A raft is made of 14 logs lashed together. Each is 29.5 cm in diameter and has a length of 6.50 m. How many people (whole number) can the raft hold before they start getting their feet wet, assuming the average person has a mass of 75.0 kg? Do not neglect the weight of the logs. Assume the density of wood is 600 kg/m³
Vr = 14 * pi*r^2 * h = 3.14*(0.295/2)^2 * 6.5 = 6.22 m^3 = Volume of the raft.
Vr*Dr = 6.22m^3 * 600kg/m^3 = 3732 kg =
Mass of the raft.
Vw*Dw = 6.22m^3 * 1000kg/m^3 = 6220 kg =
Mass of the water displaced.
6220 - 3732 = 2488 kg can be added to the raft.
2488kg * 1person/75kg = 33 People.
To determine the number of people the raft can hold before they start getting their feet wet, we need to consider the buoyant force exerted by the raft and compare it to the total weight of the logs and people.
1. First, let's calculate the total weight of the logs:
Each log has a diameter of 29.5 cm, which converts to a radius of 14.75 cm (0.1475 m).
The volume of each log is given by V = πr²h, where r is the radius and h is the length.
V = π × (0.1475 m)² × 6.50 m = 0.218 m³ (rounded to 3 decimal places)
The mass of each log can be calculated using the density formula: density = mass/volume.
Since the density of wood is given as 600 kg/m³, we have: mass = density × volume.
Mass of each log = 600 kg/m³ × 0.218 m³ = 130.8 kg (rounded to 2 decimal places)
2. Next, let's calculate the total weight of the logs by multiplying the mass of each log by the number of logs:
Total weight of the logs = 14 logs × 130.8 kg = 1821.6 kg (rounded to 1 decimal place)
3. Now, let's calculate the buoyant force exerted by the raft:
Buoyant force = Density of water × Volume of the submerged part of the raft × g
The volume of the raft depends on the submerged part, which is affected by the number of people on the raft.
Since we have 14 logs lashed together, the length of each log (6.50 m) represents the maximum length of the raft.
The submerged part of the raft is the length of the raft (6.50 m) multiplied by the number of logs that are submerged.
Submerged length = 6.50 m × (14 - n) logs, where n is the number of logs above the water surface.
Total volume of submerged part = π × (0.1475 m)² × (6.50 m) × (14 - n)
Buoyant force = (Density of water) × (Total volume of submerged part) × g
For simplicity, we will approximate the density of water as 1000 kg/m³, and g as 9.8 m/s².
4. Lastly, let's determine the maximum number of people the raft can hold without submerging:
To find this, we need to compare the weight of the logs and the weight of the people to the buoyant force.
Weight of the logs = 1821.6 kg (rounded to 1 decimal place)
Weight of each person = 75.0 kg
Total weight of people = (75.0 kg) × n, where n is the number of people on the raft.
We want the buoyant force to be greater than or equal to the combined weight of the logs and people:
Buoyant force ≥ Weight of logs + Total weight of people
Now, let's solve for n:
(Density of water) × (Total volume of submerged part) × g ≥ Weight of logs + Total weight of people
Setting the values in the above equation:
(1000 kg/m³) × [π × (0.1475 m)² × (6.50 m) × (14 - n)] × (9.8 m/s²) ≥ 1821.6 kg + (75.0 kg) × n
Simplifying the equation and solving for n:
[π × (0.021615625 m²) × (90.1 - 6.5n)] ≥ 1821.6 + 75n
Now, we can solve this equation step-by-step:
1. Distribute π to each term inside the brackets: π × (0.021615625 m²) × (90.1 - 6.5n) ≥ 1821.6 + 75n
2. Multiply the terms inside the brackets: 0.692735781125 π - 0.141649140625 π n ≥ 1821.6 + 75n
3. Rearrange the terms: -0.141649140625 π n - 75n ≥ 1821.6 - 0.692735781125 π
4. Combine like terms: (-0.141649140625 π - 75)n ≥ 1821.6 - 0.692735781125 π
5. Divide both sides of the inequality by (-0.141649140625 π - 75): n ≤ (1821.6 - 0.692735781125 π) / (-0.141649140625 π - 75)
Now, calculating this expression will give us the maximum number of whole people the raft can hold without submerging.
To determine the maximum number of people the raft can hold before they start getting their feet wet, we need to consider the buoyancy and weight of the raft.
First, let's calculate the weight of a single log:
Density of wood = 600 kg/m³
Volume of a log = πr²h, where r is the radius (diameter/2) and h is the height (length of the log)
Given: Diameter = 29.5 cm = 0.295 m, Length = 6.50 m
Radius (r) = 0.295 m / 2 = 0.1475 m
Volume of a log = π * (0.1475 m)² * 6.50 m
Now, we can calculate the weight of a single log:
Weight of a single log = Density * Volume
Weight of a single log = 600 kg/m³ * (π * (0.1475 m)² * 6.50 m)
Next, we need to calculate the total weight of the logs:
Total weight of the logs = Weight of a single log * Number of logs
Given: Number of logs = 14
Total weight of the logs = Weight of a single log * 14
Now, let's calculate the buoyant force:
Buoyant force = Density of water * Volume of water displaced
Given: Density of water = 1000 kg/m³ (approximate value)
Volume of water displaced can be calculated as the volume of the logs.
Now, let's calculate the total volume of the logs:
Total volume of the logs = Volume of a single log * Number of logs
Finally, we can calculate the maximum number of people the raft can hold:
Maximum number of people = (Buoyant force - Total weight of the logs) / Weight of an average person
Given: Mass of an average person = 75.0 kg
Using this formula, we can now substitute the values and calculate the maximum number of people.