To solve this problem, we need to determine the resultant force $R$ of the forces applied by the three tugboats. The steps involve calculating the components of each force, finding the resultant of the two forces at point A, and then combining it with the force at point B. Here's how to proceed:

Step 1: Resolve the forces at point A

We have two forces at point A:

1. 98.1 kN at 35° from the positive $x$-axis.
2. 78.5 kN at 20° from the negative $x$-axis.

To find the components of these forces in the $x$ and $y$ directions:

Force 1 (98.1 kN at 35°):

• $F_{1x} = 98.1 \cdot \cos(35^\circ)$
• $F_{1y} = 98.1 \cdot \sin(35^\circ)$

Force 2 (78.5 kN at 20°):

• $F_{2x} = -78.5 \cdot \cos(20^\circ)$ (since it's acting to the left)
• $F_{2y} = 78.5 \cdot \sin(20^\circ)$

Step 2: Determine the resultant force at point A

Sum the components from Force 1 and Force 2 to get the resultant force at point A:

• $R_{Ax} = F_{1x} + F_{2x}$
• $R_{Ay} = F_{1y} + F_{2y}$

The magnitude of the resultant force at point A is:

$R_A = \sqrt{R_{Ax}^2 + R_{Ay}^2}$

The direction of the resultant force at point A is:

$\theta_A = \tan^{-1}\left(\frac{R_{Ay}}{R_{Ax}}\right)$

Step 3: Combine with the force at point B

There is a force of 78.5 kN at point B acting along the $y$-axis. To find the total resultant $R$, add the force components from point A to the force at point B.

Since the force at point B acts only in the $y$-direction:

• $F_{Bx} = 0$
• $F_{By} = 78.5 \, \text{kN}$

Now, combine these with the resultant components from point A:

• $R_x = R_{Ax}$
• $R_y = R_{Ay} + F_{By}$

The magnitude of the final resultant force $R$ is:

$R = \sqrt{R_x^2 + R_y^2}$

The direction of the resultant force $R$ is:

$\theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Step 4: Determine the location where the resultant passes through the x-axis

To find the point on the $x$-axis where the resultant force passes, use the moment equilibrium about point B:

$x_R = \frac{M_A}{R_y}$

Where $M_A$ is the moment created by the resultant force at point A, calculated as:

$M_A = R_A \times 54.9 \, \text{m}$

Let's calculate these now.

I'll compute the values for you!### Results:

1. Resultant force at point A:

   • $R_{Ax} = 6.59 \, \text{kN}$
   • $R_{Ay} = 83.12 \, \text{kN}$
   • Magnitude of $R_A = 83.38 \, \text{kN}$
   • Direction $\theta_A = 85.46^\circ$ (from the positive $x$-axis)

2. Final resultant force:

   • Magnitude $R = 161.75 \, \text{kN}$
   • Direction $\theta_R = 87.66^\circ$ (from the positive $x$-axis)

3. Location on the $x$-axis where the resultant passes**:

   • $x_R = 28.32 \, \text{m}$ from point A.

Would you like further details or any clarifications?

Here are five related questions to explore:

1. How would the resultant force change if the angle of the second tugboat at point A was altered?
2. How is the moment about point B affected by varying the force magnitude at point A?
3. What happens to the location of the resultant force on the x-axis if the distance between points A and B increases?
4. How would you compute the resultant force if a fourth tugboat was added at a new point on the barge?
5. What is the physical interpretation of the resultant force direction in this problem?

Tip: Always resolve forces into their components when dealing with forces at angles, as this simplifies combining them for resultant force calculations.