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If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

Short Answer

Expert verified
(a) 1.88 N, (b) 0.684 N, (c) 1.88\hat{\imath} + 0.684\hat{\jmath} N.

Step by step solution

01

Understanding the Problem

The problem is asking for the components of the net force acting on a 1 kg body given its acceleration and direction. We need to find the x and y components of the force in unit-vector notation.
02

Identify Given Variables

We have the mass of the body, which is 1 kg, and the acceleration, which is 2.00 m/s² at an angle of 20.0° to the x-axis. The mass acts as a scaling factor for the force, as per Newton's Second Law.
03

Calculate the Magnitude of the Net Force

Using Newton's Second Law, the net force magnitude is found as: \[ F = m imes a = 1 imes 2.00 = 2.00 ext{ N} \]
04

Determine the Force Components

We use trigonometry to find the x and y components of the force. The formulas are:\[ F_x = F \cdot \cos(\theta) \]\[ F_y = F \cdot \sin(\theta) \]
05

Calculate the x Component of the Force

The x component is:\[ F_x = 2.00 \cdot \cos(20.0°) = 2.00 \cdot 0.9397 \approx 1.88 \text{ N} \]
06

Calculate the y Component of the Force

The y component is:\[ F_y = 2.00 \cdot \sin(20.0°) = 2.00 \cdot 0.3420 \approx 0.684 \text{ N} \]
07

Express the Net Force in Unit-Vector Notation

Using the calculated components, the net force in unit-vector notation is:\[ \mathbf{F} = 1.88\hat{\imath} + 0.684\hat{\jmath} \text{ N} \]

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Key Concepts

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

Force components
When dealing with forces in physics, it's important to understand that a force exerted at an angle can be broken down into its horizontal and vertical components. These are known as the force components. This concept allows us to analyze the effects of forces acting in different directions.

To calculate these components, we use basic trigonometric functions applied to the known force vector. Consider a force vector \(F\) acting at an angle \(\theta\) with respect to a reference axis. The x-component, which is parallel to the x-axis, can be found using the cosine function, while the y-component, parallel to the y-axis, uses the sine function:
  • \( F_x = F \cdot \cos(\theta) \)
  • \( F_y = F \cdot \sin(\theta) \)
These components are crucial for resolving the overall effect of the force in different directions, particularly when multiple forces interact in a system.

Breaking down forces into components simplifies complex problems, allowing us to analyze physical scenarios more effectively, such as calculating the net force on an object.
Trigonometry in physics
Trigonometry is a powerful tool in physics for solving problems involving angles and lengths. It becomes particularly effective in resolving forces, such as gravitational or applied forces, into components.

By using trigonometric ratios, namely sine, cosine, and tangent, we can relate the angle of a force to its projection on the coordinate axes. When a force acts at an angle \(\theta\) to the horizontal x-axis, the following relations hold true:
  • The adjacent side (x component) relates to cosine through \( \cos(\theta) \)
  • The opposite side (y component) relates to sine through \( \sin(\theta) \)

This way, trigonometry allows us to decompose a vector into its horizontal and vertical components, facilitating easier calculations and better understanding of the net effects of forces. Remember, trigonometry isn't just confined to mechanics; it is ubiquitous in various branches of physics, such as optics, electromagnetism, and quantum mechanics.

Developing a strong grasp of these concepts can greatly enhance your ability to solve physical problems efficiently.
Unit-vector notation
Unit-vector notation is a streamlined way to express vectors in physics, with minimal ambiguity. Vectors, such as force, velocity, or displacement, are conventionally represented by magnitude and direction. Using unit vectors simplifies vector mathematics and expressions.

In the context of Cartesian coordinates, unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\) represent unit vectors along the x-axis and y-axis, respectively. Any vector \(\mathbf{A}\) in two-dimensional space can then be expressed as:
  • \( \mathbf{A} = A_x \hat{\imath} + A_y \hat{\jmath} \)
This format makes it clear which component of the vector lies in each directional axis, allowing for easier computation and visualization.

For the exercise, the net force vector is given as \( \mathbf{F} = 1.88\hat{\imath} + 0.684\hat{\jmath} \text{ N} \). Here, \(1.88 \hat{\imath}\) indicates force in the x-direction, and \(0.684 \hat{\jmath}\) indicates force in the y-direction, providing a precise representation of the force's orientation and magnitude.

Unit-vector notation is invaluable for solving complex problems, facilitating straightforward manipulation of vectors and easing the understanding of their physical interpretations.

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Most popular questions from this chapter

Only two horizontal forces act on a \(3.0 \mathrm{~kg}\) body that can move over a frictionless floor. One force is \(9.0 \mathrm{~N}\), acting due east, and the other is \(8.0 \mathrm{~N}\), acting \(62^{\circ}\) north of west. What is the magnitude of the body's acceleration?

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

An \(80 \mathrm{~kg}\) person is parachuting and experiencing a downward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the parachute is \(5.0 \mathrm{~kg}\). (a) What is the upward force on the open parachute from the air? (b) What is the downward force on the parachute from the person?

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times\) \(10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg}\). Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

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