A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W). What is the direction of the resultant acceleration?Select one:a.62° N of Wb.30° N of Ec.60° N of Ed.74° N of W

To solve this problem, we need to break down the forces into their components and then use these components to find the resultant force.

1. The eastward force of 2000 N by the ocean tide can be represented as a force of 2000 N in the positive x-direction.

2. The wind force of 6000 N directed toward the northwest (45° N of W) can be broken down into its components. The x-component (westward) is -6000 cos(45°) = -4242.64 N and the y-component (northward) is 6000 sin(45°) = 4242.64 N.

3. The total force in the x-direction is the sum of the x-components of the two forces, which is 2000 N - 4242.64 N = -2242.64 N.

4. The total force in the y-direction is just the y-component of the wind force, which is 4242.64 N.

5. The resultant force can be found using the Pythagorean theorem: sqrt((-2242.64 N)^2 + (4242.64 N)^2) = 4781.92 N.

6. The direction of the resultant force can be found using the arctan function: arctan(4242.64 N / -2242.64 N) = -62°.

However, because the resultant force is in the second quadrant (northwest), we need to add 180° to this angle to get the correct direction, which is 118°.

This angle is measured from the positive x-axis (east), so to express it as an angle north of west, we subtract it from 180°: 180° - 118° = 62°.

So, the direction of the resultant acceleration is 62° N of W. Therefore, the correct answer is a. 62° N of W.