A boat is constructed in the shape of a rectangular cube. The base of the boat is 2m by 3m and the side of the boat is 2m high. The boat has a total mass of 4500kg. Hint-The density of the water is 1000kg/m^3.
a. What is the buoyant force on the boat? Explain answer.
b. Calculate the length of the side of the boat sticking out of the water.
c. What is the pressure on the bottom of the boat? Calculate this in two different ways and compare answer.
a) buoyant force equal total mass, assuming the boat floats.
b) From Archimedes principle, the total mass equals the mass of liquid displaced by the boat, so
if h=height immersed, then
2*3*h*1000 kg = 4500 kg.
Solve for h, and then subtract from 2 m to get the height out of water.
c)
Method 1:
Pressure = buoyant force / area of bottom
Method 2:
Hydrostatic pressure at depth h
= h*1000 kg/m^2
The values obtained by both methods should be equal.
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a. The buoyant force on an object is equal to the weight of the water displaced by that object. To calculate the buoyant force on the boat, we need to determine the volume of the boat that is submerged in water. Since the boat is shaped like a rectangular cube, the volume of the boat can be calculated using the formula V = lwh, where l is the length, w is the width, and h is the height of the boat.
Given:
Base length (l) = 2m
Base width (w) = 3m
Side height (h) = 2m
Volume of the boat (V) = (2m)(3m)(2m) = 12m³
Since the boat is completely submerged, the volume of water displaced is equal to the volume of the boat itself. Therefore, the buoyant force is given by the formula F = ρVg, where ρ is the density of the water, V is the volume of water displaced, and g is the acceleration due to gravity.
Given:
Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Buoyant Force (F) = (1000 kg/m³)(12m³)(9.8 m/s²) = 117600 N
Therefore, the buoyant force on the boat is 117600 Newtons.
b. To calculate the length of the side of the boat sticking out of the water, we can use the concept that the buoyant force is equal to the weight of the water displaced. The weight of the submerged portion of the boat is equal to the buoyant force acting on it. The weight of this portion can be calculated using the formula W = mg, where m is the mass of the boat and g is the acceleration due to gravity.
Given:
Mass of the boat (m) = 4500 kg
Acceleration due to gravity (g) = 9.8 m/s²
Weight of the submerged portion of the boat (W) = (4500 kg)(9.8 m/s²) = 44100 N
Since the weight of the submerged portion is equal to the buoyant force, we can equate these two values and solve for the length of the side sticking out of the water.
Weight of submerged portion (W) = Buoyant force (F) = 117600 N
Length of the side sticking out of the water = W / (density of water * acceleration due to gravity)
Length = 44100 N / (1000 kg/m³ * 9.8 m/s²) = 4.5 m
Therefore, the length of the side of the boat sticking out of the water is 4.5 meters.
c. To calculate the pressure on the bottom of the boat, we can use two different approaches:
Approach 1:
The pressure at a certain depth in a fluid can be calculated using the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Given:
Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Height of the boat (h) = 2m
Pressure on the bottom of the boat (P) = (1000 kg/m³)(9.8 m/s²)(2m) = 19600 Pa (or N/m²)
Approach 2:
Since pressure is force per unit area, we can also calculate the pressure by dividing the force by the area. The force can be represented as the weight of the boat, and the area can be represented as the base area of the boat.
Given:
Weight of the boat (W) = 4500 kg * 9.8 m/s² = 44100 N
Base area of the boat (A) = 2m * 3m = 6 m²
Pressure on the bottom of the boat (P) = Weight of the boat / Base area of the boat = 44100 N / 6 m² = 7350 Pa (or N/m²)
By comparing these two approaches, we can conclude that the pressure on the bottom of the boat is either 19600 Pa or 7350 Pa, depending on the approach used.