Two tugboats pull a barge by means of 2 horizontal ropes at the same elevation, as shown. The tensions
in these ropes are F1 = 280 kN and F2 = 336 kN
respectively which makes an angle of 60o between them. If the direction of the motion of the barge should be to the east direction,
1. Compute the direction of the pull F1.
2. Compute the direction of the pull F2.
3. Compute the resultant pull on the barge

To solve this problem, we will break down the forces into their components and then determine the resultant force vector. Here's the step-by-step approach:

### Step 1: Represent the Forces as Vectors
Given:
- Force \( F_1 = 280 \, \text{kN} \)
- Force \( F_2 = 336 \, \text{kN} \)
- The angle between \( F_1 \) and \( F_2 \) is \( 60^\circ \).

Let:
- \( \theta_1 \) be the angle of \( F_1 \) with respect to the east (x-axis).
- \( \theta_2 \) be the angle of \( F_2 \) with respect to the east (x-axis).
  
We are required to determine \( \theta_1 \), \( \theta_2 \), and the resultant force \( R \).

### Step 2: Resolve the Forces into Components
Assume:
- \( F_1 \) is at an angle \( \theta_1 \) to the x-axis.
- \( F_2 \) is at an angle \( \theta_2 \) to the x-axis.

#### Components of \( F_1 \):
\[
F_{1x} = F_1 \cos(\theta_1)
\]
\[
F_{1y} = F_1 \sin(\theta_1)
\]

#### Components of \( F_2 \):
\[
F_{2x} = F_2 \cos(\theta_2)
\]
\[
F_{2y} = F_2 \sin(\theta_2)
\]

### Step 3: Apply the Conditions
1. The barge moves directly east, so the total y-component of the resultant force should be zero:
   \[
   F_{1y} + F_{2y} = 0.
   \]
   
2. Solve for \( \theta_1 \) and \( \theta_2 \) using the condition above.

### Step 4: Compute the Resultant Force
The resultant force \( R \) can be found by summing the x-components of both forces:
\[
R_x = F_{1x} + F_{2x}.
\]
  
Now, let's calculate these values:Here are the results of the calculations:

1. **Direction of the pull \( F_1 \)**: \( \theta_1 \approx 68.95^\circ \) (counterclockwise from the east direction).
2. **Direction of the pull \( F_2 \)**: \( \theta_2 \approx 128.95^\circ \) (counterclockwise from the east direction).
3. **Resultant pull on the barge**: \( R \approx 534.21 \, \text{kN} \).

These angles and the resultant force ensure that the barge moves directly eastward.

Would you like more details on the calculations or have any questions? Here are some related questions that might help:

1. How would the result change if the angle between the forces was different?
2. What would happen if one of the forces was increased or decreased?
3. How can the equilibrium conditions be applied to find unknown forces?
4. What role does vector resolution play in real-world physics problems?
5. How can you determine the angle between two forces using their components?

**Tip:** Always ensure the equilibrium of forces in both x and y directions when dealing with such vector problems to simplify calculations.

Two tugboats pull a barge by means of 2 horizontal ropes at the same elevation. The tensions in these ropes are F1 = 280 kN and F2 = 336 kN respectively, making an angle of 60 degrees between them. If the direction of the motion of the barge should be to the east direction:
1. Compute the direction of the pull F1.
2. Compute the direction of the pull F2.
3. Compute the resultant pull on the barge.

Solve a problem involving two tugboats pulling a barge with forces of 280 kN and 336 kN at a 60-degree angle. Learn how to compute the direction of forces and the resultant force ensuring eastward motion using vector resolution and trigonometry. Suitable for students in Grades 10-12.

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