To find the sum of the two forces on the barge, we need to decompose the forces into components along the centerline of the canal (horizontal direction) and perpendicular to the centerline (vertical direction).

Step 1: Horizontal Components

Each horse pulls with a force of 827 N at an angle of 15° with respect to the centerline of the canal. The horizontal component of each force (along the centerline) is given by:

$F_{\text{horizontal}} = 827 \, \text{N} \times \cos(15^\circ)$

The total horizontal force from both horses is twice this value:

$F_{\text{total horizontal}} = 2 \times 827 \, \text{N} \times \cos(15^\circ)$

Step 2: Vertical Components

The vertical component of each force (perpendicular to the centerline) is:

$F_{\text{vertical}} = 827 \, \text{N} \times \sin(15^\circ)$

Since the horses are on opposite sides of the canal, their vertical forces will cancel each other out (one is pulling up, and the other is pulling down). So, the total vertical force is:

$F_{\text{total vertical}} = 0$

Step 3: Calculate the Total Force

Thus, the total force on the barge is simply the sum of the horizontal components:

$F_{\text{total}} = F_{\text{total horizontal}} = 2 \times 827 \, \text{N} \times \cos(15^\circ)$

Now let's compute this value.The total force exerted by both horses on the barge is approximately $1597.64 \, \text{N}$ along the centerline of the canal.

Would you like further details on the process, or have any additional questions?

Here are five related questions you might explore:

1. What would the total force be if the angle was increased to 20°?
2. How would the total force change if each horse pulled with a different force?
3. What if the canal had some resistance? How would that affect the total force?
4. How can we calculate the work done by the horses if the barge moves a certain distance?
5. What would happen if the horses pulled with different angles?

Tip: When dealing with forces in two dimensions, always break them into components to simplify calculations.