A 19000 kg sailboat experiences an eastward
force 16400 N due to the tide pushing its hull
while the wind pushes the sails with a force of
55200 N directed toward the northwest (45
◦
westward of North or 45
◦
northward of West).
What is the direction of the acceleration? Which is 2.37 m/s^2
To find the direction of acceleration, we need to calculate the net force acting on the sailboat. We'll consider the forces in the east-west (horizontal) and north-south (vertical) directions separately.
Horizontal Forces:
The sailboat experiences an eastward force due to the tide of 16400 N. The wind pushes the sails with a force of 55200 N directed toward the northwest. To find the horizontal net force, we need to resolve the wind force into its east and west components.
The northwest direction is 45° westward of north or 45° northward of west. This means the angle between the wind force and the westward direction is 45°.
Using trigonometry, we can calculate the westward component of the wind force:
Westward component = Wind force * cos(angle)
= 55200 N * cos(45°)
= 55200 N * 0.7071
= 39000 N (approx.)
The eastward component of the wind force will be the same in magnitude but opposite in direction:
Eastward component = -39000 N
The net horizontal force is the sum of the forces:
Net horizontal force = Eastward force + Eastward component + Westward component
= 16400 N + (-39000 N) + 39000 N
= 16400 N
Since the net horizontal force is positive (eastward), the sailboat will experience an eastward acceleration.
Vertical Forces:
There is no vertical force mentioned in the problem statement. Hence, the net vertical force is zero, resulting in no acceleration in the vertical direction.
Direction of Acceleration:
Based on the analysis above, we found that the horizontal acceleration is eastward, and the vertical acceleration is zero.
So, the direction of the total acceleration is purely eastward.
Whoever wrote this question knows nothing about the physics of sailing.
The tidal current does not push the hull in general unless the boat is anchored to something. The current is just a velocity component of the boat. It moves with the water at the speed of the water.
All I can do is guess that there are only the two forces given in the problem acting on the boat and calculate the magnitude and direction of the acceleration.
north component of force = 55.2^103 cos 45 = 39*10^3 N north
east component of force = 16.4*10^3 - 39*10^3 = -22.6*10^3 Newtons East
magnitude of force = 10^3 sqrt(39^2+22.6^2)
= 45*10^3 Newtons
a = F/m = 45*10^3/19*10^3 = 2.37 agreed
tan of angle north of west = 39/22.6
angle north of west = 60 degrees north of west or 30 degrees west of north or 330 degrees on a compass clockwise from north.
By the way the f=ma has the wrong m if we were really doing a boat in water. When you accelerate a boat in water, you also accelerate some water around it leading to what is called "added mass" in hydrodynamics. As I said, whoever wrote this question knows nothing about the physics of boats.