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Chapter 4: Problem 12

At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or \(x\) component of the ball's acceleration is 810 \(\mathrm{m} / \mathrm{s}^{2}\) and the vertical or \(y\) component of its acceleration is 1100 \(\mathrm{m} / \mathrm{s}^{2}\) . The ball's mass is 0.43 \(\mathrm{kg}\) . What is the magnitude of the net force acting on the soccer ball at this instant?

Short Answer

Expert verified
The net force on the ball is approximately 587.33 N.

Step by step solution

01

Understand the Problem

We are given the components of acceleration of a soccer ball and its mass. We need to find the magnitude of the net force acting on the ball. To solve this, we'll use Newton's second law and the Pythagorean theorem.
02

Identify Known Values

The horizontal component of the ball's acceleration \(a_x\) is 810 \(\mathrm{m/s^2}\), the vertical component \(a_y\) is 1100 \(\mathrm{m/s^2}\), and the mass of the ball \(m\) is 0.43 \(\mathrm{kg}\).
03

Apply Newton's Second Law

Newton's second law states that \( \vec{F} = m \cdot \vec{a} \). We will use \(F_{x} = m \cdot a_x\) and \(F_{y} = m \cdot a_y\) to find the force components. Here, \(F_x = 0.43 \, \mathrm{kg} \times 810 \, \mathrm{m/s^2}\) and \(F_y = 0.43 \, \mathrm{kg} \times 1100 \, \mathrm{m/s^2}\).
04

Calculate the Force Components

Calculate \(F_x\):\[ F_x = 0.43 \cdot 810 = 348.3 \, \mathrm{N} \]Calculate \(F_y\):\[ F_y = 0.43 \cdot 1100 = 473 \, \mathrm{N} \]
05

Calculate the Magnitude of the Net Force

Use the Pythagorean theorem to find the magnitude of the net force:\[ F = \sqrt{F_x^2 + F_y^2} \]Plug in the values:\[ F = \sqrt{348.3^2 + 473^2} \]
06

Solve for the Magnitude of the Net Force

Calculate inside the square root:\[ F = \sqrt{121305.69 + 223729} = \sqrt{345034.69} \]Finally, calculate \(F\):\[ F \approx 587.33 \, \mathrm{N} \]
07

Conclude with the Final Answer

The magnitude of the net force acting on the soccer ball at this instant is approximately 587.33 N.

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Key Concepts

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

Acceleration Components
When analyzing the motion of an object, such as a soccer ball being kicked, we often use **acceleration components** to break down and better understand its motion. Acceleration components are the individual parts of an object's acceleration in specific directions:
  • The **horizontal component** (denoted as \(a_x\)) refers to the acceleration along the horizontal axis. In our example, it's given as 810 m/s虏.

  • The **vertical component** (denoted as \(a_y\)) refers to the acceleration along the vertical axis. Here, we have 1100 m/s虏.
These components allow us to separately analyze forces acting in the horizontal and vertical directions. This separation is pivotal because each force can affect motion differently depending on its direction. By focusing on acceleration components, we can calculate how much force is acting in each direction, which is crucial for solving motion-related problems.
Net Force Calculation
In order to determine the magnitude of the overall force acting on an object, we perform a **net force calculation**. Using Newton's Second Law of Motion, which states that \(\vec{F} = m \cdot \vec{a}\), we calculate the force for each direction.
  • **Horizontal Force** \(F_x\): We calculate this by multiplying the ball's mass (0.43 kg) with the horizontal acceleration component (810 m/s虏). Thus, \(F_x = 0.43 \times 810 = 348.3\,\mathrm{N}\).

  • **Vertical Force** \(F_y\): Similarly, this is found by multiplying the mass with the vertical acceleration component (1100 m/s虏). Therefore, \(F_y = 0.43 \times 1100 = 473\,\mathrm{N}\).
By calculating these components separately, we prepare the data needed to find the net force. This process breaks down complex systems into simpler parts that can be easily solved.
Pythagorean Theorem
To find the **magnitude of the net force**, regardless of its direction, the **Pythagorean theorem** comes into play. The theorem states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
For our example:
  • Use the equation \(F = \sqrt{F_x^2 + F_y^2}\)

  • Plug in the values: \(F = \sqrt{348.3^2 + 473^2}\)

  • Calculate inside the square root: \(F = \sqrt{121305.69 + 223729} = \sqrt{345034.69}\)
Finally, by solving \(F\), we find \(F \approx 587.33\,\mathrm{N}\). This calculation provides us the complete picture of the net force acting on the soccer ball at that instant. These steps illustrate how breaking down forces and using geometric principles can solve real-world physical problems.

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Most popular questions from this chapter

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes \(28 s\) to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

ssm The mass of a robot is 5450 \(\mathrm{kg}\) . This robot weighs 3620 \(\mathrm{N}\) more \(\mathrm{e}\) e on planet \(\mathrm{A}\) than it does on planet \(\mathrm{B}\) . Both planets have the same radius of \(1.33 \times 10^{7} \mathrm{m}\) . What is the difference \(M_{\mathrm{A}} M_{\mathrm{B}}\) in the masses of these planets?

A damp washcloth is hung over the edge of a table to dry. Thus, part (mass \(=m_{\mathrm{ca}} )\) of the washcloth rests on the table and part (mass \(=m_{\text { oft }} )\) does not. The coefficient of static friction between the table and the washcloth is 0.40 . Determine the maximum fraction \(\left[m_{\mathrm{oft}} /\left(m_{\mathrm{cn}}+m_{\mathrm{off}}\right)\right]\) that can hang over the edge without causing the whole washcloth to slide off the table.

The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\overline{\mathbf{F}}_{\text { SM }}\)that the sun exerts on the moon is perpendicular to the force \(\overrightarrow{\mathbf{F}}_{\mathrm{EM}}\) that the earth exerts on the moon. The masses are: mass of sun \(=1.99 \times 10^{30} \mathrm{kg}\) , mass of earth \(=5.98 \times 10^{24} \mathrm{kg}\) , mass of moon \(=7.35 \times 10^{22} \mathrm{kg}\) . The distances shown in the drawing are \(r_{\mathrm{SM}}=1.50 \times 10^{11} \mathrm{m}\) and \(r_{\mathrm{EM}}=3.85 \times 10^{8} \mathrm{m} .\) Determine the magnitude of the net gravitational force on the moon.

A cup of coffee is on a table in an airplane flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates forward, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?
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