Archimedes principle and floating slab

In summary, the first problem involves using Archimedes principle to calculate the area of a styrofoam slab floating in water with a swimmer on top. The second problem involves using Bernouilli's equation and projectile motion to calculate the height of the water level in a tank based on the velocity and pressure of two water jets emerging from the tank. It is necessary to first calculate the velocity of the water jets and then use that to find the pressure difference inside the tank.
  • #1
ness9660
35
0
A styrofoam slab has a thickness of 12.6 cm
and a density of 470 kg/m 3 . What is the area
of the slab if it floats just awash (top of slab
is even with the water surface) in fresh water
when a 65.7 kg swimmer is aboard? Answer
in units of m 2 .

im trying to use pressure=density of water times thickness of slab times 9.8 and mass=density of water times thickness of slab times area

i think I am on the correct path here but I am sure as how to account for the force of the person on the slab in these equations




A tall water cooler tank is standing on the
floor. Some fool punched two small holes
in the tank's wall, one hole at a height of
34 cm above the floor and the other hole
60 cm directly above the first hole and 94 cm
above the floor. Each hole produces a jet of
water that emerges in a horizontal direction
but eventually hits the floor at some distance
from the tank.
If the two water jets (emerging from each
hole) hit the floor at exactly the same spot,
how high H is the water level in the tank
(relative to the room's floor)? Answer in
units of cm.


i can't even find the correct equations to use for this problem, can anyone offer some help
 
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  • #2
ness9660 said:
A styrofoam slab has a thickness of 12.6 cm
and a density of 470 kg/m 3 . What is the area
of the slab if it floats just awash (top of slab
is even with the water surface) in fresh water
when a 65.7 kg swimmer is aboard? Answer
in units of m 2 .

im trying to use pressure=density of water times thickness of slab times 9.8 and mass=density of water times thickness of slab times area

i think I am on the correct path here but I am sure as how to account for the force of the person on the slab in these equations

Yours is not the easiest approach. Use Archimiedes principle
 
  • #3
ness9660 said:
A tall water cooler tank is standing on the
floor. Some fool punched two small holes
in the tank's wall, one hole at a height of
34 cm above the floor and the other hole
60 cm directly above the first hole and 94 cm
above the floor. Each hole produces a jet of
water that emerges in a horizontal direction
but eventually hits the floor at some distance
from the tank.
If the two water jets (emerging from each
hole) hit the floor at exactly the same spot,
how high H is the water level in the tank
(relative to the room's floor)? Answer in
units of cm.


i can't even find the correct equations to use for this problem, can anyone offer some help

Do you have Bernouilli's Equation relating the velocity change of a fluid to the pressure difference?
 
  • #4
OlderDan said:
Yours is not the easiest approach. Use Archimiedes principle


ah, I've figured it out, area = Mass / Thickness * (density of water - density of slab)


im still pretty lost on the second problem.
 
  • #5
OlderDan said:
Do you have Bernouilli's Equation relating the velocity change of a fluid to the pressure difference?


P1 + qgh1 + ½qv1 = P2 + qgh2 + 1/2qv2


im just confused on how to use it relating to this problem. p1 and p2 are the initial pressures on the holes, i think, so p1 = 1000 times 9.8 times 34 and p2= 1000 times 9.8 times 60

but then you plug these in and solve for h, correct, where h would be the height of the water level in the tank

edit; ah you solve for v instead in the above equation
 
  • #6
You should be able to use this equation to get the initial velocity of each water jet leaving the tank. For each jet individually you have no height difference, a pressure difference between the inside and outside of the tank, and essentially zero velocity in the tank. When you get your two initial velocities for the two holes you have two projectiles hitting the ground in the same place.

While inside the tank, you have essentially no velocity and the pressure difference is related to the height difference
 
Last edited:
  • #7
OlderDan said:
You should be able to use this equation to get the initial velocity of each water jet leaving the tank. For each jet individually you have no height difference, a pressure difference between the inside and outside of the tank, and essentially zero velocity in the tank. When you get your two initial velocities for the two holes you have two projectiles hitting the ground in the same place.

While inside the tank, you have essentially no velocity and the pressure difference is related to the height difference

ugh, nothing I am doing is coming out right for this. for p1 and p2 I am putting in h times g times 1000, filling in ½qv1 with 1/2 times 1000 times v1, and qgh with 1000 times 9,8 times height on tank.

this is just not coming out right at all
 
  • #8
It sounds to me like you are working in the wrong direction. The first thing you need to do is figure out how long it takes a drop of water to reach the floor from the height of each hole. From that you can calculate the relative horizontal velocity of the two jets of water. From there I think you can find the pressure ratio inside the tank at the two holes, and I know you can find the pressure difference. Knowing both of those will get you to the answer.
 

1. What is Archimedes' principle?

Archimedes' principle states that an object immersed in a fluid experiences an upward force equal to the weight of the fluid it displaces. This explains why objects float or sink in a fluid.

2. How does Archimedes' principle apply to floating slabs?

Archimedes' principle is the basis for understanding the concept of buoyancy, which is what allows floating slabs to stay afloat. The weight of the slab is equal to the weight of the fluid it displaces, so as long as the weight of the slab is less than the weight of the fluid, it will float.

3. What factors affect the amount of buoyancy a floating slab experiences?

The amount of buoyancy a floating slab experiences is influenced by its weight and the density of the fluid it is submerged in. The denser the fluid, the more buoyant the slab will be. The shape and size of the slab also play a role in its buoyancy.

4. How can Archimedes' principle be applied to real-world scenarios?

Archimedes' principle is used in many applications, such as shipbuilding, designing submarines, and determining the density of objects. It is also the principle behind hydraulic systems, which use fluids to transmit and amplify forces.

5. What are some common misconceptions about Archimedes' principle and floating slabs?

One common misconception is that an object will only float if it is less dense than the fluid it is submerged in. In reality, the weight of the object and the density of the fluid both play a role in determining buoyancy. Another misconception is that the entire object needs to be submerged for Archimedes' principle to apply, but it actually only applies to the portion of the object that is submerged.

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