A 343-kg boat is sailing 16.2° north of east at a speed of 1.98 m/s. Thirty seconds later, it is sailing 37.5° north of east at a speed of 3.76 m/s. During this time, three forces act on the boat: a 30.3-N force directed 16.2° north of east (due to an auxiliary engine), a 22.3-N force directed 16.2° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east. Magnitue in Newtons and and direction
To find the magnitude and direction of the force W, we can use the fact that the net force acting on an object is equal to its mass times its acceleration.
First, let's calculate the acceleration of the boat. We can find the change in velocity by subtracting the initial velocity from the final velocity:
Change in velocity = Final velocity - Initial velocity
For the eastward component of velocity:
Δvᵢₑ = vfᵢₑ - vᵢₑ = (3.76 m/s * cos(37.5°)) - (1.98 m/s * cos(16.2°))
For the northward component of velocity:
Δvₙ = vfₙ - vₙ = (3.76 m/s * sin(37.5°)) - (1.98 m/s * sin(16.2°))
Next, we can calculate the acceleration using the formula:
a = Δv / t
For the eastward component of acceleration:
aᵢₑ = Δvᵢₑ / t = Δvᵢₑ / (30 s)
For the northward component of acceleration:
aₙ = Δvₙ / t = Δvₙ / (30 s)
Now we can find the net force using Newton's second law:
ΣF = ma
For the eastward component of force:
Fᵢₑ = maᵢₑ = (343 kg) * aᵢₑ
For the northward component of force:
Fₙ = maₙ = (343 kg) * aₙ
Now, we need to find the components of the force W. Let the force W have components F𝑤ᵢₑ and F𝑤ₙ along the eastward and northward directions, respectively.
Using vector addition, we can write the net force equation in terms of components:
Fᵢₑᵢ + Fₙᵢ = Fᵢₑ + Fₙ + F𝑤ᵢₑ
Fᵢₑₙ + Fₙₙ = Fₙ - Fᵢₑ + F𝑤ₙ
We can rewrite the equations as:
ΣFᵢₑ = Fᵢₑ + F𝑤ᵢₑ,
ΣFₙ = Fₙ + F𝑤ₙ
We already know the values for Fᵢₑ and Fₙ (auxiliary engine and water resistance) and we can find the values for ΣFᵢₑ and ΣFₙ using the equations obtained above.
Finally, we can calculate the magnitude and direction of W:
Magnitude of W = √(F𝑤ᵢₑ² + F𝑤ₙ²)
Direction of W = atan(F𝑤ᵢₑ / F𝑤ₙ) + 90° (with respect to due east)
To find the magnitude and direction of force W, we need to analyze the net force acting on the boat.
Let's break down the given information:
The boat experiences three forces:
1. Auxiliary engine force: 30.3 N, directed 16.2° north of east.
2. Water resistance force: 22.3 N, directed 16.2° south of west.
3. Wind force (W): unknown.
We can represent these forces using vector notation:
Auxiliary engine force: F_aux = 30.3 N at 16.2° north of east.
Water resistance force: F_res = 22.3 N at 196.2° (180° + 16.2°) south of east.
Wind force: F_wind = W N at θ° east of north.
Now let's analyze the net force acting on the boat before and after the 30-second time interval:
Before:
The boat is initially moving at a velocity of 1.98 m/s at an angle of 16.2° north of east. The net force acting on the boat is zero (since it is sailing at a constant velocity), so the sum of all the forces acting on the boat must be zero:
F_net = F_aux + F_res + F_wind = 0
After:
The boat is sailing at a new velocity of 3.76 m/s at an angle of 37.5° north of east. Now, we can calculate the net force:
F_net = F_aux + F_res + F_wind
To find the magnitude of force W, we need to set up equations for the x and y components of the net force in terms of W:
x-component:
Sum of horizontal forces = F_aux * cos(16.2°) - F_res * cos(196.2°) + W * cos(θ°) = m * (change in x-velocity) / change in time
y-component:
Sum of vertical forces = F_aux * sin(16.2°) + F_res * sin(196.2°) + W * sin(θ°) = m * (change in y-velocity) / change in time
Using the given values and solving these equations will give us the magnitude and direction of force W.