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Problems with Fluids

  1. Nov 30, 2005 #1
    Fluid Problems

    1) The gauge pressure in each of the four tires of a car is 240kPa. If each tire has a "footprint" of 220cm^2, estimate the mass of the car. answer: 2.2 x 10^3 kg pressure = 240 kPa, Area = 2.2 x 10^3 kg

    The first thing I assume i need to do is convert kPa but I have not found that in my book... I know that pressure = Force/surface area = mg/A. I could use that formula to solve for mass?

    240kPa = (m x 9.8 m/s^2)/2.2 m^2

    2) A balloon has a radius of 7.35 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon has a mass of 930 kg. (Neglect the buoyant force on the cargo volume itself).920 kg

    r = 7.35 m, m = 930 kg, cargo mass = ? Again I have no idea how to attempt this. This chapter has been extremely hard for me.

    3) What is the likely idenity of a metal if a sample has a mass of 63.5g when measured in air and an apparent mass of 55.4 g when submerged in water? (It also gives us a table and let's us know the densities of several metals but the important one is steel/iron which is 7.8 x 10^2 kg/m^3)

    How do we use the two masses given to find out that it has that density. I have no idea's about this one.

    The answer is iron/steel
     
    Last edited: Nov 30, 2005
  2. jcsd
  3. Nov 30, 2005 #2
    for 1)
    kPa = Kilo Pascal
    Pascal = 1 Newton/meter^2
    so you have your pressure in N/m^2 and your area is in cm^2 so you can easily change area from cm^2 to m^2
     
  4. Nov 30, 2005 #3
    Yes I know but I mean the kPa part. I know that 220 cm^2 = 2.2 m^2.

    Would 240kPa = 240/1000 = .24 Pa = .24 N/m^2
     
    Last edited: Nov 30, 2005
  5. Nov 30, 2005 #4
    So would it be .24 N/m^2 x 2.2 m^2 = m(9.8) ?

    because that does not work...
     
  6. Nov 30, 2005 #5
    ok well... kilo means thousand so you dont divide bya 1000... you multiply by a thousand.... and you also have to remember there are 4 tires supporting a car so you have to account for all 4 tires... but im unsure of whether you converted your area correctly because its squared... so you cant just divide by 100 if i am not mistaken
     
  7. Nov 30, 2005 #6
    Actually there is an example in my book and it has m^2.

    24,0000 Pa x 2.2 m^s = 4m x 9.8 m/s^s

    so if I divide both sides by 9.8 and 4 m I get 13,469... that is not correct.

    * 1 Pa = 1 N/m^2
     
  8. Nov 30, 2005 #7

    Fermat

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    1 Pa = 1 N/m²
    1 kPa = 1 kN/m² = 1000 N/m²
    ======================

    1 cm = 1/100 of 1 m
    1 cm² = 1/100² of 1 m²
    220 cm² = 220/100² of 1 m²
    220 cm² = 0.022 m²
    ===============

    You know the pressure, and you can work out the total area from the 4 footprints, so using F = PA, you can get the total force provided by the pressure in the tyres, hence the weight of the car.
     
  9. Nov 30, 2005 #8

    mezarashi

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    As Fermat's noted, understand the units conversion first. Then apply

    Fgrav = 4PA (since there are 4 tires)

    For your other problems, apply the concept of the buoyant force. The buoyant force acting on any object submerged in a fluid (including air) is the weight of the displaced fluid.

    [tex]F_B = \rho V g[/tex]
    where rho indicates the fluid's density, and V is the volume of the object.

    You can use the buoyant force in a force diagram like you would any other. For example in your third question, you have the condition:

    [tex]F_{net} = 0 = F_{grav} - T - F_B[/tex]
     
  10. Nov 30, 2005 #9

    Integral

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    1.

    You know that
    [tex] P= \frac {m g} A [/tex]
    The first step is to solve for the quanity you need, that is mass. so:

    [tex] m = \frac {PA} g [/tex]

    you have P= 240kPa

    and A = 220 cm2 for EACH TIRE. How many tires are on the average car?

    How many cm2 in 1 m2?
     
  11. Dec 1, 2005 #10
    So taking acrea of number 1 - m=PA/g = (4 x 240,000)(.022m)/(9.8)

    = 2155 = 2.2 x 10^3 N!

    Wow. I cannot believe it was that simple.

    #2) I use [tex]F_B = \rho V g[/tex] some how.

    I can figure out the volume by the equation (4/3)(pie)(r)^3
    = 1662

    But where do I go from there?
     
  12. Dec 1, 2005 #11

    Doc Al

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    You calculate the buoyant force, using that formula. (You'll need the density of air.) That buoyant force must support the weight of the balloon structure, the helium inside the balloon (what's the density of helium?), plus any added cargo. Set up the equation to solve for the maximum cargo that the balloon can lift.
     
  13. Dec 1, 2005 #12
    [tex]F_B = \rho V g[/tex]

    airs denisty = 1.29, helium's = 0.179

    so I add those denisties together and then solve for Fb?
     
  14. Dec 1, 2005 #13

    Doc Al

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    Why would you add the densities? The buoyant force is the weight of the displaced fluid, which, in this case, is air.
     
  15. Dec 1, 2005 #14
    Okay so...

    (1.29)(1662)(9.8) = Fb
    = 21011 N for a boyancy force.

    But how does thi shelp me with the mass of the cargo?
     
  16. Dec 1, 2005 #15

    Doc Al

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    The buoyant force is the upward force on the balloon. When maximally loaded, that upward force will be balanced by the downward forces. One of those downward forces is the weight of the cargo. (See post #11.)
     
  17. Dec 1, 2005 #16
    So I do something w/ the denisty of helium? Correct?

    Find it's force? but how? And then once I have heliums I can minus the two forces to get the cargo mass?
     
  18. Dec 1, 2005 #17

    Doc Al

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    There are three "parts" to the balloon/cargo system: the helium, the balloon structure, and the cargo. It is their weight that pulls the system down. You'll need the density of helium to find the weight of the helium.
     
  19. Dec 2, 2005 #18
    As I previously stated the denisty of helium is 0.179

    weight of balloon = (930 kg)(9.8) = 9114 N

    F(boyancy) = 21011 N

    How do I change helium, by using denisty, into a mass?
     
  20. Dec 2, 2005 #19

    Doc Al

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    What's the definition of density?
     
  21. Dec 2, 2005 #20
    Mass/volume.
     
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