We are given a structural system with two forces applied to a beam supported at points $A$ (a roller support) and $B$ (a pin support). To solve for the horizontal reaction at point $B$, we will follow these steps:

Given:

• A $12 \, \text{kN}$ force inclined at $60^\circ$ to the horizontal.
• A vertical $6 \, \text{kN}$ force.
• Distances: $10 \, \text{m}$ between $A$ and $B$, and $5 \, \text{m}$ from $B$ to the vertical force application point.

Steps:

1. Resolve the inclined force into horizontal and vertical components:

   • Horizontal component of $12 \, \text{kN}$ force: $F_{\text{horizontal}} = 12 \, \text{kN} \times \cos(60^\circ) = 12 \times 0.5 = 6 \, \text{kN}$
   • Vertical component of $12 \, \text{kN}$ force: $F_{\text{vertical}} = 12 \, \text{kN} \times \sin(60^\circ) = 12 \times \frac{\sqrt{3}}{2} \approx 10.39 \, \text{kN}$

2. Summing forces in the horizontal direction: Since $A$ is a roller support, it does not resist horizontal forces. So, the horizontal reaction at $B$ must balance the horizontal component of the $12 \, \text{kN}$ force: $R_{Bx} = F_{\text{horizontal}} = 6 \, \text{kN}$

Thus, the horizontal component of the reaction force at $B$ is $\boxed{6 \, \text{kN}}$.

Would you like more details or have any questions on this?

Here are 5 related questions for further study:

1. How would you calculate the vertical reaction at point $B$?
2. What is the role of equilibrium conditions in this problem?
3. How would the reactions change if the force angle was different?
4. Can you explain why there is no horizontal reaction at point $A$?
5. How would the reaction forces change if point $A$ were fixed?

Tip: Always resolve forces into horizontal and vertical components when working with angled forces in statics problems.