Fluids


by Daniel E. Myers and Brett Hamliton, June 1998

Table of Contents

Here's the quantities you can know:
These quantities are defined and explained on other pages except for

 Hydrostatic pressure and Buoyant Force which is explained below

New Quantities

And here are the formulas that we have so far: 
    

  1. D = m/V  Where m = mass and V = volume


  2. P = F/A   Where F = force and A = surface area

Definig Density

The density of an object is defined as its mass per unit volume where m is the mass of the object and V is its volume. Density is a characteristic property of any pure substance. The SI unit for density is kg/m^3.

Defining Pressure

Pressure is defined as force per unit area where the force F is understood to be acting perpendicular to the surface area A. The SI unit of pressure is N/m^2 called pascal (pa).

Defining Hydrostatic Pressure

Defining Hydrostatic Pressure is easy, since we already know the principles normal pressure. Hydrostatic pressure is easily explained by stating the experimental fact that a fluid exerts pressure in all directions. For example, when swimming, we feel pressure surrounding our entire bodies, this pressure is hydrostatic pressure. The amount of pressure depends on the density of the liquid D, the depth of the liquid h, and the acceleration of gravity g
So we have a new formula:

Ph = F/A = Dgh

Which works only with liquids.

Defining Buoyant Force


Buoyant Force occurs because the pressure in a fluid increases with depth. Thus the upward pressure on the bottom surface of a submerged object is greater than the downward pressure on tits top surface. In example, a floating piece of wood has the force of gravity acting downward, but the buoyant force is exerted upward by the liquid. So in essence, the buoyant force is the net force due to the fluid pressure. The buoyant force depends on the fluid density Df, the force of gravity g and the volume of the object V, where:
So we have another new formula
B = DfgV

Formulas

The formulas we already have:

Ph = Dgh

B = PhgV

D = m/V

P = F/A

And . . .

The density of water is 1.00 x 10^3 kg/m^3 just for future reference this will come in handy later so remember it, since water will be the main fluid we are dealing with.
So now we have all the formulas we need for solving fluid problems:
Go back to: Table of Contents

General Problem Solving Strategy:
  1. Read the problem.
  2. Go through the problem and figure out what is given or implied
    Make a list, and identify the quantities you know.
  3. Find any formula that will allow you to calculate
    anything that you don't know, and apply it.
  4. Add what you just found in the last step to your list of knowns.
  5. Check to see if you have found the answer. If not, repeat the
    previous two steps until you are done.

Go back to: Table of Contents

Example problem 1

What is the hydrostatic pressure within a bottle 10 cm in height filled with water. Here is what you start with:
So, if you look at the formulas, the obvious one that fits is P = Dgh




We can imput 1.0 x 10^3 for D, 9.8 for g and .1 for h.

Multiply these together to get 980 N/m^2 or 980 pascales.



Go back to: Table of Contents

Example problem 2

An brick with a Buoyant force of 2500 N rests in a swampy lake where the density is 1.2 x 10^3 kg/m^3. What its Volume?
Here is what we start with:

B = 2500 D = 1.2E3 g = 9.8 V = ?
To find the Volume, simply substitute the numerical values of B, D, and G and then solve for V. So . . .
2500 = (1.2E3)(9.8)(V)
V = 2500/[(1.2E3)(9.8)]
V = 0.2126
Go back to: Table of Contents


Sample Problems

The answers to each problem follow it in parentheses. They also link to a solution to the problem. Try the problem, check your answer, and go to the solution if you do not understand.

1.

 A piece of wood has a volume of .024 m3 and a mass of 12.4 kg. What is its density? (517 Kg/m3)

2.

 Steel has a density of 7800 Kg/m3. What volume of it has a mass of 40 kg? (.0051 m3)

3.

 Mercury has a density of 14,000 kg/m3. If you fill a cup that is 3.2 cm in radius and 12.5 com in height, (it's a perfect cylinder) with mercury, what mass of mercury do you have to put in it? (5.6 kg)

4.

 What is the mass of the air in a room full of air with a density of 1.3 kg/m3 that measures 20 m x v.0 m x 10 m? (780 kg)

5.

 What is the force of atmospheric pressure (1 atm = 14.7 lbs/in2) on a pane of glass that measures 24" x 42"? (1.48E4 lbs) Go back to: Table of Contents

6.

 A person exerts 520 N on an area that measures 2.5 cm x 2.5 cm. What is the pressure in Pa. (8.32E5 pa)

7.

 A hydraulic lift must lift 6000 lbs using oil that is under a pressure of 120 lbs/in2. What must be the surface area of the lift cylinder? (50 in2)

8.

