Vector Addition - Free Printable

Since I can't view images directly, I’ll help you solve the vector addition problems based on your description. You've provided three scenarios involving force vectors (in Newtons, N), and you're asked to find the resultant vector in each case.

Let’s interpret each one step by step using vector addition principles.

---

Problem 1:

```
10 N (up and to the right) + 10 N (up and to the left) = ???
```

Assuming:
- The first vector is 10 N at 45° (diagonally up-right).
- The second vector is 10 N at 135° (diagonally up-left).

These two vectors are symmetric about the vertical axis.

#### Step-by-step:
- Break both into x and y components.
- First vector (45°):
$$
F_{x1} = 10 \cos(45^\circ) = 10 \cdot \frac{\sqrt{2}}{2} \approx 7.07\, \text{N} \\
F_{y1} = 10 \sin(45^\circ) = 10 \cdot \frac{\sqrt{2}}{2} \approx 7.07\, \text{N}
$$

- Second vector (135°):
$$
F_{x2} = 10 \cos(135^\circ) = 10 \cdot (-\frac{\sqrt{2}}{2}) \approx -7.07\, \text{N} \\
F_{y2} = 10 \sin(135^\circ) = 10 \cdot \frac{\sqrt{2}}{2} \approx 7.07\, \text{N}
$$

Now add:
- Total x-component: $ 7.07 + (-7.07) = 0 $
- Total y-component: $ 7.07 + 7.07 = 14.14\, \text{N} $

So the resultant is:
$$
\boxed{14.14\, \text{N upward}} \quad (\text{along the positive y-axis})
$$

---

Problem 2:

```
10 N (up and to the right) + 10 N (leftward) = ???
```

Assume:
- First vector: 10 N at 45° → same as before: $ F_x = 7.07\, \text{N}, F_y = 7.07\, \text{N} $
- Second vector: 10 N to the left → $ F_x = -10\, \text{N}, F_y = 0 $

Add components:
- $ F_x = 7.07 + (-10) = -2.93\, \text{N} $
- $ F_y = 7.07 + 0 = 7.07\, \text{N} $

Now find magnitude:
$$
R = \sqrt{(-2.93)^2 + (7.07)^2} = \sqrt{8.58 + 50.0} \approx \sqrt{58.58} \approx 7.65\, \text{N}
$$

Direction:
$$
\theta = \tan^{-1}\left(\frac{7.07}{-2.93}\right) \approx \tan^{-1}(-2.41) \approx 116^\circ \text{ from positive x-axis}
$$

So the resultant is approximately:
$$
\boxed{7.65\, \text{N at } 116^\circ} \quad (\text{up and to the left})
$$

---

Problem 3:

```
10 N (up and to the right) + 5 N (rightward) = ???
```

First vector: 10 N at 45° → $ F_x = 7.07\, \text{N}, F_y = 7.07\, \text{N} $

Second vector: 5 N to the right → $ F_x = 5\, \text{N}, F_y = 0 $

Add:
- $ F_x = 7.07 + 5 = 12.07\, \text{N} $
- $ F_y = 7.07 + 0 = 7.07\, \text{N} $

Magnitude:
$$
R = \sqrt{(12.07)^2 + (7.07)^2} = \sqrt{145.68 + 50.0} \approx \sqrt{195.68} \approx 14.0\, \text{N}
$$

Direction:
$$
\theta = \tan^{-1}\left(\frac{7.07}{12.07}\right) \approx \tan^{-1}(0.585) \approx 30.3^\circ \text{ above x-axis}
$$

So the resultant is approximately:
$$
\boxed{14.0\, \text{N at } 30.3^\circ} \quad (\text{up and to the right})
$$

---

✔ Final Answers:

1. 14.14 N upward
2. 7.65 N at 116° (or ~26° above negative x-axis)
3. 14.0 N at 30.3° (above positive x-axis)

> ⚠️ Note: These assume the diagonal vectors are at 45°. If the angles differ (e.g., not exactly 45°), please specify. But for typical textbook diagrams, this assumption is standard.

Let me know if you'd like a diagram or explanation using graphical methods!