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Let's solve this vector problem step by step.

The problem asks to **find the magnitude and direction of the vector sum** of three forces given in Figure 1-14. The forces are:

- **300 N** at an angle of **45°** (from the positive \(y\)-axis, in the second quadrant)
- **200 N** at **30°** (from the positive \(x\)-axis, in the first quadrant)
- **155 N** at **53°** (from the negative \(x\)-axis, in the third quadrant)

We'll solve this both analytically and conceptually. Let’s break each force into its components along the \(x\)- and \(y\)-axes. 

---

### 1. Force Components

#### \(F_1 = 300 \, \text{N}\) at 45° (above negative \(x\)-axis, second quadrant):
\[
F_{1x} = 300 \cos(135^\circ) = 300 \cos(180^\circ - 45^\circ) = 300 (-0.707) = -212.1 \, \text{N}
\]
\[
F_{1y} = 300 \sin(135^\circ) = 300 \sin(180^\circ - 45^\circ) = 300 (0.707) = 212.1 \, \text{N}
\]

---

#### \(F_2 = 200 \, \text{N}\) at 30° (first quadrant):
\[
F_{2x} = 200 \cos(30^\circ) = 200 \cdot 0.866 = 173.2 \, \text{N}
\]
\[
F_{2y} = 200 \sin(30^\circ) = 200 \cdot 0.5 = 100 \, \text{N}
\]

---

#### \(F_3 = 155 \, \text{N}\) at 53° (third quadrant):
\[
F_{3x} = 155 \cos(180^\circ + 53^\circ) = 155 \cdot (-0.602) = -93.3 \, \text{N}
\]
\[
F_{3y} = 155 \sin(180^\circ + 53^\circ) = 155 \cdot (-0.799) = -123.9 \, \text{N}
\]

---

### 2. Sum of Components

Now, sum the components along the \(x\)-axis and the \(y\)-axis.

\[
R_x = F_{1x} + F_{2x} + F_{3x} = -212.1 + 173.2 - 93.3 = -132.2 \, \text{N}
\]
\[
R_y = F_{1y} + F_{2y} + F_{3y} = 212.1 + 100 - 123.9 = 188.2 \, \text{N}
\]

---

### 3. Magnitude of the Resultant

The magnitude of the resultant vector \(R\) is given by:

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

\[
R = \sqrt{(-132.2)^2 + (188.2)^2} = \sqrt{17481.8 + 35417.6} = \sqrt{52899.4} \approx 230 \, \text{N}
\]

---

### 4. Direction of the Resultant

The direction \( \theta \) of the resultant vector (measured from the positive \(x\)-axis) is:

\[
\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)
\]

\[
\theta = \tan^{-1} \left( \frac{188.2}{-132.2} \right) = \tan^{-1}(-1.424)
\]

Since the vector lies in the **second quadrant**, the angle from the positive \(x\)-axis will be:

\[
\theta \approx 180^\circ - 55^\circ = 125^\circ
\]

---

### 5. Final Result

- **Magnitude of the resultant vector**: \( R \approx 230 \, \text{N} \)
- **Direction of the resultant vector**: \( \theta \approx 125^\circ \) (from the positive \(x\)-axis)

---

Would you like further details on any step? Here are five follow-up questions:

1. How do we find components of vectors graphically?
2. Why is the angle adjusted when moving between quadrants?
3. What happens if one of the forces is changed in magnitude or direction?
4. How would you solve this using a polygon method?
5. Can we represent vector addition in three dimensions?

**Tip:** Always break vectors into components to simplify the problem, especially for forces acting at arbitrary angles.

Find graphically the magnitude and direction of the vector sum of the three forces (300 N at 45°, 200 N at 30°, and 155 N at 53°).

Solve a vector problem involving three forces of 300 N, 200 N, and 155 N at different angles using trigonometric vector components. Detailed step-by-step instructions for calculating the magnitude and direction of the resultant vector.

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