# Homework Help: Pressure and Buoyant Force problem

1. Nov 13, 2011

### ChunkymonkeyI

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution
Fb=F(bottom)-F(top)
480=18 minus f(top)
F(top)=-462 and we cant have it negative so idk what im 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!!

2. Nov 13, 2011

### BruceW

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. Nov 13, 2011

### ChunkymonkeyI

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: Nov 13, 2011
4. Nov 13, 2011

### ChunkymonkeyI

Do I solve for the sum of the forces by (480 plus 18 plus x)=0?

5. Nov 13, 2011

### BruceW

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. Nov 13, 2011

### ChunkymonkeyI

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. Nov 13, 2011

### BruceW

You're right that the buoyancy force is 498N. And this is what you must plug in to F=mg. (Not the apparent force).

8. Nov 13, 2011

### ChunkymonkeyI

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. Nov 13, 2011

### ChunkymonkeyI

Wait I meant 2 use 480N/.0508 m^3 but I still dont get 964 but instead 9448.8189 so idk what Im doing wrong

10. Nov 14, 2011

### BruceW

Right. You did this calculation:
$$V = \frac{m_w}{ \rho_w}$$
(where$m_w$ 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.