A 2,000-kg sailboat experiences an eastward force of 2,500 N by the ocean tide and a wind force against its sails with magnitude of 6,000 N directed toward the northwest (45° N of W).What is the direction of the resultant acceleration?
A) 23° N of W
B) 12° N of W
C) 60° N of E
D) 68° N of W
E) 68° N of E

Question

A 2,000-kg sailboat experiences an eastward force of 2,500 N by the ocean tide and a wind force against its sails with magnitude of 6,000 N directed toward the northwest (45° N of W).What is the direction of the resultant acceleration?
A) 23° N of W
B) 12° N of W
C) 60° N of E
D) 68° N of W
E) ​68° N of E

Quizplus · Accepted Answer

First, we need to find the vector sum of the two forces acting on the sailboat. To do this, we can break the wind force into its x and y components:
Fx = - 6000 cos(45°) = -4243 N (westward)
Fy = 6000 sin(45°) = 4243 N (northward)
Next, we can add up the x and y components of all the forces acting on the sailboat:
ΣFx = -4243 N + 0 N = -4243 N (westward)
ΣFy = 2500 N + 4243 N = 6743 N (northward)
Now we can use the Pythagorean theorem to find the magnitude of the resultant force:
FR = sqrt((-4243 N)^2 + (6743 N)^2) = 7977 N
Finally, we can use the inverse tangent function to find the direction of the resultant force:
θ = tan^-1(6743 N / 4243 N) = 60.11°
Since the northward force is greater than the westward force, the direction of the resultant force will be slightly north of east. D is the closest answer choice to this direction.

A 2,000-Kg Sailboat Experiences an Eastward Force of 2,500 N