Register to reply 
Pressure over an areaby abpandanguyen
Tags: pressure 
Share this thread: 
#1
Jan510, 03:53 PM

P: 33

1. The problem statement, all variables and given/known data
A swimming pool has dimensions 28.0 m 12.0 m and a flat bottom. When the pool is filled to a depth of 2.10 m with fresh water, what is the force caused by the water on the bottom? On each end? (The ends are 12.0 m.) On each side? (The sides are 28.0 m.) 2. Relevant equations d = density P = dgV P = F/A 3. The attempt at a solution I got my first answer and it was solved by using the above equation (28.0)(12.0)(2.10)m^{3} x (1000) kg/m^{3} x 9.80 m/s^{2} = 6914880 N for each end I am using (12.0)(2.10)m^{2} x (9.80)m/s^{2} x (1000) kg/m^{3} = which comes around to 246960 It wants the answers in kN and it says my answer is within 10% of the correct answer. I did the same for the third part of the problem and got the same response. I even looked at the solutions manual and they seemed to follow the same procedure so I am not sure what I am doing wrong. 


#2
Jan510, 04:13 PM

Sci Advisor
HW Helper
P: 8,955

Thats the correc tmethod for the bottom.
But think about the pressure on the sides and the ends. Whats the water pressure at the bottom of the sides? What about the top (at the surface of the water) ? 


#3
Jan510, 04:20 PM

P: 33

Sorry... I am not really following you. I think I'm having a hard time picturing the setting in question. I am thinking of this question as it was asking for the pressure the water had on the side as in a rectangle. Am I thinking of this right?



#4
Jan510, 04:36 PM

Sci Advisor
HW Helper
P: 8,955

Pressure over an area
Yes, but on the bottom you have the full depth (and mass) of water pressing down.
On the sides you have a varying height of water, how much water is pressing down (and out) at the bottom compared to the surface? 


#5
Jan510, 04:44 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,672

The units of your answer for the sides are wrong. It doesn't come out to N/m^2.
What mgb is getting at is that the pressure varies with depth, so you generally can't just take the pressure at an arbitrary depth and multiply it by the area to get the total force. 


#6
Jan510, 05:03 PM

Sci Advisor
HW Helper
P: 8,955

The question is force not pressure
so the answer should be width (m)* depth(m) * height (m) * g (m/s^2) * density (kg/m^3) = kg m/s^2 = N (remember, F = ma = kg m/s^2 = N ) 


#7
Jan510, 05:12 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,672




#8
Jan510, 05:30 PM

P: 33

Okay so now I am doing this
1000 x 9.80 x 2.10 to get Pascals (N/m^{2} = kg/(m)(s)^{2}) I am then multiplying this number by the respective areas of the sides and ends to get Newtons but if my original answers were within 10% of the correct answer, then these are definitely way out of the ballpark. 


#9
Jan510, 08:13 PM

P: 33

I am still slightly confused ;_; I think of my idea of how pressure is distributed is skewed.



#10
Jan510, 09:14 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,672

You calculated the pressure at the bottom of the pool, where the pressure is the highest, and the pressure at the top is 0. So your calculation is going to give you an overestimate of the force on the wall.
Think of a horizontal sliver of area dA at a depth h. Calculate the force dF on it, and then integrate from the top to the bottom of the pool. 


#11
Jan510, 10:18 PM

Sci Advisor
HW Helper
P: 8,955

Another way to think of it (if you haven't done calculus)
At the top of the side (just at the water surface) the pressure is zero At the bottom the pressure is 2.1m as you worked out Now, what would the average pressure be. 


#12
Jan510, 10:33 PM

P: 33

Okay, so I'm basically adding the pressure at the top and the bottom, then dividing it by two (averaging it)? That makes sense if its right >.>.



#13
Jan510, 10:36 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,672

Yes. It works because the pressure varies linearly with depth.



#14
Jan510, 10:47 PM

P: 33

Okay, that makes sense now. Thank you very much!



Register to reply 
Related Discussions  
Pressure/volume/area problem!  Introductory Physics Homework  1  
Pressure based on Force/Area  Introductory Physics Homework  4  
Pressure change and area ratio help  Engineering Systems & Design  4  
Pressure = Force/Area  Introductory Physics Homework  9  
Water pressure vs. area of an object  Introductory Physics Homework  1 