2/3 SS The slope of the 6.5-kN force F is specified as shown in the figure. Express F as a vector in terms of the unit vectors i and j. F = 6.5 kN 5 12 PROBLEM 2/3
2/3 SS The slope of the 6.5-kN force F is specified as shown in the figure. Express F as a vector in terms of the unit vectors i and j. F = 6.5 kN 5 12 PROBLEM 2/3
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Title: Vector Representation of Forces
#### Problem 2/3
The following diagram illustrates a force \( F \) of 6.5 kN acting in a specified direction. The force is directed at an angle where the vector slope is given by a rise of 5 units and a run of 12 units.
### Diagram Description
The diagram displays a coordinate system with X and Y axes. The force \( F \), denoted as \( F = 6.5 \text{ kN} \), is represented by a vector positioned in the first quadrant of the Cartesian plane. This vector forms a right triangle with the x-axis, where the vertical leg (rise) is 5 units, and the horizontal leg (run) is 12 units.
To express the force \( F \) as a vector in terms of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), we need to determine its components along the x and y axes.
#### Task
**Express \( F \) as a vector in terms of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \).**
The force \( F \) can be decomposed into its horizontal (x-direction) and vertical (y-direction) components using the given slope values.
Given the slope:
- Rise = 5
- Run = 12
The slope ratio \(\frac{\text{rise}}{\text{run}} = \frac{5}{12}\).
Let's denote the force components as \( F_x \) and \( F_y \).
Using trigonometry:
\[ F_x = F \cos(\theta) \]
\[ F_y = F \sin(\theta) \]
Where \(\theta \) is the angle between the force vector and the x-axis.
From the slope:
\[ \cos(\theta) = \frac{12}{\sqrt{12^2 + 5^2}} \]
\[ \sin(\theta) = \frac{5}{\sqrt{12^2 + 5^2}} \]
The hypotenuse (vector magnitude):
\[ \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
Thus:
\[ \cos(\theta) = \frac{12}{13} \]
\[ \sin(\theta) = \frac{5}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf950827-7d70-472f-a52e-1d3c26fa2196%2Fa9dbbb18-1107-4263-9bc4-df74cf7e5059%2F4w08z3f_processed.png&w=3840&q=75)
Transcribed Image Text:### Title: Vector Representation of Forces
#### Problem 2/3
The following diagram illustrates a force \( F \) of 6.5 kN acting in a specified direction. The force is directed at an angle where the vector slope is given by a rise of 5 units and a run of 12 units.
### Diagram Description
The diagram displays a coordinate system with X and Y axes. The force \( F \), denoted as \( F = 6.5 \text{ kN} \), is represented by a vector positioned in the first quadrant of the Cartesian plane. This vector forms a right triangle with the x-axis, where the vertical leg (rise) is 5 units, and the horizontal leg (run) is 12 units.
To express the force \( F \) as a vector in terms of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), we need to determine its components along the x and y axes.
#### Task
**Express \( F \) as a vector in terms of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \).**
The force \( F \) can be decomposed into its horizontal (x-direction) and vertical (y-direction) components using the given slope values.
Given the slope:
- Rise = 5
- Run = 12
The slope ratio \(\frac{\text{rise}}{\text{run}} = \frac{5}{12}\).
Let's denote the force components as \( F_x \) and \( F_y \).
Using trigonometry:
\[ F_x = F \cos(\theta) \]
\[ F_y = F \sin(\theta) \]
Where \(\theta \) is the angle between the force vector and the x-axis.
From the slope:
\[ \cos(\theta) = \frac{12}{\sqrt{12^2 + 5^2}} \]
\[ \sin(\theta) = \frac{5}{\sqrt{12^2 + 5^2}} \]
The hypotenuse (vector magnitude):
\[ \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
Thus:
\[ \cos(\theta) = \frac{12}{13} \]
\[ \sin(\theta) = \frac{5}{
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