The man pulls. the boy up to the tree limb C by walking backward. If he starts from rest when x = 0 and moves backward with a constant acceleration a = 0.2 m/s², determine the speed of the boy at the instant y = 4 m. Neglect the size of the limb. When x₁ = 0, YB 8 m, so that XA A and B are coincident, i.e., the rope is 16 m long. = -XA- C B Ув 8 m Ans. VB 1.41 m/s ↑
The man pulls. the boy up to the tree limb C by walking backward. If he starts from rest when x = 0 and moves backward with a constant acceleration a = 0.2 m/s², determine the speed of the boy at the instant y = 4 m. Neglect the size of the limb. When x₁ = 0, YB 8 m, so that XA A and B are coincident, i.e., the rope is 16 m long. = -XA- C B Ув 8 m Ans. VB 1.41 m/s ↑
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
need help with step by step
![### Problem Description:
A man pulls a boy up to the tree limb \( C \) by walking backward. He starts from rest when \( x_A = 0 \) and moves backward with a constant acceleration \( a_A = 0.2 \, \text{m/s}^2 \). You are required to determine the speed of the boy at the instant \( y_B = 4 \, \text{m} \). Neglect the size of the limb. When \( x_A = 0 \), \( y_B = 8 \, \text{m} \), indicating that points \( A \) and \( B \) are coincident and the rope is 16 meters long.
### Diagram Explanation:
- **Diagram Elements**:
- A man (at point \( A \)) is holding a rope.
- A boy is seated and is being pulled upwards (at point \( B \)).
- The rope passes over a tree limb (at point \( C \)).
- Labels like \( x_A \) and \( y_B \) indicate horizontal and vertical displacements, respectively.
- **Diagram Annotations**:
- \( x_A \): The horizontal displacement of the man.
- \( y_B \): The vertical displacement of the boy, initially 8 meters.
- The rope length is constant at 16 meters.
### Calculation:
To find the speed \( v_B \) of the boy at \( y_B = 4 \, \text{m} \):
1. **Use the length of the rope**:
The total length of the rope is fixed at 16 meters.
\[
\sqrt{x_A^2 + y_B^2} = 16
\]
2. **Differentiate with respect to time**:
\[
x_A \frac{dx_A}{dt} + y_B \frac{dy_B}{dt} = 0
\]
3. **Use given information**:
Given \( a_A = 0.2 \, \text{m/s}^2 \).
4. **Find speed \( v_B \)**:
By solving the equations, the answer for the boy's speed at the given instant \( y_B = 4 \) is:
\[
v_B = 1.41 \, \text{m/s}
\]
This solution illustrates](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fbf329d-7944-4f84-8902-b0b326aaed6b%2Fec1ce1fd-4cb6-43f9-8fe9-8f77514e6848%2F8qc07uo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Description:
A man pulls a boy up to the tree limb \( C \) by walking backward. He starts from rest when \( x_A = 0 \) and moves backward with a constant acceleration \( a_A = 0.2 \, \text{m/s}^2 \). You are required to determine the speed of the boy at the instant \( y_B = 4 \, \text{m} \). Neglect the size of the limb. When \( x_A = 0 \), \( y_B = 8 \, \text{m} \), indicating that points \( A \) and \( B \) are coincident and the rope is 16 meters long.
### Diagram Explanation:
- **Diagram Elements**:
- A man (at point \( A \)) is holding a rope.
- A boy is seated and is being pulled upwards (at point \( B \)).
- The rope passes over a tree limb (at point \( C \)).
- Labels like \( x_A \) and \( y_B \) indicate horizontal and vertical displacements, respectively.
- **Diagram Annotations**:
- \( x_A \): The horizontal displacement of the man.
- \( y_B \): The vertical displacement of the boy, initially 8 meters.
- The rope length is constant at 16 meters.
### Calculation:
To find the speed \( v_B \) of the boy at \( y_B = 4 \, \text{m} \):
1. **Use the length of the rope**:
The total length of the rope is fixed at 16 meters.
\[
\sqrt{x_A^2 + y_B^2} = 16
\]
2. **Differentiate with respect to time**:
\[
x_A \frac{dx_A}{dt} + y_B \frac{dy_B}{dt} = 0
\]
3. **Use given information**:
Given \( a_A = 0.2 \, \text{m/s}^2 \).
4. **Find speed \( v_B \)**:
By solving the equations, the answer for the boy's speed at the given instant \( y_B = 4 \) is:
\[
v_B = 1.41 \, \text{m/s}
\]
This solution illustrates
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY