a rectangular barge, 5 meters long and 2 meters wide, floats in fresh water. suppose that a 400kg crate of auto parts is loaded onto the barge. show that the barge floats 4cm deeper.
new volume submerged = 5 * 2 * h
so new mass of water displaced = 5 * 2 * h * 1000 kg/m^3 = 10,000 h kg
so Archimedes says
10,000 h = 400
h = 4/100 meters = 4 cm
Well, that's quite a heavy load for a little barge! But let's work through it.
To show that the barge floats 4cm deeper, we need to calculate the change in displacement.
First, let's calculate the original displacement of the barge:
Displacement = length x width x height
Displacement = 5m x 2m x h (where h is the original height or depth)
Now, let's calculate the new displacement after the crate is loaded onto the barge:
Displacement_after = length x width x (h + 0.04m) (where h + 0.04m is the new height or depth)
To show that the barge floats 4cm deeper, we need to compare the two displacements:
Displacement_after - Displacement = (5m x 2m x (h + 0.04m)) - (5m x 2m x h)
Displacement_after - Displacement = 10m^2 x 0.04m
Displacement_after - Displacement = 0.4m^3
So, the barge floats 0.4m^3 deeper.
Now, let's have a little fun with this calculation:
If you think about it, a 400kg crate of auto parts doesn't exactly sound like a clown-friendly cargo. Maybe the barge would float a lot more cheerfully if it was carrying cotton candy or squeaky rubber ducks! But that's just my clown's perspective on cargo choices. Keep the laughter coming, my friend!
To solve this problem, we will use the principle of buoyancy. The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
1. Calculate the volume of the barge:
The volume of a rectangular barge is given by the formula: Volume = Length × Width × Height. Since the barge is floating, the height is equal to the depth it sinks in the water.
Volume = 5 meters × 2 meters × Depth
Volume = 10 square meters × Depth
2. Calculate the initial weight of the barge:
The weight of an object is equal to its mass multiplied by the acceleration due to gravity.
Weight = Mass × Gravity
Weight = Density × Volume × Gravity
Weight = Density × 10 square meters × Depth × Gravity
3. Calculate the final weight of the barge with the added crate:
Weight with crate = Mass of barge + Mass of crate
Weight with crate = Density × 10 square meters × Depth × Gravity + 400 kg × Gravity
4. Since the barge and crate are floating, the weight of the water displaced by the combination must equal the weight of the combined system:
Weight with crate = Density of water × Volume of barge with crate × Gravity
5. Substitute the values:
Density × 10 square meters × Depth × Gravity + 400 kg × Gravity = Density of water × Volume of barge with crate × Gravity
6. Solve for the depth:
Density × 10 square meters × Depth + 400 kg = Density of water × Volume of barge with crate
7. Rearrange the equation:
Density × 10 square meters × Depth = Density of water × Volume of barge with crate - 400 kg
8. Divide both sides by (Density × 10 square meters):
Depth = (Density of water × Volume of barge with crate - 400 kg) / (Density × 10 square meters)
9. Substitute the given values:
Depth = (1000 kg/m^3 × 10 square meters × Depth - 400 kg) / (Density × 10 square meters)
10. Simplify and solve for Depth:
Depth = (10000 kg - 400 kg) / Density
Depth = 9600 kg / Density
Since we are interested in finding the change in depth, we can subtract the initial depth from the final depth:
Change in Depth = Final Depth - Initial Depth
Change in Depth = 9600 kg / Density - 10000 kg / Density
Change in Depth = -400 kg / Density
Given that the change in depth is 4 cm, we can set up the following equation:
-400 kg / Density = 0.04 meters
To isolate the density, we can divide both sides by -0.04 meters:
Density = -400 kg / -0.04 meters
Density = 10000 kg/m^3
Therefore, the density of the fresh water is 10000 kg/m^3.
To calculate the change in depth of a floating object due to an added weight, we can use Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Given:
- Length of the barge (L) = 5 meters
- Width of the barge (W) = 2 meters
- Mass of the crate (m) = 400 kg
- Initial depth of the barge (H_initial) = ?
- Change in depth of the barge (ΔH) = 4 cm
Let's break down the problem into steps:
Step 1: Calculate the area of the barge's base (A_base):
A_base = Length × Width = L × W
Step 2: Calculate the volume of water displaced by the barge without the crate (V_initial):
V_initial = A_base × H_initial
Step 3: Calculate the weight of the displaced water without the crate (W_water_initial):
W_water_initial = V_initial × density_water × g
density_water = 1000 kg/m^3 (density of fresh water)
g = 9.8 m/s^2 (acceleration due to gravity)
Step 4: Calculate the volume of water displaced by the barge with the crate (V_final):
V_final = A_base × (H_initial + ΔH)
Step 5: Calculate the weight of the displaced water with the crate (W_water_final):
W_water_final = V_final × density_water × g
Step 6: Calculate the weight of the crate (W_crate):
W_crate = m × g
Using Archimedes' principle, we can determine that the change in depth of the barge is equal to the ratio between the weight of the crate and the weight of the displaced water:
ΔH = (W_crate / W_water_initial) × ΔH
Now let's substitute the respective values into the equation and calculate the change in depth:
Step 7: Substitute the values into the equation:
ΔH = (W_crate / W_water_initial) × ΔH
ΔH = (W_crate / (V_initial × density_water × g)) × ΔH
Step 8: Calculate the change in depth (ΔH):
ΔH = (W_crate / (V_initial × density_water × g)) × ΔH
ΔH = (m × g / (A_base × H_initial × density_water × g)) × ΔH
Now you can substitute the given values to calculate the change in depth of the barge in meters.