 What is the pressure at a depth of 10.0 m below the ocean where D = 1030 kg/m3 (1.01E5 Pa)

9.

 What is the pressure at a depth of 74 cm beneath mercury there D = 13,900 kg/m3 (1.01E5 pa)

10.

 To what depth must you go in meters to be at 79 atmospheres,(1 atm = 1.013E5 Pa) where D = 1030? (793 m, about 10m/atmospheres)

11.

 A submarine is at a depth of 390 m in the ocean where D = 1030 kg/m3. What is the force on a circular hatch 50 cm in radius? (Find the pressure first) (690,000 lbs or 347 tons)

12.

 A window in the Nautilus measures 24 cm x 50 cm and can withstand a force of 3.6E5 N. How deep can the Nautilus go without breaking that window? (297 m)

13.

 What is the Buoyant force on a log with a volume of 2.4 m3 when it is submerged in water? D = 1000 kg/m3 (2.35E4 N)

14.

 What is the Buoyant force on a box that measures 42 cm x 54 cm x 120 cm, when it is submerged in air? D = 1.3 kg/m3 (3.5 N)

15.

 A plane weighs 20,000 n. What volume of water must its pontoons displace in order for it to float? D = 1000 kg/m3 (2.0 m3)

16.

 A block of wood measures 2.0 cm x 3.0 cm x 60 cm and weighs .30 N in air. A) What is the density of the wood? (find the mass first) b) What is the density of a liquid if the block weighs .082 n when it is immersed in it? (figure out what the Buoyant force is first) (850 kg/m3, 618 kg/m3)

17.

 A block of metal has a density of 2,700 kg/m3 and measures 12.5 cm x 15 cm x 2.5 cm A)What is the block's mass? B) what is the block's weight? C) what is the block's weight when it is submerged in water? D = 1000 kg/m3 (1.27, 12.4, 7.8)
Go back to: Table of Contents



Solutions to Sample Problems

1.

 A piece of wood has a volume of .024 m3 and a mass of 12.4 kg. What is its density? (517 kg/m3) Here is what you start with: Use the equation D = m/V Input 12.4 for V and .024 for V and divide to get the answer of 517 kg/m3 for the volume Go to: Problem Formulas Table of Contents

2.

 Steel has a density of 7800 Kg/m3. What volume of it has a mass of 40 kg? (.0051 m3) Here is what you start with: Use the equation D = m/V and input the values for Density and volume. Then solve algebraically to give you (40)/(7800) = V = .075 Go to: Problem Formulas Table of Contents
> Mercury has a density of 14,000 kg/m3. If you fill a cup that is 3.2 cm in radius and 12.5 cm in height, (it's a perfect cylinder) with mercury, what mass of mercury do you have to put in it?

3.

 (5.6 kg) Here is what you start with: First find the volume converting cm to m with the formula (pi)r2h so volume = 4.02E-4 m3 Go to: Problem Formulas Table of Contents

4.

 What is the mass of the air in a room full of air with a density of 1.3 kg/m3 that measures 20 m x 3.0 m x 10 m? (780 kg) Here is what you start with: First find the volume by multiplying the room dimensions of 20 x 3 x 10 to get 600 m3 Plug the numbers into D = m/V and solve for the mass to get (1.3)(600) = m = 780 kg Go to: Problem Formulas Table of Contents

5.

 What is the force of atmospheric pressure (1 atm = 14.7 lbs/in2) on a pane of glass that measures 24" x 42"? (1.48E4) Here is what you start with: First find the area by multiplying 24 x 42 to get an area of 1008 Use the pressure formula P = F/A solve algebraically for F to get (14.7)(1008) = F = 1.48E4 Go to: Problem Formulas Table of Contents

6.

 A person exerts 520 N on an area that measures 2.5 cm x 2.5 cm. What is the pressure in Pa. (8.32E5 Pa) Here is what you start with: Start by converting to meters, then multiplying .025 x .025 to find the area and get .000625 m2 Use the equation P = F/A to get 520/.000625 = P = 8.32E5 Pa Go to: Problem Formulas Table of Contents

7.

 A hydraulic lift must lift 6000 lbs using oil that is under a pressure of 120 lbs/in2. What must be the surface area of the lift cylinder? (50 in2) Here is what you start with: Use the equation P = F / A and input what you know. Algebraically solve for A to get A = 6000 / 120 which equals 50 in2 Go to: Problem Formulas Table of Contents

8.

 What is the pressure at a depth of 10.0 m below the ocean where D = 1030 kg/m3 (1.01E5) Here is what you start with: Use the equation P = Dgh and input what you know to get (1030)(10)(9.8) = 1.01E5 Pa Go to: Problem Formulas Table of Contents

9.

