A 15700 kg sailboat experiences an eastward force 22000 N due to the tide pushing its hull while the wind pushes the sails with a force of 80900 N directed toward the northwest (45◦ westward of North or 45◦ northward
of West).
A) What is the magnitude of the resultant acceleration of the sailboat?Answer in units of m/s^2
B.) What is the direction of the boat’s acceleration? Answer in units of ◦ (N of West)
where did you get 135 degrees when finding the resultant force?
A) To find the magnitude of the resultant acceleration, we need to use the forces acting on the sailboat and divide it by the mass of the boat.
The eastward force due to the tide is 22000 N, which we can consider as positive. The northwest force due to the wind is 80900 N. To find the eastward component of the wind force, we can use the cosine of the angle between the northwest direction and the eastward direction, which is 45°. The eastward component of the wind force is 80900 N * cos(45°) = 80900 N * (√2/2) = 80900 N * 0.707 ≈ 57194 N.
Now we can calculate the resultant force in the east-west direction:
Resultant force = Eastward force + Eastward component of the wind force
= 22000 N + 57194 N
= 79194 N
To find the acceleration, we divide the resultant force by the mass of the sailboat:
Acceleration = Resultant force / Mass
= 79194 N / 15700 kg
≈ 5.04 m/s^2
Therefore, the magnitude of the resultant acceleration of the sailboat is approximately 5.04 m/s^2.
B) To find the direction of the boat's acceleration, we can use the vector sum of the forces. We have an eastward force of 22000 N and a northwest force of 80900 N.
To find the northward component of the wind force, we can use the sine of the angle between the northwest direction and the northward direction, which is 45°. The northward component of the wind force is 80900 N * sin(45°) = 80900 N * (√2/2) = 80900 N * 0.707 ≈ 57194 N.
Therefore, the northward force on the boat is equal to the northward component of the wind force, which is approximately 57194 N.
Since the eastward force due to the tide is much smaller than the wind force, the boat's acceleration will be in the northwest direction, close to the direction of the wind force.
Hence, the direction of the boat's acceleration is approximately 45° northward of West.
To find the magnitude of the resultant acceleration of the sailboat, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Given:
Mass of sailboat (m) = 15700 kg
Force due to tide (F_tide) = 22000 N
Force due to wind (F_wind) = 80900 N
A) To find the magnitude of the resultant acceleration (a), we need to find the net force acting on the sailboat:
Net force (F_net) = F_tide + F_wind
F_net = 22000 N + 80900 N
F_net = 102900 N
Using the formula F = m × a, we can find the acceleration:
102900 N = 15700 kg × a
a = 102900 N / 15700 kg
a ≈ 6.56 m/s^2
Therefore, the magnitude of the resultant acceleration of the sailboat is approximately 6.56 m/s^2.
B) The direction of the boat's acceleration can be determined by analyzing the given forces.
The force due to the tide is acting eastward, while the force due to the wind is acting northwest (45 degrees westward of North or 45 degrees northward of West).
Since the force due to the wind is greater than the force due to the tide, the resultant force will be in the direction of the wind.
Since the wind is blowing towards the northwest, 45 degrees westward of North or 45 degrees northward of West, the direction of the boat's acceleration will also be in that direction.
Therefore, the direction of the boat's acceleration is 45 degrees northwest (45 degrees northward of West).
To find the magnitude of the resultant acceleration of the sailboat, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
A) To calculate the magnitude of the resultant acceleration, we need to sum up the forces acting on the sailboat in the x and y directions separately. The x direction is taken to be eastward, and the y direction is taken to be northward.
Let's break down the forces:
1. Eastward force due to tide = 22000 N
2. Northwest force due to wind = 80900 N
We can resolve the northwest force into its north and west components using trigonometry. The angle with the north direction is 45 degrees.
North component = 80900 N * sin(45°)
West component = 80900 N * cos(45°)
Now, let's calculate the net force in the x direction (eastward):
Net force in the x direction = Eastward force - West component
= 22000 N - (80900 N * cos(45°))
Similarly, let's calculate the net force in the y direction (northward):
Net force in the y direction = North component
= 80900 N * sin(45°)
Now we can apply Newton's second law to find the resultant acceleration:
Resultant acceleration = Net force magnitude / Mass
= √((Net force in the x direction)^2 + (Net force in the y direction)^2) / Mass
= √((22000 N - (80900 N * cos(45°)))^2 + (80900 N * sin(45°))^2) / 15700 kg
Evaluate this expression to find the magnitude of the resultant acceleration in units of m/s^2.
B) To find the direction of the boat's acceleration, we need to determine the angle that the resultant acceleration vector makes with the west direction.
Angle = arctan((Net force in the y direction) / (Net force in the x direction))
= arctan((80900 N * sin(45°)) / (22000 N - (80900 N * cos(45°))))
Evaluate this expression to find the angle in units of ◦ (N of West).
Fr = 22,000N @ 0o+80,900N @ 135o =
Resultant force.
A. X = 22000 + 80900*cos135 = -35,205 N.
Y = 80900*sin135 = 57,205 N.
Fr^2=X^2 + Y^2 = (-35205)^2 + (57205)^2
Fr = 67,170 N = Resultant force.
a = Fr/m = 67170/15700 = 4.3 m/s^2.
B. tanAr = Y/X = 57205/-35205=-1.6249
Ar = -58.4o.
A = 180 - _58.4 = 122o, CCW = 58o N of West = Direction.