Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 58
The position of a \(2.75 \times 10^{5} \mathrm{N}\) training helicopter under test is given by \(\vec{r}=\left(0.020 \mathrm{m} / \mathrm{s}^{3}\right) t^{3} \hat{\imath}+(2.2 \mathrm{m} / \mathrm{s}) f \hat{\jmath}-\left(0.060 \mathrm{m} / \mathrm{s}^{2}\right) t^{2} \hat{k}\) Find the net force on the helicopter at \(t=5.0 \mathrm{s}\)
Short Answer
Expert verified
\( \vec{F} = 16836.73 \hat{\imath} - 3367.35 \hat{k} \text{ N} \) at \( t = 5.0 \text{ s} \).
Step by step solution
01
Identify the Position Vector Components
The position vector \( \vec{r} \) is given by three components: \( x = (0.020 \text{ m/s}^3) t^3 \), \( y = (2.2 \text{ m/s}) t \), and \( z = -(0.060 \text{ m/s}^2) t^2 \). These will be differentiated to find the velocity and acceleration.
02
Differentiate Position to Find Velocity
Differentiate each component of the position vector with respect to time \( t \) to find the velocity vector \( \vec{v} \):\( v_x = \frac{d}{dt}[(0.020 \text{ m/s}^3) t^3] = (0.060 \text{ m/s}^3) t^2 \ v_y = \frac{d}{dt}[(2.2 \text{ m/s}) t] = 2.2 \text{ m/s} \ v_z = \frac{d}{dt}[-(0.060 \text{ m/s}^2) t^2] = -0.12 \text{ m/s}^2 t \).
03
Differentiate Velocity to Find Acceleration
Differentiate each component of the velocity vector to find the acceleration vector \( \vec{a} \):\( a_x = \frac{d}{dt}[(0.060 \text{ m/s}^3) t^2] = (0.12 \text{ m/s}^3) t \ a_y = \frac{d}{dt}[2.2 \text{ m/s}] = 0 \text{ m/s}^2 \ a_z = \frac{d}{dt}[-0.12 \text{ m/s}^2 t] = -0.12 \text{ m/s}^2 \).
04
Evaluate Acceleration at t = 5.0 s
Substitute \( t = 5.0 \) seconds into each component of the acceleration vector:\( a_x = (0.12 \text{ m/s}^3)(5.0) = 0.60 \text{ m/s}^2 \ a_y = 0 \text{ m/s}^2 \ a_z = -0.12 \text{ m/s}^2 \). The acceleration vector at \( t = 5.0 \) s is \( \vec{a} = 0.60 \hat{\imath} \text{ m/s}^2 - 0.12 \hat{k} \text{ m/s}^2 \).
05
Calculate Net Force Using Newton's Second Law
Newton's Second Law states \( \vec{F} = m \vec{a} \). First, calculate the helicopter's mass: \( m = \frac{2.75 \times 10^5 \text{ N}}{9.8 \text{ m/s}^2} = 28061.22 \text{ kg} \).The net force is then:\( \vec{F} = 28061.22 \text{ kg} \times (0.60 \hat{\imath} - 0.12 \hat{k}) \text{ m/s}^2 \ = 16836.73 \hat{\imath} - 3367.35 \hat{k} \text{ N} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Vector
In physics, a position vector is a fundamental concept that describes the position of a point in space relative to a reference origin. It is usually denoted as \( \vec{r} \) and expressed in terms of its components along the Cartesian coordinates: x, y, and z.
The example given in this exercise specifies the position vector \( \vec{r} = (0.020 \, \text{m/s}^3) \, t^3 \hat{\imath} + (2.2 \, \text{m/s}) \, t \hat{\jmath} - (0.060 \, \text{m/s}^2) \, t^2 \hat{k} \).
The example given in this exercise specifies the position vector \( \vec{r} = (0.020 \, \text{m/s}^3) \, t^3 \hat{\imath} + (2.2 \, \text{m/s}) \, t \hat{\jmath} - (0.060 \, \text{m/s}^2) \, t^2 \hat{k} \).
- The coefficient terms (like \(0.020 \, \text{m/s}^3\)) scale the dimensions of the vector components.
- The powers of \( t \) indicate how the position changes over time.
- Each unit vector \( \hat{\imath}, \hat{\jmath}, \hat{k} \) represents direction along the x, y, and z axes, respectively.
Velocity Vector
The velocity vector provides the rate of change of an object's position with respect to time. It essentially tells us how fast and in which direction an object is moving.
To find the velocity vector, we differentiate the position vector with respect to time. For the exercise:
To find the velocity vector, we differentiate the position vector with respect to time. For the exercise:
- \( v_x = (0.060 \, \text{m/s}^3) \, t^2 \) from differentiating \( x \)-component.
- \( v_y = 2.2 \, \text{m/s} \) from differentiating \( y \)-component as it is linear with respect to time.
- \( v_z = -0.12 \, \text{m/s}^2 \, t \) from differentiating \( z \)-component.
Acceleration Vector
An acceleration vector quantifies how an object's velocity changes over time. It's the derivative of the velocity vector with respect to time, reflecting both magnitude and direction changes.
To calculate the acceleration for each component:
To calculate the acceleration for each component:
- \( a_x = (0.12 \, \text{m/s}^3) \, t \) is derived from the \( x \)-component of velocity.
- \( a_y = 0 \, \text{m/s}^2 \) as \( v_y \) is constant and independent of \( t \).
- \( a_z = -0.12 \, \text{m/s}^2 \) comes from the \( z \)-component of velocity.
Mass Calculation
Mass is a measure of the amount of matter in an object and plays a crucial role in mechanics, as it influences how forces affect motion.
The exercise determines mass from the weight equation:\( m = \frac{\text{Weight}}{g} \)
Here,
The exercise determines mass from the weight equation:\( m = \frac{\text{Weight}}{g} \)
Here,
- The weight of the helicopter is \( 2.75 \times 10^5 \, \text{N} \).
- The acceleration due to gravity \( g \) is \( 9.8 \, \text{m/s}^2 \).
Net Force
Newton's Second Law relates force to mass and acceleration with the equation \( \vec{F} = m \vec{a} \).
In the exercise setup:
This results in a net force of \( 16836.73 \hat{\imath} - 3367.35 \hat{k} \,\text{N} \), indicating forces applied in the x and z directions. This concept demonstrates how forces interact with an object's inertia to produce motion.
In the exercise setup:
- Mass was calculated to be approximately \( 28061.22 \, \text{kg} \).
- Acceleration vector at \( t = 5.0 \, \text{s} \) is \( \vec{a} = 0.60 \, \text{m/s}^2 \hat{\imath} - 0.12 \, \text{m/s}^2 \hat{k} \).
This results in a net force of \( 16836.73 \hat{\imath} - 3367.35 \hat{k} \,\text{N} \), indicating forces applied in the x and z directions. This concept demonstrates how forces interact with an object's inertia to produce motion.
