Diagrams illustrating net force calculations for various force vectors acting on objects.

To solve the problem, we need to calculate the net force (\( F_{\text{net}} \)) for each scenario and determine its direction. The net force is calculated by summing all the forces acting on an object, taking into account their directions. Forces in the same direction are added, while forces in opposite directions are subtracted.

Let's analyze each case step by step:

---

Row 1:

#### Column 1 (Pink Square)
- Forces: \( 4 \, \text{N} \) (left), \( 10 \, \text{N} \) (right)
- Net force:
\[
F_{\text{net}} = 10 \, \text{N} - 4 \, \text{N} = 6 \, \text{N}
\]
- Direction: Right (since the rightward force is greater)

Answer:
\[
F_{\text{net}} = 6 \, \text{N}, \quad \text{Direction: Right}
\]

#### Column 2 (Red Square)
- Forces: \( 15 \, \text{N} \) (left), \( 25 \, \text{N} \) (right)
- Net force:
\[
F_{\text{net}} = 25 \, \text{N} - 15 \, \text{N} = 10 \, \text{N}
\]
- Direction: Right (since the rightward force is greater)

Answer:
\[
F_{\text{net}} = 10 \, \text{N}, \quad \text{Direction: Right}
\]

---

Row 2:

#### Column 1 (Blue Square)
- Forces: \( 10 \, \text{N} \) (left), \( 10 \, \text{N} \) (right), \( 7 \, \text{N} \) (down)
- Net force in the horizontal direction:
\[
F_{\text{horizontal}} = 10 \, \text{N} - 10 \, \text{N} = 0 \, \text{N}
\]
- Net force in the vertical direction:
\[
F_{\text{vertical}} = 0 \, \text{N} - 7 \, \text{N} = -7 \, \text{N} \quad (\text{downward})
\]
- Resultant net force:
\[
F_{\text{net}} = \sqrt{(F_{\text{horizontal}})^2 + (F_{\text{vertical}})^2} = \sqrt{0^2 + (-7)^2} = 7 \, \text{N}
\]
- Direction: Down

Answer:
\[
F_{\text{net}} = 7 \, \text{N}, \quad \text{Direction: Down}
\]

#### Column 2 (Yellow Square)
- Forces: \( 5 \, \text{N} \) (left), \( 5 \, \text{N} \) (right), \( 3 \, \text{N} \) (up), \( 3 \, \text{N} \) (down)
- Net force in the horizontal direction:
\[
F_{\text{horizontal}} = 5 \, \text{N} - 5 \, \text{N} = 0 \, \text{N}
\]
- Net force in the vertical direction:
\[
F_{\text{vertical}} = 3 \, \text{N} - 3 \, \text{N} = 0 \, \text{N}
\]
- Resultant net force:
\[
F_{\text{net}} = \sqrt{(F_{\text{horizontal}})^2 + (F_{\text{vertical}})^2} = \sqrt{0^2 + 0^2} = 0 \, \text{N}
\]
- Direction: None (no net force)

Answer:
\[
F_{\text{net}} = 0 \, \text{N}, \quad \text{Direction: None}
\]

---

Row 3:

#### Column 1 (Green Square)
- Forces: \( 10 \, \text{N} \) (left), \( 5 \, \text{N} \) (right), \( 5 \, \text{N} \) (down)
- Net force in the horizontal direction:
\[
F_{\text{horizontal}} = 10 \, \text{N} - 5 \, \text{N} = 5 \, \text{N} \quad (\text{leftward})
\]
- Net force in the vertical direction:
\[
F_{\text{vertical}} = 0 \, \text{N} - 5 \, \text{N} = -5 \, \text{N} \quad (\text{downward})
\]
- Resultant net force:
\[
F_{\text{net}} = \sqrt{(F_{\text{horizontal}})^2 + (F_{\text{vertical}})^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{N}
\]
- Direction: Diagonal (use tangent to find the angle):
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{5} = 1 \implies \theta = 45^\circ
\]
The direction is \( 45^\circ \) below the horizontal (leftward).

Answer:
\[
F_{\text{net}} = 5\sqrt{2} \, \text{N}, \quad \text{Direction: } 45^\circ \, \text{below horizontal (leftward)}
\]

#### Column 2 (Light Blue Square)
- Forces: \( 12 \, \text{N} \) (left), \( 5 \, \text{N} \) (right)
- Net force:
\[
F_{\text{net}} = 12 \, \text{N} - 5 \, \text{N} = 7 \, \text{N}
\]
- Direction: Left (since the leftward force is greater)

Answer:
\[
F_{\text{net}} = 7 \, \text{N}, \quad \text{Direction: Left}
\]

---

Row 4:

#### Column 1 (Red Square)
- Forces: \( 15 \, \text{N} \) (left), \( 5 \, \text{N} \) (right)
- Net force:
\[
F_{\text{net}} = 15 \, \text{N} - 5 \, \text{N} = 10 \, \text{N}
\]
- Direction: Left (since the leftward force is greater)

Answer:
\[
F_{\text{net}} = 10 \, \text{N}, \quad \text{Direction: Left}
\]

#### Column 2 (Dark Red Square)
- Forces: \( 24 \, \text{N} \) (up), \( 13 \, \text{N} \) (down)
- Net force:
\[
F_{\text{net}} = 24 \, \text{N} - 13 \, \text{N} = 11 \, \text{N}
\]
- Direction: Up (since the upward force is greater)

Answer:
\[
F_{\text{net}} = 11 \, \text{N}, \quad \text{Direction: Up}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{|c|c|}
\hline
\text{Row 1, Col 1} & F_{\text{net}} = 6 \, \text{N}, \, \text{Direction: Right} \\
\text{Row 1, Col 2} & F_{\text{net}} = 10 \, \text{N}, \, \text{Direction: Right} \\
\hline
\text{Row 2, Col 1} & F_{\text{net}} = 7 \, \text{N}, \, \text{Direction: Down} \\
\text{Row 2, Col 2} & F_{\text{net}} = 0 \, \text{N}, \, \text{Direction: None} \\
\hline
\text{Row 3, Col 1} & F_{\text{net}} = 5\sqrt{2} \, \text{N}, \, \text{Direction: } 45^\circ \, \text{below horizontal (leftward)} \\
\text{Row 3, Col 2} & F_{\text{net}} = 7 \, \text{N}, \, \text{Direction: Left} \\
\hline
\text{Row 4, Col 1} & F_{\text{net}} = 10 \, \text{N}, \, \text{Direction: Left} \\
\text{Row 4, Col 2} & F_{\text{net}} = 11 \, \text{N}, \, \text{Direction: Up} \\
\hline
\end{array}
}
\]