When an open-faced boat has a mass of 5750kg, including its cargo and passengers, it floats with the water just up to the side on a freshwater lake. (a)What is the volume of this boat? (b) The captain decides that it is too dangerous and wants to throw some cargo overboard so that 20% of the boat's volume will be above water. How much mass should he throw out?
I think I am overthinking this ?
For part a, is it just the V=D*M?
(a) No, you have density (D) in the wrong place. For freshwater, D = 1000 kg/m^3
Volume *(water density) = Mass of boat
(b) (reduction in mass) = density * (reduction in volume). 20% of the original mass should be thrown out.
For part a, to find the volume of the boat, you can use the equation V = D * M, where V represents the volume, D represents the density of water, and M represents the mass of the boat + cargo + passengers.
In this case, the boat floats with the water just up to the side, which means it is in equilibrium, and the weight of the boat is equal to the buoyant force acting on it. The buoyant force is given by the equation B = D * V * g, where B represents the buoyant force, D represents the density of water, V represents the volume of the submerged part of the boat, and g represents the acceleration due to gravity.
Since the boat floats with the water just up to the side, it means that the weight of the boat (including cargo and passengers) is equal to the buoyant force. Therefore, we can write the equation W = B, where W represents the weight of the boat (including cargo and passengers).
We know that the mass of the boat (including cargo and passengers) is 5750 kg. Since weight is the product of mass and acceleration due to gravity (W = M * g), we can rearrange the equation to calculate the weight. Assuming the acceleration due to gravity is approximately 9.8 m/s², we can find that W = 5750 kg * 9.8 m/s².
Now, equating W to B, we have 5750 * 9.8 = D * V * g. Rearranging the equation, we get V = (5750 * 9.8) / (D * g).
To find the volume, we need to know the density of the water. In this case, let's assume it is approximately 1000 kg/m³ (typical density of freshwater).
Substituting the values into the equation, V = (5750 * 9.8) / (1000 * 9.8), cancels out the gravity terms, leaving you with V = 5750 / 1000.
Therefore, the volume of the boat is 5.75 m³.
For part b, the captain wants to throw some cargo overboard so that 20% of the boat's volume will be above water. To determine how much mass the captain should throw out, we need to find the volume of the boat that needs to be above water.
The captain wants 20% of the boat's volume to be above water, which means 80% of the boat's volume should be submerged. To calculate the volume above water, we can multiply the total volume by 20% (0.2) and subtract it from the total volume.
Volume above water = Total volume - (0.2 * Total volume)
Substituting the known value for the total volume (5.75 m³), we get Volume above water = 5.75 - (0.2 * 5.75).
This will give you the volume above water. To calculate the mass that needs to be thrown out, you can use the equation M = D * V, where D represents the density of the cargo.
Remember to use the same units for density (kg/m³) and volume (m³) in the equation.
Once you have the volume above water, substitute the known value for the volume, and solve for the mass to determine how much cargo the captain needs to throw overboard.