 What is the pressure at a depth of 74 cm beneath mercury where D = 13,900 kg/m3? Here is what you start with: Once again, use P = Dgh and plug in the values we know to get (13900)(.74)(9.8) which equals 1.01E5 Pa Go to: Problem Formulas Table of Contents

10.

 To what depth must you go in meters to be at 79 atmospheres,(1 atm = 1.013E5 Pa) where D = 1030? (793 m) First find the total pressure by multiplying 79 X 1.013E5 to get 8E6. Here is what you start with: Use P = pgh and plug in what you know to get 8E6 = 1030(9.8)h Then algebraically solve for h to get 793 m Go to: Problem Formulas Table of Contents

11.

 A submarine is at a depth of 390 m in the ocean where D = 1030 kg/m3. What is the force on a circular hatch 50 cm in radius? (Find the pressure first) (3.1E6 N) First find the area of the hatch with (pi)(.5)^2 = .785 cm2 Here is what you start with: First, find pressure using P = Dgh and plug in numbers to get (1030)(9.8)(390) = 3.9E6 Now, remember that P = F/A where we can input what we know to get 3.9E6 = F / .785 which can be solved algebraically to get 3.1E6 N Go to: Problem Formulas Table of Contents

12.

 A window in the Nautilus measures 24 cm x 50 cm and can withstand a force of 3.6E5 N. How deep can the Nautilus go without breaking that window? (297 m) First, convert to meters and find area (.24)(.50) = .12 m2 Here is what you start with: Use P = F / A and plug in numbers to get 3.6E5 / .12 = 3E6 Pa Then use P = Dgh = (3E6)(9.8)(1030) = 297 m Go to: Problem Formulas Table of Contents

13.

 What is the Buoyant force on a log with a volume of 2.4 m3 when it is submerged in water? D = 1000 kg/m3 (2.35E4) Here is what you start with: Use the equation F = DgV and plug in what we know which will make it (2.4)(9.8)(1000) = 2.35E4 Go to: Problem Formulas Table of Contents

14.

 What is the Buoyant force on a box that measures 42 cm x 54 cm x 120 cm, when it is submerged in air? D = 1.3 kg/m3 (3.5 N) First convert the lengths to meters and find volume to get (.42)(.54)(1.2) = .27 Here is what you start with: Use F = DgV and plug in what we know to get F = (1.3)(9.8)(.27) = 3.5 N Go to: Problem Formulas Table of Contents

15.

 A plane weighs 20,000 n. What volume of water must its pontoons displace in order for it to float? D = 1000 kg/m3 (2.0 m3) Here is what you start with: Again, use F = DgV and plug in what is known to get 20000 = (1000)(9.8)V Solve algebraically for V to get 2.0 m3 Go to: Problem Formulas Table of Contents

16.

 A block of wood measures 2.0 cm x 3.0 cm x 6.0 cm and weighs .30 N in air. A) What is the density of the wood? (find the mass first) b) What is the density of a liquid if the block weighs .082 N when it is immersed in it? (figure out what the Buoyant force is first) (850 kgm3, 618 kgm3) Part A First, convert to meters and find the volume to get (.02)(.03)(.06) = 3.6E-5 Also, convert N to kg as follows .3 / 9.8 = .0306 Here is what you start with: Use density formula D = m / V to get .0306 / 3.6E-5 = 850 kg/m3 Part B To find the Buoyant force subtract the force while submerged in the air from the force while submerged in the liquid, so that .3 N - .082 N where B = .218 So then we start with: Then we solve for the density using B = Dgh where D = .218/(9.8 x 3.6E-5)= 618 Go to: Problem Formulas Table of Contents

17.

 A block of metal has a density of 2,700 kg/m3 and measures 12.5 cm x 15 cm x 2.5 cm A)What is the block's mass? B) what is the block's weight? C) what is the block's weight when it is submerged in water? D = 1000 kg/m3 (1.27 kg, 12.4 N, 7.8 N) Part A First convert the dimensions to volume using the usual method to get V = 4.68E-4 Here is what you start with: Solve algebraically using D = m/V to get m = DV where m = (4.68E-4)(2700) = 1.27 kg Part B Simply use the mass from part A, which is 1.27 and multiply by g which is 9.8 to get 12.4 N Part C Here is what you start with Subtracting Fs from F is equal to DgV. So we can use F - Fs = DgV and solve algebraically for Fs which is equal to - DgV + F. So -(1000)(9.8)(4.68E-4) + 12.4 = Fs = 7.8 N Go to: Problem Formulas Table of Contents