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Chapter 6: Problem 1

Standard Body If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20^{\circ}\) to the positive direction of the \(x\) axis, then what are (a) the \(x\) -component and (b) the \(y\) -component of the net force on it, and (c) what is the net force in unit- vector notation?

Short Answer

Expert verified
a) 1.8794 N, b) 0.6840 N, c) 1.8794 \hat{i} + 0.6840 \hat{j} \mathrm{ N}.

Step by step solution

01

Understand the Problem

We need to find the components of the net force acting on a 1 kg body with a given acceleration at a certain angle to the x-axis.
02

Use Newton's Second Law

According to Newton's Second Law, the net force acting on an object is given by \( \textbf{F} = m \textbf{a} \) where \(m\) is the mass and \(\textbf{a}\) is the acceleration.
03

Calculate the Magnitude of the Net Force

Given, mass \(m = 1 \mathrm{~kg}\) and acceleration \(a = 2.00 \mathrm{~m} / \mathrm{s}^{2}\). Thus, the net force magnitude is \( F = m \times a = 1 \mathrm{~kg} \times 2.00 \mathrm{~m} / \mathrm{s}^{2} = 2.00 \mathrm{~N} \).
04

Find the x-component of the Net Force

The x-component of the force can be found using \( F_x = F \times \cos(\theta) \) where \( \theta = 20^{\circ} \). \( F_x = 2.00 \mathrm{~N} \times \cos(20^{\circ}) \). Using a calculator, \( \cos(20^{\circ}) \approx 0.9397 \). Thus, \( F_x = 2.00 \times 0.9397 = 1.8794 \mathrm{~N} \).
05

Find the y-component of the Net Force

The y-component of the force can be found using \( F_y = F \times \sin(\theta) \) where \( \theta = 20^{\circ} \). \( F_y = 2.00 \mathrm{~N} \times \sin(20^{\circ}) \). Using a calculator, \( \sin(20^{\circ}) \approx 0.3420 \). Thus, \( F_y = 2.00 \times 0.3420 = 0.6840 \mathrm{~N} \).
06

Write the Net Force in Unit-Vector Notation

Using unit-vector notation, the net force can be written as \( \textbf{F} = F_x \hat{i} + F_y \hat{j} \). From Steps 4 and 5, \( F_x = 1.8794 \mathrm{~N} \) and \( F_y = 0.6840 \mathrm{~N} \). Thus, the net force in unit-vector notation is \( \textbf{F} = 1.8794 \hat{i} + 0.6840 \hat{j} \mathrm{~N} \).

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Key Concepts

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

Net Force Calculation
Newton's Second Law tells us that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This relationship is expressed as: \( \textbf{F} = m \textbf{a} \).
  • Net force is the sum of all forces acting on an object.
  • This force determines the acceleration of the object.
  • For our problem, the body mass is 1 kg and the acceleration is 2 m/s虏.
So, to find the net force, you simply multiply the mass by the acceleration:\( F = 1 \text{ kg} \times 2 \text{ m/s}^2 = 2 \text{ N} \).Once you have the net force, it's important to understand and break it into components to analyze its effects in different directions.
Vector Components
Any force acting at an angle can be broken down into its horizontal and vertical components. These are known as the x-component and y-component of the force.
  • Components help us understand the force's effects in each direction.
  • To find these components, we use trigonometric functions.
The formulas to find the components are:
  • \( F_x = F \cos(\theta) \)
  • \( F_y = F \sin(\theta) \)
For the given problem, the angle \( \theta \) is 20 degrees. Using these formulas:
  • \( F_x = 2 \text{ N} \times \cos(20掳) \).
  • \( \cos(20掳) \approx 0.9397 \).
  • \( F_x = 2 \times 0.9397 = 1.8794 \text{ N} \).
  • Similarly, for \( F_y = 2 \text{ N} \times \sin(20掳) \).
  • \( \sin(20掳) \approx 0.3420 \).
  • \( F_y = 2 \times 0.3420 = 0.6840 \text{ N} \).
So, now we know the force's components in each direction!
Trigonometric Functions
Trigonometric functions are essential tools in breaking down forces. They relate the angles to the lengths of sides in right-angled triangles.
  • \( \cos(\theta) \) is used to find the adjacent side (x-component).
  • \( \sin(\theta) \) is used to find the opposite side (y-component).
In our example, these functions help convert the force into directions that parallel the coordinate axes.
When you apply these to our problem, we got:
  • \( \cos(20掳) \approx 0.9397 \)
  • \( \sin(20掳) \approx 0.3420 \)
For accuracy, always use a calculator when finding these values. Plugging these into our component formulas, we found:
  • \( F_x = 2 \text{ N} \times 0.9397 = 1.8794 \text{ N} \)
  • \( F_y = 2 \text{ N} \times 0.3420 = 0.6840 \text{ N} \)
Once we know the components, we can fully describe the force in our coordinate system, using unit vector notation. The final net force is \( \textbf{F} = 1.8794 \hat{i} + 0.6840 \hat{j} \text{ N} \).

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Most popular questions from this chapter

A \(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu^{\text {kin }}\) \(=0.35\), what is the magnitude of the initial acceleration of the crate?

We know that as an object passes through the air, the air exerts a resistive force on it. Suppose we have a spherical object of radius \(R\) and mass \(m\). What might the force plausibly depend on? \- It might depend on the properties of the object. The only ones that seem relevant are \(m\) and \(R\). \- It might depend on the object's coordinate and its derivatives: \(\vec{r}, \vec{v}, \vec{a}, \ldots\) \- It might depend on the properties of the air, such as the density, \(\rho\). (a) Explain why it is plausible that the force the air exerts on a sphere depends on \(R\) but implausible that it depends on \(m\). (b) Explain why it is plausible that the force the air exerts depends on the object's speed through it, \(|\vec{v}|\), but not on its position, \(\vec{r}\), or acceleration, \(\vec{a}\). (c) Dimensional analysis is the use of units (e.g., meters, seconds, or newtons) associated with quantities to reason about the relationship between the quantities. Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.

A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the total force on the bicycle from the road.

A block is projected up a frictionless inclined plane with initial speed \(v_{1}=3.50 \mathrm{~m} / \mathrm{s}\). The angle of incline is \(\theta=32.0^{\circ}\). (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. (a) What is the magnitude of the net force acting on the rider? (b) What is the magnitude of the force on the rider from the motorcycle?
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