# Can someone please check my homework on fluid dynamics?

Hi there, I just would like if someone could check over my work on my current homework assignment involving fluid dynamics.

## Homework Statement

1) Calculate the absolute pressure at an ocean depth of two kilometers. Assume the density of seawater is 1000 kg/m3, the density of air is 1.3 kg/m3, and that the depth of the atmosphere above the ocean is 10 km.

2) The small piston of a hydraulic lift has a cross sectional area of 3.0 cm2, and its large piston has a cross sectional area of 200 cm2. What downward force of magnitude F1 must be applied to the small piston for the lift to raise a load whose mass is 1000 kg?

3) The mass of an object is 3 kg. When the object is suspended from a spring and submerged in water, the scale reads 2 N. Find the density of the object.

## Homework Equations

1) P=Po+ρgh

2)F=mg, F1= $\frac{(F2)(A1)}{(A2)}$

3)F=mg, ρ= $\frac{m}{v}$

## The Attempt at a Solution

1) So this one I thought I should add the density and distance values together into the formula as following and also using the value of air pressure as Po=1 atm=1.013x105 kg/m*s2. I'm not very confident in this one though:

P=Po+(ρseaair)g(h+y), where h=10 km and y=2 km.
P=1.013x105kg/m*s2+(1001.3 kg/m3)(9.81 m/s2)(12 km)=2.19x105 kg/m*s2

2)
F2=(1000 kg)(9.81 m/s2)=9810 N
F1=$\frac{(9810 N)(0.03 m)}{(2 m)}$=147.2 N

3)
Force exerted by gravity on the object:
F=(3 kg)(9.81 m/s2)= 29.43 N

Difference between submerged and not:
29.43 N-2 N= 27.43 N

Volume displaced by liquid and therefore the volume of the object (I'm not sure if this part is right):
$\frac{(27.43 N)}{(9.81 m/s^{2})}$ = 2.796 kg ∴ V= 2.796 m2

ρ=$\frac{(3 kg)}{(2.796 m^{3})}$ = 1.073 kg/m3

I'm quite dyslexic and new to putting in formulas on here so I apologize for any typos that may be in here. Thank you.

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SteamKing
Staff Emeritus
Homework Helper
Please post individual problems in the future.

1. Wrong.
(x1 + x2)*(y1 + y2) is not equal to x1*y1 + x2*y2. This is basic algebra.

2. Wrong.
If a meter is equal to 100 cm, how many square centimeters are there in 1 sq. meter?

3. Wrong.
Clearly, you don't know/understand what the density of water is.

Sorry about that. Could you possibly help me questions 1 & 3. I'm struggling with the concepts of this chapter and would like to understand.

SteamKing
Staff Emeritus
Homework Helper

Okay, I'll show you what I've done so far. Just give me a minute to type it all up :P...

So number one I'm still quite clueless about the whole "absolute pressure" thing but I think I fixed number 2 and 3. I'm terrified that my answers are still similar. Thank you for pointing out I forgot to convert cm2 to m2. I have a feeling this year is going to be very tough for me because there's a lack of practice problems and other resources for me to use. The internet's going to be my best friend :P...

2) Weight of the mass: ƩFy=mg=(1000 kg)(9.81 m/s2)= 9810 N
$\frac{F1}{A1}$=$\frac{F2}{A2}$
∴ F2=$\frac{F1*A2}{A1}$=$\frac{9810N(3x10^{-4}m^{2})}{0.02m^{2}}$=147.15 N

3) This one I did some research on and found out that I can use a Newtonian approach to these questions. So this is how I did.

OUT OF WATER:
ƩFy1=T1-mg=0 ∴ T1=mg=(3kg)(9.81m/s2)=29.43 N
AND m=ρoV, ρo is the density of the object.
∴ T1oVg

IN WATER:
ƩFy2=T2+B-mg=0
where B=ρVg and ρ is the density of water
∴ T2+ρVg=T1
V=$\frac{T_{1}-T_{2}}{ρg}$
T1o($\frac{T_{1}-T_{2}}{ρg}$)g
∴ ρo=ρ($\frac{T_{1}}{T_{1}-T_{2}}$)
ρo=(1000 kg/m3)($\frac{29.43N}{29.43N-2N}$)=1072.9 kg/m3

I really hope I'm on the right track now. I think I'm having some really bad conceptual issues.

SteamKing
Staff Emeritus
Homework Helper
Probs. 2 & 3 look OK now. Your answer didn't change in 2. because the ratio of your areas was OK, although the individual areas were incorrect.

In Prob. 1, why do you think that the total pressure at the seabed is equal to the sum of the densities multiplied by the sum of the heights? As I showed you with my hint, (x1 + x2)*(y1 + y2) is not equal to x1*y1 + x2*y2. The absolute pressure is atmospheric pressure + hydrostatic pressure. That's your formula.

Now that you put it that way, what I did does not make any sense. So I think I understand it but I'm not quite confident.

So I'll use h=2km and y=10km

P=ρseagh+ρairgy

In the equation I was given for absolute pressure it has a Po which is the pressure for the atmosphere... Now in this case since I'm given the actual depth of the atmosphere, is it necessary for me to use Po or not?

SteamKing
Staff Emeritus
Homework Helper
Given the fact that the density of the atmosphere is not constant, and varies with height, I would use Po.

• 1 person
Good point. So then it would be,
P=Poseagh+ρwatergy
P=1.013x105 N/m2+(1000 kg/m3)(9.81 m/s2)(2000 m)+(1.3 kg/m3)(9.81 m/s2)(10000 m)=1.985x107 N/m2

Thank you so much, SteamKing. I can not express to you how much I appreciate your help. I will post the mark I got and any corrections when I get my assignment back... If that's appropriate to do on here.

SteamKing
Staff Emeritus
Homework Helper
Good point. So then it would be,
P=Poseagh+ρwatergy
P=1.013x105 N/m2+(1000 kg/m3)(9.81 m/s2)(2000 m)+(1.3 kg/m3)(9.81 m/s2)(10000 m)=1.985x107 N/m2

Thank you so much, SteamKing. I can not express to you how much I appreciate your help. I will post the mark I got and any corrections when I get my assignment back... If that's appropriate to do on here.
I'm confused by this work. You seem to think that you have a $\rho$sea and a $\rho$water to make $\rho$seawater.

In the OP, the $\rho$ for seawater was given as 1000 kg/m^3. This is the value to be used. Granted, it really represents the density of fresh water, as sea water is slightly denser than fresh. Similarly, the OP has also specified that the atmosphere is to be assumed to be 10 km deep with a constant density of 1.3 kg/m^3. You should use these values instead of Po. I apologize for the earlier confusion.

Oops sorry. The subscript on water was supposed to be air.