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Determine the resultant of the four forced acting on the body shown

To determine the resultant of the four forces acting on the body, we’ll resolve each force into its x and y components and then add up these components.

### Step 1: Resolving Each Force into Components
For simplicity, let’s assume:
- \( F_1 = 4000 \) lb (left at 30° above the x-axis)
- \( F_2 = 5000 \) lb (left at 30° below the x-axis)
- \( F_3 = 2000 \) lb (upward along y-axis)
- \( F_4 = 3000 \) lb (right at 30° below the x-axis)
- \( F_5 = 1000 \) lb (downward along y-axis)

#### Components:
1. **\( F_1 = 4000 \) lb at 30° above x-axis**:
   - \( F_{1x} = 4000 \cos(30^\circ) = 4000 \times 0.866 = 3464 \) lb (negative x-direction)
   - \( F_{1y} = 4000 \sin(30^\circ) = 4000 \times 0.5 = 2000 \) lb (positive y-direction)

2. **\( F_2 = 5000 \) lb at 30° below x-axis**:
   - \( F_{2x} = 5000 \cos(30^\circ) = 5000 \times 0.866 = 4330 \) lb (negative x-direction)
   - \( F_{2y} = -5000 \sin(30^\circ) = -5000 \times 0.5 = -2500 \) lb (negative y-direction)

3. **\( F_3 = 2000 \) lb along y-axis**:
   - \( F_{3x} = 0 \)
   - \( F_{3y} = 2000 \) lb (positive y-direction)

4. **\( F_4 = 3000 \) lb at 30° below x-axis**:
   - \( F_{4x} = 3000 \cos(30^\circ) = 3000 \times 0.866 = 2598 \) lb (positive x-direction)
   - \( F_{4y} = -3000 \sin(30^\circ) = -3000 \times 0.5 = -1500 \) lb (negative y-direction)

5. **\( F_5 = 1000 \) lb along y-axis**:
   - \( F_{5x} = 0 \)
   - \( F_{5y} = -1000 \) lb (negative y-direction)

### Step 2: Summing Up Components
#### Total \( x \)-components:
\[
\sum F_x = -3464 - 4330 + 2598 = -5196 \text{ lb}
\]

#### Total \( y \)-components:
\[
\sum F_y = 2000 - 2500 + 2000 - 1500 - 1000 = -1000 \text{ lb}
\]

### Step 3: Calculating the Resultant
The resultant force \( R \) is given by:
\[
R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}
\]
\[
R = \sqrt{(-5196)^2 + (-1000)^2} = \sqrt{26971616 + 1000000} = \sqrt{27971616} \approx 5290 \text{ lb}
\]

### Step 4: Direction of the Resultant
The direction \( \theta \) of \( R \) relative to the x-axis is:
\[
\theta = \tan^{-1} \left(\frac{\sum F_y}{\sum F_x}\right) = \tan^{-1} \left(\frac{-1000}{-5196}\right) \approx 10.89^\circ
\]
So, the resultant force is approximately **5290 lb at 10.89° above the negative x-axis**.

Would you like further clarification on any of these steps?

---

1. How are x and y components calculated for forces at an angle?
2. Why do we use trigonometric functions to resolve forces?
3. How does the resultant magnitude relate to individual forces?
4. How would adding a new force impact the resultant direction?
5. What effect does the angle have on the component values?

**Tip:** Breaking down forces into components simplifies calculating the net force direction and magnitude.

Determine the resultant of the four forces acting on the body shown.

Learn how to find the resultant force of four forces acting at various angles on a body. This solution demonstrates vector resolution, force summation, and trigonometric calculations for engineering students.

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