We are asked to determine the angle $\theta$ so that the resultant force of $F_A$ and $F_B$ is directed horizontally to the right, as well as to find the magnitude of the resultant force. Here's a breakdown of how we can solve this problem step by step:

Step 1: Resolve forces into components

The forces $F_A = 8 \, \text{kN}$ and $F_B = 6 \, \text{kN}$ are acting at different angles, so we need to resolve them into their horizontal and vertical components.

Force $F_A$:

• Horizontal component: $F_{A_x} = F_A \cos(\theta)$
• Vertical component: $F_{A_y} = F_A \sin(\theta)$

Force $F_B$ (which is at 40° from the horizontal):

• Horizontal component: $F_{B_x} = F_B \cos(40^\circ)$
• Vertical component: $F_{B_y} = F_B \sin(40^\circ)$

Step 2: Conditions for the resultant force

We want the resultant force to be directed purely horizontally, which means the total vertical component of the forces must sum to zero. Therefore, we set up the equation:

$F_{A_y} + F_{B_y} = 0$

Substitute the components:

$F_A \sin(\theta) + F_B \sin(40^\circ) = 0$

Substitute the values of $F_A = 8 \, \text{kN}$ and $F_B = 6 \, \text{kN}$:

$8 \sin(\theta) + 6 \sin(40^\circ) = 0$

Now solve for $\theta$.

Step 3: Solve for $\theta$

$8 \sin(\theta) = -6 \sin(40^\circ)$

Calculate $\sin(40^\circ)$:

$\sin(40^\circ) \approx 0.6428$

$8 \sin(\theta) = -6 \times 0.6428$ $8 \sin(\theta) = -3.8568$ $\sin(\theta) = \frac{-3.8568}{8} \approx -0.4821$

Now, find $\theta$ using $\sin^{-1}$:

$\theta = \sin^{-1}(-0.4821)$ $\theta \approx -28.9^\circ$

However, since $\theta$ is an angle, we can express this as:

$\theta \approx 28.9^\circ \, \text{below the horizontal}$

Step 4: Find the magnitude of the resultant force

Now that we have $\theta$, we can calculate the magnitude of the resultant force $F_R$, which is the sum of the horizontal components of the two forces:

$F_R = F_{A_x} + F_{B_x}$ $F_R = F_A \cos(\theta) + F_B \cos(40^\circ)$ Substitute the known values:

$F_R = 8 \cos(28.9^\circ) + 6 \cos(40^\circ)$

Calculate the cosines:

$F_R = 8 \times 0.8746 + 6 \times 0.7660$ $F_R = 6.9968 + 4.5960$ $F_R \approx 11.59 \, \text{kN}$

Final Answer:

• The angle $\theta \approx 28.9^\circ$ below the horizontal.
• The magnitude of the resultant force is approximately $F_R \approx 11.59 \, \text{kN}$.

Let me know if you would like more details or have any further questions!

Here are 5 related questions for further practice:

1. What would the resultant force be if the angle $\theta$ was different?
2. How would the solution change if $F_A$ were decreased to 6 kN?
3. Can you solve for $\theta$ if both forces $F_A$ and $F_B$ were known to be equal?
4. How would you calculate the direction of the resultant force if it was not strictly horizontal?
5. How do the vertical and horizontal components of forces affect the direction of the resultant?

Tip: Always resolve forces into components before trying to find resultant magnitudes or directions. This approach simplifies complex force systems into manageable parts.