This problem involves vector analysis to determine the force $F_B$ and its direction $\theta$ such that the resultant force acts along the positive x-axis and has a magnitude of 5 kN.

Step-by-Step Solution:

1. Breakdown of forces:

We have two forces acting on the system:

• $F_A = 3 \, \text{kN}$ acting at an angle of $30^\circ$ from the horizontal (x-axis).
• $F_B$ with an unknown magnitude and direction $\theta$.

The goal is to ensure that the resultant force acts entirely along the x-axis with a magnitude of 5 kN. This means that the y-component of the resultant force must be zero, and the x-component must equal 5 kN.

2. Force Components:

The force components of $F_A$ and $F_B$ can be expressed as:

• For $F_A$ (given): $F_{A_x} = F_A \cos 30^\circ = 3 \times \cos 30^\circ = 3 \times 0.866 = 2.598 \, \text{kN}$ $F_{A_y} = F_A \sin 30^\circ = 3 \times \sin 30^\circ = 3 \times 0.5 = 1.5 \, \text{kN}$

• For $F_B$ (unknown): $F_{B_x} = F_B \cos \theta$ $F_{B_y} = F_B \sin \theta$

3. Condition for the resultant force:

Since the resultant force is along the x-axis:

• The sum of the y-components must be zero: $F_{A_y} + F_{B_y} = 0$ This gives: $1.5 + F_B \sin \theta = 0$ $F_B \sin \theta = -1.5$

• The sum of the x-components must be 5 kN: $F_{A_x} + F_{B_x} = 5$ This gives: $2.598 + F_B \cos \theta = 5$ $F_B \cos \theta = 5 - 2.598 = 2.402$

4. Solving for $F_B$ and $\theta$:

Now we have two equations:

• $F_B \sin \theta = -1.5$
• $F_B \cos \theta = 2.402$

We can square both equations and add them to find $F_B$: $(F_B \sin \theta)^2 + (F_B \cos \theta)^2 = (-1.5)^2 + (2.402)^2$ $F_B^2 (\sin^2 \theta + \cos^2 \theta) = 1.5^2 + 2.402^2$ $F_B^2 = 2.25 + 5.77 = 8.02$ $F_B = \sqrt{8.02} \approx 2.83 \, \text{kN}$

Now, to find $\theta$, we use one of the original equations: $F_B \sin \theta = -1.5$ $2.83 \sin \theta = -1.5$ $\sin \theta = \frac{-1.5}{2.83} \approx -0.53$ $\theta = \arcsin(-0.53) \approx -32^\circ$

Thus, the magnitude of $F_B$ is approximately $2.83 \, \text{kN}$, and its direction $\theta$ is approximately $-32^\circ$.

Final Answers:

• Magnitude of $F_B$: 2.83 kN
• Direction of $F_B$: $-32^\circ$

Would you like further clarification or details? Here are some related questions for further practice:

1. How would the solution change if the resultant force was required to act along the y-axis instead?
2. What if $F_A$ was acting at a different angle, say $45^\circ$? How would this affect the result?
3. How do the trigonometric identities help simplify force component calculations?
4. Can we solve this problem using vector addition graphically? If so, how?
5. How would the system behave if the magnitude of $F_B$ was fixed and only $\theta$ was allowed to vary?

Tip: Always check whether forces acting in opposite directions balance each other when a resultant force is aligned along one axis!