Archimedes’ Principle Examples
Most recent answer: 10/22/2007
Q:
A flat bottomed rectangular boat is 4 meters long and is 1.5 meters wide. if the load is 2000 kg, including the mass of the boat, how much of the boat will be submerged when it floats in a lake?
And......
A solid cube of iron 1.0 m on each side is made into sheets, from these sheets, to make a hollow cube that will not sink, what should the minimum lenth of the sides be?
A steel cube 0.25 meters on each side is suspended from a scale and immersed in water. What will the scale read?
- Dave (age 23)
Mesa AZ
- Dave (age 23)
Mesa AZ
A:
We’re not going to do your homework for you, but here’s the general approach.
Archimedes’ Principle states that the buoyant force on something is equal to the weight of the water that is displaced. In some of your problems you are given a weight which is to be canceled by a buoyant force and asked to calculate the volume of water displaced, and in others you are give the volume displaced and asked to calculate the buoyant force.
The sum of the forces on the objects in your problems will be that of gravity, that of the buoyant force pushing upwards, and any other forces, like that of the cable suspending your steel cube from the scale.
In your first problem, the volume of water displaced is just the area of the boat’s cross-section times the depth below water the bottom is. You’ll need the density of water to find out how much volume corresponds to 2000 kg.
The others are just variations on this theme.
Tom
Archimedes’ Principle states that the buoyant force on something is equal to the weight of the water that is displaced. In some of your problems you are given a weight which is to be canceled by a buoyant force and asked to calculate the volume of water displaced, and in others you are give the volume displaced and asked to calculate the buoyant force.
The sum of the forces on the objects in your problems will be that of gravity, that of the buoyant force pushing upwards, and any other forces, like that of the cable suspending your steel cube from the scale.
In your first problem, the volume of water displaced is just the area of the boat’s cross-section times the depth below water the bottom is. You’ll need the density of water to find out how much volume corresponds to 2000 kg.
The others are just variations on this theme.
Tom
(published on 10/22/2007)