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To solve the problem, we'll use the conditions of equilibrium for a particle. A particle is in equilibrium when the vector sum of all forces acting on it is zero. This means that the sum of the forces in the horizontal direction (x-direction) and the vertical direction (y-direction) must each be zero.

Given forces:
- 15 N force directed to the right (horizontal)
- 12 N force directed downward (vertical)
- \( F \) with unknown magnitude and direction

Let's denote the angle that \( F \) makes with the horizontal direction as \( \theta \). We can resolve \( F \) into its horizontal (\( F_x \)) and vertical (\( F_y \)) components:
\[ F_x = F \cos \theta \]
\[ F_y = F \sin \theta \]

The equilibrium conditions give us two equations:
1. Sum of forces in the x-direction:
\[ F_x - 15 = 0 \]
\[ F \cos \theta = 15 \]

2. Sum of forces in the y-direction:
\[ F_y - 12 = 0 \]
\[ F \sin \theta = 12 \]

We can solve these equations simultaneously to find the magnitude of \( F \) and the angle \( \theta \).

First, solve for \( F \):
\[ F = \sqrt{(F \cos \theta)^2 + (F \sin \theta)^2} \]
Substituting \( F \cos \theta = 15 \) and \( F \sin \theta = 12 \):
\[ F = \sqrt{15^2 + 12^2} \]
\[ F = \sqrt{225 + 144} \]
\[ F = \sqrt{369} \]
\[ F = 19.2 \, \text{N} \]

Now, find the angle \( \theta \):
\[ \cos \theta = \frac{15}{F} = \frac{15}{19.2} \]
\[ \theta = \cos^{-1}\left(\frac{15}{19.2}\right) \]
\[ \theta \approx \cos^{-1}(0.78125) \]
\[ \theta \approx 38.7^\circ \]

So, the magnitude of \( F \) is 19.2 N, and the angle it makes with the direction of the 15 N force is approximately 38.7 degrees.

Would you like more details or have any questions?

Here are five related questions for further exploration:
1. How would the equilibrium conditions change if the forces were not coplanar?
2. What would happen if an additional force was added to the system?
3. Can you determine the components of a force if only its magnitude and angle are given?
4. How would the system be analyzed if the angle of the force \( F \) were given instead of the 12 N force?
5. How can vector addition be used to find the resultant force in a non-equilibrium situation?

**Tip:** When dealing with equilibrium problems, always start by resolving all forces into their components and applying the equilibrium conditions to solve for unknowns.

A particle is in equilibrium under the action of three forces: 15 N to the right, 12 N downward, and an unknown force F. Find the magnitude of F and the angle it makes with the direction of the 15 N force.

Solve a 3-force equilibrium problem involving a 15 N, 12 N, and unknown force F. Learn how to find the magnitude of F and the angle it makes with the 15 N force using vector resolution and trigonometry. Ideal for Grades 10-12.

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