A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.0-N force directed 15.0° north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° south of west (resistance due to the water), and (due to the wind). Find the magnitude and direction of the force . Express the direction as an angle with respect to due east.
Well, isn't this boat going through quite the adventure! I hope it doesn't get seasick from all these forces.
To find the magnitude and direction of the force due to the wind, we need to break it down into its x and y components. Let's get into some serious sailor math!
First, let's break down the initial velocity of the boat. Given that it is sailing 15.0° north of east at a speed of 2.00 m/s, we can find the x and y components of the velocity as follows:
Vx = 2.00 m/s * cos(15.0°)
Vy = 2.00 m/s * sin(15.0°)
Now, let's find the final velocity of the boat. Given that it is sailing 35.0° north of east at a speed of 4.00 m/s, we can find the x and y components of the velocity as follows:
V'x = 4.00 m/s * cos(35.0°)
V'y = 4.00 m/s * sin(35.0°)
To find the change in velocity, we subtract the initial velocity from the final velocity:
ΔVx = V'x - Vx
ΔVy = V'y - Vy
Now, it's time to consider the net force acting on the boat. The three forces are given as follows:
Force 1: 31.0 N directed 15.0° north of east
Force 2: 23.0 N directed 15.0° south of west
The net force can be calculated using the x and y components of the forces:
Net Force in the x-direction = Force 1x - Force 2x
Net Force in the y-direction = Force 1y + Force 2y
Finally, we can find the magnitude and direction of the force due to the wind using the Pythagorean theorem and the inverse tangent function:
Magnitude of the force = √(Net Force in the x-direction)^2 + (Net Force in the y-direction)^2
Direction of the force = arctan(Net Force in the y-direction / Net Force in the x-direction) + 90°
So, if you do all these calculations, you can find the magnitude and direction of the force due to the wind. Trust me, it's a real shipshape answer!
To find the magnitude and direction of force due to the wind, we need to use the principle of vector addition.
Step 1: Determine the x-components of the forces acting on the boat.
The 31.0-N force directed 15.0° north of east can be resolved into its x-component:
Fx1 = 31.0 N * cos(15.0°) = 29.888 N (positive direction)
The 23.0-N force directed 15.0° south of west can be resolved into its x-component:
Fx2 = 23.0 N * cos(180° - 15.0°) = -22.319 N (negative direction)
Step 2: Determine the y-components of the forces acting on the boat.
The 31.0-N force directed 15.0° north of east can be resolved into its y-component:
Fy1 = 31.0 N * sin(15.0°) = 8.079 N (positive direction)
The 23.0-N force directed 15.0° south of west can be resolved into its y-component:
Fy2 = 23.0 N * sin(180° - 15.0°) = -5.939 N (negative direction)
Step 3: Calculate the net force acting on the boat.
To find the net force in the x-direction, we add the x-components of the forces:
Fx = Fx1 + Fx2 = 29.888 N + (-22.319 N) = 7.569 N (in the positive direction)
To find the net force in the y-direction, we add the y-components of the forces:
Fy = Fy1 + Fy2 = 8.079 N + (-5.939 N) = 2.140 N (in the positive direction)
Step 4: Calculate the magnitude and direction of the net force.
The magnitude of the net force can be found using the Pythagorean theorem:
|F| = √(Fx^2 + Fy^2) = √(7.569 N^2 + 2.140 N^2) = √(57.308 N^2) = 7.570 N
The direction of the net force can be found using the inverse tangent function:
θ = tan^(-1)(Fy/Fx) = tan^(-1)(2.140 N/7.569 N) = 16.101°
Step 5: Convert the direction angle to a direction with respect to due east.
Since the boat is sailing north of east, the direction angle with respect to due east is 90° - θ:
Direction angle = 90° - 16.101° = 73.899°
Therefore, the magnitude of the net force is 7.570 N and its direction with respect to due east is 73.899°.
To find the magnitude and direction of the force due to the wind, we can use the principles of vector addition.
Step 1: Draw a vector diagram
Start by drawing a vector diagram to represent the forces acting on the boat. Each force will be represented by an arrow with its magnitude and direction.
Step 2: Resolve the forces into components
Since the forces are not all in the same direction, we need to resolve them into components along the east-west (x-axis) and north-south (y-axis) directions. Let's consider the east-west direction as the x-axis and the north-south direction as the y-axis.
The 31.0 N force directed 15.0° north of east can be resolved into x and y components as follows:
Fx = 31.0 N * cos(15°)
Fy = 31.0 N * sin(15°)
Similarly, for the 23.0 N force directed 15.0° south of west:
Fx = 23.0 N * cos(195°)
Fy = 23.0 N * sin(195°)
We'll also have the force due to the wind, represented by Fw, which we need to find.
Step 3: Apply the principle of vector addition
Using the principle of vector addition, the net force in the x-direction (east-west) can be found by summing the x-components of all the forces:
Net Fx = Fx1 + Fx2 + Fwx = 0 (since there is no acceleration in the x-direction)
Similarly, the net force in the y-direction (north-south) can be found by summing the y-components of all the forces:
Net Fy = Fy1 + Fy2 + Fwy = ma (where m = mass of the boat, a = acceleration of the boat)
The only unknown in these equations is the force due to the wind, Fwy.
Step 4: Solving the equations
We know the mass of the boat is 325 kg, as given in the problem. We also know the boat's initial velocity is 2.00 m/s, and it increases to 4.00 m/s after 30 seconds. We can calculate the acceleration by using the formula:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (4.00 m/s - 2.00 m/s) / 30 s
Substitute the known values into the equation Net Fy = ma, and solve for Fwy:
325 kg * Acceleration = (Fy1 + Fy2 + Fwy)
Step 5: Convert the magnitude and direction to the desired form
Once you find the magnitude and direction of the force due to the wind, you can convert the direction to an angle with respect to due east by applying basic trigonometry.
Magnitude = √(Fxw² + Fyw²)
Direction = arctan(Fyw / Fxw)
These calculations will provide you with the magnitude and direction of the force due to the wind acting on the boat.
Fnet=Fb+Fr+Fw
Fnet=ma
a=change v/change t = 2/30=0.0667
fnet=325(0.0667)=21.67N
21.67=31+(-23)+Fw
Fw=13.7N 15 North of East
35.0 is an irrelivant detail - fw can be calculated with the given times and masses