A fire helicopter carries a 555 kg empty water bucket at the end of a cable 20.9 m long. As the aircraft flies back from a fire at a constant speed of 38.4 m/s, the cable makes an angle of 38.3° with respect to the vertical.

A) After filling the bucket with sea water, the pilot returns to the fire at the same speed with the bucket now making an angle of 8.55° with the vertical. What is the mass of the water in the bucket?

To find the mass of the water in the bucket, we need to analyze the forces acting on the helicopter and the bucket.

Let's break down the forces involved:

1. The weight of the empty bucket: This force can be represented by the formula Fg = m * g, where m is the mass of the bucket and g is the acceleration due to gravity (9.8 m/s²).
Fg = 555 kg * 9.8 m/s² = 5439 N (downward force)

2. The tension in the cable: The tension in the cable is equal to the net force acting on the bucket. Its vertical component balances the bucket's weight, and its horizontal component provides centripetal force for rotating the bucket.
Tension = Weight of the bucket = 5439 N

3. The centripetal force: The centripetal force acting on the bucket is provided by the horizontal component of the tension in the cable.
Fcentripetal = Tension * sin(38.3°)

Now, let's calculate the centripetal force using the given angle of 38.3°.

Fcentripetal = 5439 N * sin(38.3°) = 3387 N

4. The centripetal force is also given by the equation Fcentripetal = (mass of water + mass of bucket) * v^2 / r, where v is the speed of the helicopter and r is the length of the cable.

Since the helicopter returns to the fire at the same speed (38.4 m/s) but with the bucket at a different angle (8.55°), the length of the cable (r) remains the same, and we can equate the centripetal forces before and after filling the bucket with water.

Fcentripetal (before) = Fcentripetal (after)

5. Since the only change in the system is the addition of water to the bucket, we can assume that the mass of the bucket (555 kg) remains the same.

Fcentripetal (before) = Fcentripetal (after)
(mass of water + mass of bucket) * v^2 / r = mass of bucket * v^2 / r

By canceling out the r and v^2 terms on both sides and then solving for the mass of water, we get:

mass of water = mass of bucket * sin(angle before) / sin(angle after)

Now we can substitute the given values into the equation to find the mass of the water in the bucket.

mass of water = 555 kg * sin(38.3°) / sin(8.55°)
mass of water ≈ 679.7 kg

Therefore, the mass of the water in the bucket is approximately 679.7 kg.

To find the mass of the water in the bucket, we need to analyze the forces acting on the helicopter and the bucket.

First, let's consider the forces acting on the bucket when it is empty:
1. The weight of the bucket: This is given by the equation W(bucket) = m(bucket) * g, where m(bucket) is the mass of the bucket and g is the acceleration due to gravity.

Next, let's consider the forces acting on the bucket when it is filled with water:
1. The weight of the bucket: This remains the same as before, given by the equation W(bucket) = m(bucket) * g.
2. The buoyant force: This is a force exerted on the bucket by the water, opposing the weight of the bucket. It is given by the equation B = ρ * V * g, where ρ is the density of water, V is the volume of the water, and g is the acceleration due to gravity. Since the bucket is fully submerged in water, the volume of the water is the same as the volume of the bucket. So, V = m(water) / ρ, where m(water) is the mass of the water.

Let's now analyze the forces acting on the helicopter:
1. The tension in the cable: This force pulls the bucket upward and is the same throughout the cable. It can be found using the equation T = m(total) * a, where m(total) is the total mass (bucket + water) and a is the acceleration of the helicopter.
2. The vertical component of the tension: This component helps to counteract the weight of the bucket and the water. It can be found using the equation T_vertical = T * cos(θ), where θ is the angle of the cable with respect to the vertical.
3. The horizontal component of the tension: This component helps to keep the helicopter moving at a constant speed. It can be found using the equation T_horizontal = T * sin(θ).

In this problem, we are given the following information:
- The empty bucket has a mass of 555 kg.
- The cable length is 20.9 m.
- When the helicopter flies back from the fire, the angle of the cable with the vertical is 8.55°.
- The speed of the helicopter is constant at 38.4 m/s.

To find the mass of the water in the bucket, we need to find the tension in the cable when the bucket is empty and when it is filled with water. We can equate the vertical component of the tension in both cases.

1. When the bucket is empty:
- T_vertical(empty) = m(bucket) * g
- T_vertical(empty) = 555 kg * g * cos(38.3°)

2. When the bucket is filled with water:
- T_vertical(filled) = m(total) * g
- T_vertical(filled) = (555 kg + m(water)) * g * cos(8.55°)

Since the speed is constant, the tension remains the same in both cases. Therefore,
T_vertical(empty) = T_vertical(filled)

Now, we can set up the equation and solve for m(water):

555 kg * g * cos(38.3°) = (555 kg + m(water)) * g * cos(8.55°)

We can simplify the equation by canceling out the acceleration due to gravity:
555 kg * cos(38.3°) = (555 kg + m(water)) * cos(8.55°)

Now, we can rearrange the equation to solve for m(water):

m(water) = (555 kg * cos(38.3°) / cos(8.55°)) - 555 kg

By plugging in the values for the given angles and calculating the result, you will find the mass of the water in the bucket.