ņ Three objects of mass 1 kg, 2 kg, and 3 kg are placed at the vertices of a right triangle, at the points (0, 0), (2, 0), and (0, 1), respectively. The center of mass of the system is (а) (1/3, 1/2). (b) (1/2, 2/3). (c) (2/3, 1/2).
ņ Three objects of mass 1 kg, 2 kg, and 3 kg are placed at the vertices of a right triangle, at the points (0, 0), (2, 0), and (0, 1), respectively. The center of mass of the system is (а) (1/3, 1/2). (b) (1/2, 2/3). (c) (2/3, 1/2).
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement:**
Three objects of mass 1 kg, 2 kg, and 3 kg are placed at the vertices of a right triangle, at the points (0, 0), (2, 0), and (0, 1), respectively. The center of mass of the system is:
(a) (1/3, 1/2)
(b) (1/2, 2/3)
(c) (2/3, 1/2)
**Explanation:**
This problem is focused on determining the center of mass for a system of three masses arranged in a right triangle configuration. The positions of the masses are given in a Cartesian coordinate system. We need to calculate the center of mass based on the mass and position of each object.
No graphs or diagrams are provided in this problem statement. However, you can imagine the points plotted on a Cartesian plane to visualize the problem better.
**Solution Approach (not provided in text):**
1. Use the formula for the center of mass:
\[
\text{Center of Mass (x-coordinate)} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}
\]
\[
\text{Center of Mass (y-coordinate)} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}
\]
2. Substitute the masses and coordinates:
\[
\text{x-coordinate} = \frac{1 \times 0 + 2 \times 2 + 3 \times 0}{1 + 2 + 3}
\]
\[
\text{y-coordinate} = \frac{1 \times 0 + 2 \times 0 + 3 \times 1}{1 + 2 + 3}
\]
3. Calculate and compare with options provided.
This will help you find which option correctly represents the center of mass of the system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25fb0993-5d26-4149-a829-39bf7f0bd1d0%2F9982133c-761e-461c-8419-694ecccff685%2Ff9uuojr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Three objects of mass 1 kg, 2 kg, and 3 kg are placed at the vertices of a right triangle, at the points (0, 0), (2, 0), and (0, 1), respectively. The center of mass of the system is:
(a) (1/3, 1/2)
(b) (1/2, 2/3)
(c) (2/3, 1/2)
**Explanation:**
This problem is focused on determining the center of mass for a system of three masses arranged in a right triangle configuration. The positions of the masses are given in a Cartesian coordinate system. We need to calculate the center of mass based on the mass and position of each object.
No graphs or diagrams are provided in this problem statement. However, you can imagine the points plotted on a Cartesian plane to visualize the problem better.
**Solution Approach (not provided in text):**
1. Use the formula for the center of mass:
\[
\text{Center of Mass (x-coordinate)} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}
\]
\[
\text{Center of Mass (y-coordinate)} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}
\]
2. Substitute the masses and coordinates:
\[
\text{x-coordinate} = \frac{1 \times 0 + 2 \times 2 + 3 \times 0}{1 + 2 + 3}
\]
\[
\text{y-coordinate} = \frac{1 \times 0 + 2 \times 0 + 3 \times 1}{1 + 2 + 3}
\]
3. Calculate and compare with options provided.
This will help you find which option correctly represents the center of mass of the system.
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