Pressure and Buoyant Force problem

In summary: Right. You did this calculation:V = \frac{m_w}{ \rho_w}(wherem_w is the mass of the water and \rho_w is the density of water.) And this calculation gave you a volume of 0.0508 m^3 This is the correct volume.Now you've got the volume of the person, and the weight of the person, you can find the density of the person. But you have used the mass of the water again, which gave you the density of water. So you need to use the mass of the person to find the density of the... person.
  • #1
ChunkymonkeyI
35
0

Homework Statement


A downward force of 18.0 N must be applied to a woman weighing 480.0 N to keep her completely submerged in water. What is the density of her body


Homework Equations


Density=m/v
Fb=density of fluid times volume of fluid times g
Fb=F(bottom) minus F(top)

The Attempt at a Solution


Fb=F(bottom)-F(top)
480=18 minus f(top)
F(top)=-462 and we can't have it negative so idk what I am doing wrong. Also even if it was positive, I plug it in for F=mg to solve for m and then multiply it by the volume which does not get me the density answer please help me and explain the steps!
 
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  • #2
you are right that the buoyant force is equal to the weight of the water displaced. There are two other forces on the woman: her weight and the extra force pushing her under. You just need to solve for the sum of these forces to be equal to zero.

I don't understand what your F(bottom)-F(top) means... P.S. its a pretty morbid physics problem.
 
  • #3
BruceW said:
you are right that the buoyant force is equal to the weight of the water displaced. There are two other forces on the woman: her weight and the extra force pushing her under. You just need to solve for the sum of these forces to be equal to zero.

I don't understand what your F(bottom)-F(top) means... P.S. its a pretty morbid physics problem.

F(bottom) is the force that is acting at the bottom and F(top) is the force acting on the top and f(bottom)-f(top) equals the buoyant force. How do I solve for the sum of these forces to equal zero?
 
Last edited:
  • #4
Do I solve for the sum of the forces by (480 plus 18 plus x)=0?
 
  • #5
The buoyant force is equal to F(pressure at bottom) - F(pressure at top). These are not the total forces on the object, these are the forces due to the pressure of water only. And it turns out that this is equal to the weight of the water displaced.

So this is one of 3 forces on the object. The other two are its own weight and the extra force pushing down. So you would have (480 plus 18 plus x)=0 where x is the weight of the water displaced. So this gives you the weight of the water displaced. And you can use this to find the volume (and therefore density, since you have weight) of the person.
 
  • #6
BruceW said:
The buoyant force is equal to F(pressure at bottom) - F(pressure at top). These are not the total forces on the object, these are the forces due to the pressure of water only. And it turns out that this is equal to the weight of the water displaced.

So this is one of 3 forces on the object. The other two are its own weight and the extra force pushing down. So you would have (480 plus 18 plus x)=0 where x is the weight of the water displaced. So this gives you the weight of the water displaced. And you can use this to find the volume (and therefore density, since you have weight) of the person.

Ok so I got x=-498 and now I plug it into the formula Fapparent=Fweight-f(buoyant)
Fa=480+498
Fa=978N
Now I plug it into F=mg
m=F/g
m=978N/9.80 m/s^2
m=99.8 kg
V=99.8 kg/1000 kg/m^3
V=.0998 m^3
Density=m/v
Density=480N/.0998 m^3 but the answer should be 964 kg/m^3 but I didn't get that so Idk what I'm doing wrong
 
  • #7
You're right that the buoyancy force is 498N. And this is what you must plug into F=mg. (Not the apparent force).
 
  • #8
BruceW said:
You're right that the buoyancy force is 498N. And this is what you must plug into F=mg. (Not the apparent force).

Ok so I plugged it in for F=mg
m=F/g
m=498N/9.80 m/s^2
m=50.8 kg
Then I solve for the volume
V=50.8 kg/1000 kg/m^3
V=.0508 m^3
Then I used Density=m/v
Density=50.8 kg/.0508 m^3
Density=1000 kg/m^3 but Ik the answer is 964 kg/m^3 so I'm not sure where I went wrong in my step
 
  • #9
Wait I meant 2 use 480N/.0508 m^3 but I still don't get 964 but instead 9448.8189 so idk what I am doing wrong
 
  • #10
Right. You did this calculation:
[tex]V = \frac{m_w}{ \rho_w}[/tex]
(where[itex]m_w[/itex] is the mass of the water and [itex]\rho_w[/itex] is the density of water.) And this calculation gave you a volume of 0.0508 m^3 This is the correct volume.

Now you've got the volume of the person, and the weight of the person, you can find the density of the person. But you have used the mass of the water again, which gave you the density of water. So you need to use the mass of the person to find the density of the person.
 

1. What is pressure and how does it relate to buoyant force?

Pressure is the force exerted by a fluid on an object. It is directly related to buoyant force, which is the upward force exerted by a fluid on an object immersed in it. This buoyant force is equal to the weight of the fluid that the object displaces.

2. How do you calculate the buoyant force on an object?

The buoyant force on an object can be calculated using the formula Fb = ρVg, where ρ is the density of the fluid, V is the volume of the fluid displaced by the object, and g is the acceleration due to gravity.

3. What factors affect the amount of buoyant force on an object?

The amount of buoyant force on an object is affected by the density of the fluid, the volume of the fluid displaced by the object, and the acceleration due to gravity. Additionally, the shape and size of the object, as well as its depth in the fluid, can also impact the buoyant force.

4. How does the concept of buoyancy explain why objects float or sink?

According to Archimedes' principle, an object will float if the buoyant force acting on it is greater than the weight of the object. If the buoyant force is less than the weight of the object, it will sink. This is why objects with lower density than the fluid they are immersed in will float, while objects with higher density will sink.

5. Can buoyant force also act on objects in air?

Yes, buoyant force can also act on objects in air. However, the density of air is much lower than that of water, so the buoyant force is typically negligible. This is why objects that are less dense than air, such as helium balloons, will float in the air.

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