2/1 The force F has a magnitude of 600 N. Express F as a vector in terms of the unit vectors i and j. Identify the x and y scalar components of F. 40° F = 600 N PROBLEM 2/1
2/1 The force F has a magnitude of 600 N. Express F as a vector in terms of the unit vectors i and j. Identify the x and y scalar components of F. 40° F = 600 N PROBLEM 2/1
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Problem 2/1:**
The force **F** has a magnitude of 600 N. Express **F** as a vector in terms of the unit vectors **i** and **j**. Identify the **x** and **y** scalar components of **F**.
**Diagram Details:**
The diagram is a Cartesian coordinate system with the origin labeled **O**. The **x** and **y** axes are shown as dashed lines intersecting at the origin. A vector representing the force **F** is drawn from the origin, making a 40° angle with the negative **x** axis. The vector **F** is labeled as having a magnitude of 600 N, pointing towards the third quadrant of the coordinate plane.
**Solution:**
To find the **x** and **y** components of **F**, we can use trigonometric functions based on the given angle (40°) and the magnitude of the vector (600 N).
1. **x-component**:
\[ F_x = |F| \cos(\theta) \]
\[ F_x = 600 \cos(40°) \]
\[ F_x \approx 600 \times 0.766 \approx 459.6 \, \text{N} \]
Since the force is in the third quadrant, the **x**-component will be negative:
\[ F_x \approx -459.6 \, \text{N} \]
2. **y-component**:
\[ F_y = |F| \sin(\theta) \]
\[ F_y = 600 \sin(40°) \]
\[ F_y \approx 600 \times 0.643 \approx 385.8 \, \text{N} \]
Similarly, since the force is in the third quadrant, the **y**-component will be negative:
\[ F_y \approx -385.8 \, \text{N} \]
Therefore, the vector **F** in terms of the unit vectors **i** and **j** can be written as:
\[ \mathbf{F} = -459.6 \mathbf{i} - 385.8 \mathbf{j} \]
**Summary:**
- **x-component of F:** \[ -459.6 \, \text{N} \]
- **y-component of F:** \[ -385.8 \, \text{N} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf950827-7d70-472f-a52e-1d3c26fa2196%2Fc5dfdce7-a72f-4fbf-9679-9f1a75b52756%2F54cs07s_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2/1:**
The force **F** has a magnitude of 600 N. Express **F** as a vector in terms of the unit vectors **i** and **j**. Identify the **x** and **y** scalar components of **F**.
**Diagram Details:**
The diagram is a Cartesian coordinate system with the origin labeled **O**. The **x** and **y** axes are shown as dashed lines intersecting at the origin. A vector representing the force **F** is drawn from the origin, making a 40° angle with the negative **x** axis. The vector **F** is labeled as having a magnitude of 600 N, pointing towards the third quadrant of the coordinate plane.
**Solution:**
To find the **x** and **y** components of **F**, we can use trigonometric functions based on the given angle (40°) and the magnitude of the vector (600 N).
1. **x-component**:
\[ F_x = |F| \cos(\theta) \]
\[ F_x = 600 \cos(40°) \]
\[ F_x \approx 600 \times 0.766 \approx 459.6 \, \text{N} \]
Since the force is in the third quadrant, the **x**-component will be negative:
\[ F_x \approx -459.6 \, \text{N} \]
2. **y-component**:
\[ F_y = |F| \sin(\theta) \]
\[ F_y = 600 \sin(40°) \]
\[ F_y \approx 600 \times 0.643 \approx 385.8 \, \text{N} \]
Similarly, since the force is in the third quadrant, the **y**-component will be negative:
\[ F_y \approx -385.8 \, \text{N} \]
Therefore, the vector **F** in terms of the unit vectors **i** and **j** can be written as:
\[ \mathbf{F} = -459.6 \mathbf{i} - 385.8 \mathbf{j} \]
**Summary:**
- **x-component of F:** \[ -459.6 \, \text{N} \]
- **y-component of F:** \[ -385.8 \, \text{N} \
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