Question

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 magnitude of the resultant acceleration?

Answer 1

Answer:

The force exerted by the ocean tide is directed to the right (east).

The force exerted by the wind is directed to the northwest (45° N of W).

We should separate the x- and y- components of the wind force, and evaluate each component separately.

$F_x = 6000\cos(\pi/4)(-\^{x})\\F_y = 6000\sin(\pi/4)(+\^{y})$

We denote the direction to the right as the positive direction, so the x-component of the wind force is in the negative direction.

The resultant force is as follows:

$F_R = [2000 - 4242.64](\^{x}) + [4242.64](\^{y})\\F_R = -2242.64\^{x} + 4242.64\^{y}$

The resultant acceleration can be found by Newton's Second Law:

$F = ma\\\\a_R = F_R/m = -1.12\^{x} + 2.12\^{y}$

The magnitude of the resultant acceleration is

$|a_R| = \sqrt{(1.12)^2 + (2.12)^2} = 2.4 ~m/s^2$