2/4 The line of action of the 3000-lb force runs through the points A and B as shown in the figure. Determine the x and y scalar components of F. y, ft B (8,6) F=3000 lb -x, ft A (-7, -2)
2/4 The line of action of the 3000-lb force runs through the points A and B as shown in the figure. Determine the x and y scalar components of F. y, ft B (8,6) F=3000 lb -x, ft A (-7, -2)
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem 2/4: Force Components**
The figure provided shows a force vector, \( \mathbf{F} \), with a magnitude of 3000 pounds (lb) acting along a line through points \( A \) and \( B \). You are required to determine the \( x \)- and \( y \)-scalar components of \( \mathbf{F} \).
### Figure Explanation:
The given figure is a coordinate plane with an origin at point \( O \). Two points, \( A \) and \( B \), are marked on the plane with the following coordinates:
- Point \( A \) is located at \( (-7, -2) \) feet.
- Point \( B \) is located at \( (8, 6) \) feet.
The force vector \( \mathbf{F} \) has a magnitude of 3000 lb. It is represented as an arrow starting from point \( A \) and passing through point \( B \), indicating the direction of the force.
### Objective:
- **Determine the \( x \)-scalar component of \( \mathbf{F} \)**: The component of the force in the horizontal direction.
- **Determine the \( y \)-scalar component of \( \mathbf{F} \)**: The component of the force in the vertical direction.
### Solution Strategy:
1. **Find the Direction Ratios:**
Calculate the differences in the \( x \)- and \( y \)-coordinates between points \( A \) and \( B \):
\[
\Delta x = x_B - x_A = 8 - (-7) = 15 \, \text{ft}
\]
\[
\Delta y = y_B - y_A = 6 - (-2) = 8 \, \text{ft}
\]
2. **Determine the Magnitude of the Distance:**
Use the Pythagorean theorem to find the distance \( AB \):
\[
AB = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, \text{ft}
\]
3. **Calculate the Unit Vector Components:**
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{F}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf950827-7d70-472f-a52e-1d3c26fa2196%2Faef5cea7-aeaa-4e02-a628-436cfe6e65db%2Fkibqvi6_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2/4: Force Components**
The figure provided shows a force vector, \( \mathbf{F} \), with a magnitude of 3000 pounds (lb) acting along a line through points \( A \) and \( B \). You are required to determine the \( x \)- and \( y \)-scalar components of \( \mathbf{F} \).
### Figure Explanation:
The given figure is a coordinate plane with an origin at point \( O \). Two points, \( A \) and \( B \), are marked on the plane with the following coordinates:
- Point \( A \) is located at \( (-7, -2) \) feet.
- Point \( B \) is located at \( (8, 6) \) feet.
The force vector \( \mathbf{F} \) has a magnitude of 3000 lb. It is represented as an arrow starting from point \( A \) and passing through point \( B \), indicating the direction of the force.
### Objective:
- **Determine the \( x \)-scalar component of \( \mathbf{F} \)**: The component of the force in the horizontal direction.
- **Determine the \( y \)-scalar component of \( \mathbf{F} \)**: The component of the force in the vertical direction.
### Solution Strategy:
1. **Find the Direction Ratios:**
Calculate the differences in the \( x \)- and \( y \)-coordinates between points \( A \) and \( B \):
\[
\Delta x = x_B - x_A = 8 - (-7) = 15 \, \text{ft}
\]
\[
\Delta y = y_B - y_A = 6 - (-2) = 8 \, \text{ft}
\]
2. **Determine the Magnitude of the Distance:**
Use the Pythagorean theorem to find the distance \( AB \):
\[
AB = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \, \text{ft}
\]
3. **Calculate the Unit Vector Components:**
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{F}
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