A fisherman poles a boat as he searches for his next catch. He pushes parallel to the length of the light pole, exerting a force of 245 on the bottom of a shallow lake. The pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of 34.5° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 370 kg.

(a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force.
(b) Assume the forces are constant over a short interval of time. Find the velocity of the boat 0.435 s after the moment described. (Give the magnitude.)
(c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, what would happen to the buoyant force and the acceleration of the boat?

(a) The magnitude of the buoyant force can be found using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object. In this case, the boat is floating in the water, so the buoyant force is equal to the weight of the boat.

The weight of the boat can be calculated using the formula: weight = mass × gravity, where mass is the mass of the boat including its cargo and the worker, and gravity is the acceleration due to gravity.

Weight = mass × gravity
Weight = 370 kg × 9.8 m/s²
Weight = 3626 N

Therefore, the magnitude of the buoyant force is 3626 N.

(b) To find the velocity of the boat 0.435 s after the moment described, we can 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.

The net force acting on the boat can be calculated by summing the horizontal forces acting on the boat. In this case, the only horizontal force is the drag force exerted by the water, which is equal to 47.5 N.

Net force = Drag force
Net force = 47.5 N

We can rearrange Newton's second law to solve for acceleration:

Net force = mass × acceleration
47.5 N = 370 kg × acceleration
acceleration = 47.5 N / 370 kg
acceleration = 0.1284 m/s²

Now, using the equation for accelerated motion:

velocity = initial velocity + (acceleration × time)
velocity = 0.857 m/s + (0.1284 m/s² × 0.435 s)
velocity = 0.857 m/s + 0.0559 m/s
velocity = 0.912 m/s

Therefore, the magnitude of the velocity of the boat 0.435 s after the moment described is 0.912 m/s.

(c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, the buoyant force would remain the same. This is because the buoyant force depends on the weight of the boat, which is not affected by the angle of the pole.

The acceleration of the boat would also remain the same, as long as the net force acting on the boat remains constant. In this case, if the horizontal forces (such as drag force) and the exerted force against the bottom remain the same, the net force and acceleration would remain the same.

So, both the buoyant force and the acceleration would remain the same if the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same.

(a) To find the magnitude of the buoyant force exerted vertically upward on the boat, we first need to find the weight of the boat.

The weight of an object can be calculated using the formula:
Weight = mass × acceleration due to gravity

Given that the mass of the boat is 370 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the boat:

Weight = 370 kg × 9.8 m/s^2 = 3626 N

Since the buoyant force is equal in magnitude but opposite in direction to the weight of the boat, the magnitude of the buoyant force is also 3626 N.

(b) To find the velocity of the boat 0.435 s after the moment described, we can use Newton's second law of motion:

Net force = mass × acceleration

The net force is the vector sum of all the forces acting on the boat. In this case, we have the force exerted by the fisherman, the drag force, and the buoyant force. Since the buoyant force acts vertically upward, it does not contribute to the net force in the horizontal direction.

The horizontal net force is the vector sum of the force exerted by the fisherman and the drag force, and it can be calculated as follows:

Horizontal Net Force = Force exerted by fisherman - Drag force

Horizontal Net Force = 245 N - 47.5 N = 197.5 N

Using Newton's second law, we can calculate the horizontal acceleration of the boat:

Acceleration = Net force / mass

Acceleration = 197.5 N / 370 kg ≈ 0.5338 m/s^2

Since the forces are assumed to be constant over a short interval of time, we can use the kinematic equation to find the velocity of the boat:

v = u + at

where:
v = final velocity (unknown)
u = initial velocity = 0.857 m/s (given)
a = acceleration = 0.5338 m/s^2 (calculated)
t = time = 0.435 s (given)

v = 0.857 m/s + (0.5338 m/s^2)(0.435 s) ≈ 1.08 m/s

Therefore, the magnitude of the velocity of the boat 0.435 s after the moment described is approximately 1.08 m/s.

(c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, the buoyant force would still be equal to the weight of the boat. This is because the buoyant force is determined by the weight of the water displaced by the boat, which does not change as long as the mass of the boat remains constant.

However, the acceleration of the boat would change. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Since the exerted force against the bottom remained the same, the change in the angle of the pole would affect the horizontal net force acting on the boat. If the angle increased, the horizontal net force would decrease, resulting in a smaller acceleration of the boat.

To find the answers to the given questions, we need to apply the principles of Newton's laws of motion and buoyancy.

(a) To determine the magnitude of the buoyant force, we need to consider the equilibrium condition of the boat in the vertical direction. The sum of the forces in the vertical direction should be zero. Therefore, the buoyant force must balance the weight of the boat and its cargo.

The weight of the boat can be calculated using:
Weight = mass * gravity acceleration

Given:
mass of the boat (including cargo and worker) = 370 kg
gravity acceleration = 9.8 m/s^2 (approximate value)

Weight = 370 kg * 9.8 m/s^2 = 3626 N

Since the buoyant force is vertically upward, its magnitude must equal the weight of the boat to maintain equilibrium:

Magnitude of buoyant force = 3626 N

(b) To calculate the boat's velocity at a given time, we need to analyze the forces acting on the boat in the horizontal direction.

The forces acting on the boat are:
- The applied force exerted by the fisherman parallel to the length of the pole (245 N).
- The horizontal drag force exerted by the water (47.5 N).
- The frictional force opposing the boat's motion (considered negligible in this case).

Using Newton's second law of motion:
Sum of forces = mass * acceleration

The net force acting on the boat can be calculated as the difference between the applied force and the drag force:

Net force = Applied force - Drag force

Net force = 245 N - 47.5 N = 197.5 N

Now we can calculate the acceleration using the equation:
Acceleration = Net force / mass

Acceleration = 197.5 N / 370 kg = 0.5338 m/s^2 (approximate value)

To find the velocity at 0.435 s after the given moment, we can use the equation of uniformly accelerated motion:
Velocity = Initial velocity + (Acceleration * Time)

Given:
Initial velocity = 0.857 m/s
Time = 0.435 s

Velocity = 0.857 m/s + (0.5338 m/s^2 * 0.435 s) ≈ 1.078 m/s

So, the magnitude of the boat's velocity 0.435 s later is approximately 1.078 m/s.

(c) If the angle of the pole with respect to the vertical increases but the exerted force against the bottom remains the same, the buoyant force and the acceleration of the boat will remain unaffected.

The buoyant force only depends on the weight of the boat and the cargo, not on the angle of the pole. Therefore, it will remain the same.

Similarly, the acceleration of the boat, which depends on the net force applied to it (and not on the angle of the pole), will also remain the same as long as the applied force and drag force remain unchanged.