 0.4 \text{ N} \]
Understanding and solving systems of equations is crucial, not just in physics, but in any discipline that needs finding unknowns from multiple interrelated variables.

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Most popular questions from this chapter

A \(5.00\) -kg block is placed on top of a \(12.0-\mathrm{kg}\) block that rests on a frictionless table. The coefficient of static friction between the two blocks is \(0.600\). What is the maximum horizontal force that can be applied before the \(5.00-\mathrm{kg}\) block begins to slip relative to the \(12.0-\mathrm{kg}\) block, if the force is applied to (a) the more massive block and (b) the less massive block?

Problem 77 The drawing shows Robin Hood (mass \(=77.0 \mathrm{~kg}\) ) about to escape from a dangerous situation. With one hand, he is gripping the rope that holds up a chandelier (mass \(=195\) kg). When he cuts the rope where it is tied to the floor, the chandelier will fall, and he will be pulled up toward a balcony above. Ignore the friction between the rope and the beams over which it slides, and find (a) the acceleration with which Robin is pulled upward and (b) the tension in the rope while Robin escapes.

A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is \(80.0 \mathrm{~kg},\) and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

The weight of the block in the drawing is \(88.9 \mathrm{~N}\). The coefficient of static friction between the block and the vertical wall is \(0.560 .\) (a) What minimum force \(\vec{F}\) is required to prevent the block from sliding down the wall? (Hint: The static frictional force exerted on the block is directed upward, parallel to the wall.) (b) What minimum force is required to start the block moving up the wall? (Hint: The static frictional force is now directed down the wall.)

A space traveler weighs \(540 \mathrm{~N}\) on earth. What will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth?
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