To determine the magnitude and direction of force $F_3$ such that the resultant force of the system is zero, we can break down each of the given forces into their components along the $x$, $y$, and $z$ axes. Afterward, we can set up equations for equilibrium.

Given:

• $F_1 = 180 \, \text{N}$, $\alpha_1 = 30^\circ$
• $F_2 = 300 \, \text{N}$, $\alpha_2 = 40^\circ$

Let's assume the third force $F_3$ has unknown magnitude and coordinate direction angles $\alpha_3$, $\beta_3$, and $\gamma_3$.

Steps:

1. Resolve the given forces into components: For each force $F_1$ and $F_2$, we decompose them into their respective components along the $x$, $y$, and $z$ axes. We are given angles of 30° and 40°, which define their orientations.

   The force components for $F_1$ (along its angle $\alpha_1 = 30^\circ$) can be expressed as: $F_{1x} = F_1 \cos(30^\circ), \quad F_{1y} = 0, \quad F_{1z} = F_1 \sin(30^\circ)$ Since $F_1$ lies in the $xz$-plane.

   The force components for $F_2$ (along its angle $\alpha_2 = 40^\circ$) can be written as: $F_{2x} = F_2 \cos(40^\circ), \quad F_{2y} = 0, \quad F_{2z} = -F_2 \sin(40^\circ)$ As $F_2$ also lies in the $xz$-plane but is acting in the opposite direction in the $z$-axis.

2. Set up the equilibrium conditions: The problem states that the resultant of the three forces must be zero, meaning: $F_{1x} + F_{2x} + F_{3x} = 0, \quad F_{1y} + F_{2y} + F_{3y} = 0, \quad F_{1z} + F_{2z} + F_{3z} = 0$

   Using the force components calculated in step 1, and knowing that $F_{1y} = F_{2y} = 0$, we can write: $F_{3x} = -(F_{1x} + F_{2x})$ $F_{3z} = -(F_{1z} + F_{2z})$

   For the $y$-component, since both $F_1$ and $F_2$ have no $y$-component, $F_{3y} = 0$.

3. Solve for the magnitude and direction angles of $F_3$: The magnitude of $F_3$ can be found using the Pythagorean theorem: $F_3 = \sqrt{F_{3x}^2 + F_{3y}^2 + F_{3z}^2}$

   The direction angles $\alpha_3$, $\beta_3$, and $\gamma_3$ are given by: $\alpha_3 = \cos^{-1}\left(\frac{F_{3x}}{F_3}\right), \quad \beta_3 = \cos^{-1}\left(\frac{F_{3y}}{F_3}\right), \quad \gamma_3 = \cos^{-1}\left(\frac{F_{3z}}{F_3}\right)$

Let me now calculate the components of the forces and determine the unknowns for $F_3$.The magnitude and coordinate direction angles of the third force $F_3$ are as follows:

• Magnitude of $F_3$: $399.17 \, \text{N}$
• Direction angle $\alpha_3$ (with the x-axis): $165.07^\circ$
• Direction angle $\beta_3$ (with the y-axis): $90^\circ$ (since there is no y-component)
• Direction angle $\gamma_3$ (with the z-axis): $75.07^\circ$

This ensures that the resultant of all three forces is zero.

Would you like more details on the steps or any specific part of the solution?

Here are five related questions for further exploration:

1. How do force components change if one of the angles is negative?
2. What happens to the magnitude of $F_3$ if $F_2$ is doubled?
3. How would you find the resultant force if all three forces are non-coplanar?
4. Can equilibrium be achieved if one of the forces has a component along the y-axis?
5. What are the conditions for a system of forces to be in static equilibrium?

Tip: Always break forces into their components to simplify calculations, especially when dealing with non-rectangular force systems.