https://r2.erweima.ai/mathbot/image/1725936436320ehkdu7bo.jpg

To solve this problem, we need to find the resultant vector of all the forces shown in the coordinate system and its direction.

The forces given are:
- **A = 19 N** at 0 degrees (along the positive x-axis)
- **B = 15 N** at 60 degrees
- **C = 16 N** at 135 degrees
- **D = 11 N** at 210 degrees
- **E = 12 N** at 270 degrees (along the negative y-axis)

### Step 1: Resolve Each Force into Components

Let's decompose each force into its horizontal (x) and vertical (y) components. We will use the following formulas:

- \( F_x = F \cdot \cos(\theta) \)
- \( F_y = F \cdot \sin(\theta) \)

where:
- \( F \) is the magnitude of the force.
- \( \theta \) is the angle the force makes with the positive x-axis.

#### Force Components

1. **Force A (19 N at 0 degrees):**
   - \( A_x = 19 \cdot \cos(0) = 19 \, \text{N} \)
   - \( A_y = 19 \cdot \sin(0) = 0 \, \text{N} \)

2. **Force B (15 N at 60 degrees):**
   - \( B_x = 15 \cdot \cos(60) = 15 \cdot 0.5 = 7.5 \, \text{N} \)
   - \( B_y = 15 \cdot \sin(60) = 15 \cdot \frac{\sqrt{3}}{2} \approx 12.99 \, \text{N} \)

3. **Force C (16 N at 135 degrees):**
   - \( C_x = 16 \cdot \cos(135) = 16 \cdot (-\frac{\sqrt{2}}{2}) \approx -11.31 \, \text{N} \)
   - \( C_y = 16 \cdot \sin(135) = 16 \cdot \frac{\sqrt{2}}{2} \approx 11.31 \, \text{N} \)

4. **Force D (11 N at 210 degrees):**
   - \( D_x = 11 \cdot \cos(210) = 11 \cdot (-\frac{\sqrt{3}}{2}) \approx -9.53 \, \text{N} \)
   - \( D_y = 11 \cdot \sin(210) = 11 \cdot (-0.5) = -5.5 \, \text{N} \)

5. **Force E (12 N at 270 degrees):**
   - \( E_x = 12 \cdot \cos(270) = 12 \cdot 0 = 0 \, \text{N} \)
   - \( E_y = 12 \cdot \sin(270) = 12 \cdot (-1) = -12 \, \text{N} \)

### Step 2: Sum of All Components

Now, we sum all the x and y components to find the resultant vector components:

\[
R_x = A_x + B_x + C_x + D_x + E_x
\]

\[
R_y = A_y + B_y + C_y + D_y + E_y
\]

Let's calculate:

- \( R_x = 19 + 7.5 - 11.31 - 9.53 + 0 = 5.66 \, \text{N} \)
- \( R_y = 0 + 12.99 + 11.31 - 5.5 - 12 = 6.8 \, \text{N} \)

### Step 3: Magnitude of the Resultant Force

The magnitude of the resultant vector \( R \) can be found using the Pythagorean theorem:

\[
R = \sqrt{R_x^2 + R_y^2}
\]

Substitute the values:

\[
R = \sqrt{(5.66)^2 + (6.8)^2}
\]

Let's compute this value.The magnitude of the resultant force \( R \) is approximately **8.85 N**.

### Step 4: Direction of the Resultant Force

To find the direction (angle \(\theta\)) of the resultant force relative to the positive x-axis, we use the arctangent function:

\[
\theta = \arctan\left(\frac{R_y}{R_x}\right)
\]

Substitute the values:

\[
\theta = \arctan\left(\frac{6.8}{5.66}\right)
\]

Let's compute the angle \(\theta\).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still calculate the angle using a calculator. The formula to find the angle is:

\[
\theta = \arctan\left(\frac{6.8}{5.66}\right)
\]

You can use a scientific calculator to find the arctangent and convert it to degrees. The angle will give you the direction of the resultant force relative to the positive x-axis.

Would you like to ask anything else or get more details on this topic?

### Related Questions:
1. How do you convert angles from radians to degrees?
2. What are some common applications of vector addition in physics?
3. How do you handle forces in three-dimensional coordinate systems?
4. What is the difference between scalar and vector quantities?
5. How can vectors be represented graphically and numerically?

**Tip:** Remember that the direction of vectors is just as important as their magnitudes when solving physics problems!

Encuentra la resultante y su dirección de los vectores del siguiente sistema coordenado.

Learn how to find the resultant force and its direction from a set of forces applied at different angles in a coordinate system. This problem involves vector decomposition, trigonometric functions, and the Pythagorean theorem. Suitable for high school physics students.

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