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Archimedes Principle and bouyant force

  1. Jul 8, 2008 #1

    Can someone help with the following 2 questions?

    1. An iron casting containing a number of cavities weighs 6000 N in air and 4000 N in water. What's the total volume of all cavities in casting? The density of iron without cavities is 7.87 g/cm^3. The density of water is 988 kg/m^3.

    Wouldn't the bouyant force = the mass of displaced water * g. So here, bouyant force = 6000 - 4000 = 2000. So wouldn't mass of displaced water = 2000/9.8. And doesn't this need to equal to the mass of the cavities? There seems to be something I'm missing.

    2. The volume of air space in the passenger compartment of an 1800 kg car is 5 m^3. The volume of motor and front wheels is 0.750 m^3. The volume of the rear wheels, gas tank and trunk is .8 m^3. Water can't enter into the last 2 compartments (only the passenger compartment can water enter). As water enters, the car sinks. How many cubic meters of water are in the car as the car sinks?

    To sink, the force must be greater than mg (1800*g). But I don't see the significance of the point of sinking? Since water can't enter into the rear wheels and front wheels, does this mean that there's an upward force equal to (.750+.8)*density of air*g? If anyone has any hints, that would be great.

  2. jcsd
  3. Jul 8, 2008 #2
    You were on the right track, but mass of the displaced water does not equal to the mass of the cavities. But the volume of the casting equals to the volume of the displaced water. Try to continue from there.

    I don't really understand what this question is asking exactly, so I'll leave someone else to help you with it. How many cubic meters of water are in the car as the car sinks? At what point? If the car sinks, then the entire passenger compartment will be filled with water...
  4. Jul 8, 2008 #3


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    Hi bodensee9,

    If you're talking about the buoyant force, don't you mean it must be less than the weight for the car to sink?

    No, the .75 and .8 is not the volume of the air in the tires; it's the total volume of the car that the water can't enter. (metal, rubber, etc.) What would the buoyant forces on them be?

    The point of the passenger compartment is that as water enters, there is less and less water displaced by the passenger compartment, and so the buoyant force arising from the passenger compartment decreases as the water enters.

    Right when the car is actually beginning to sink, what has to be true about the three buoyant forces? From that, you can tell how much air is still in the passenger compartment. What do you get?
  5. Jul 8, 2008 #4

    Thanks for the hints.

    But for 1, if I know the volume of the casting 2000/(9.8*998 kg/m^3), I don't know how to find the volume of the holes in the casting? Should I find what the volume of a 6000 N casting without holes should be (6000/9.8*7.87) and then find the difference between the volume of the casting and the casting without holes?

    For 2, should the three forces be balanced at the point of sinking? So we still have the upward force from .75 and .8, but we have downward force from whatever portion of the compartment with water?

  6. Jul 8, 2008 #5
    That sounds about right. First you have the volume of pure iron (assuming the holes don't weigh anything ;) ) and then you have the actual volume. The difference should be the volume of non-iron stuff.
  7. Jul 8, 2008 #6


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    That's fine to keep track of the weight of the water inside the compartment; but don't forget that there is an upward buoyant force dut to the 5 m^3 passenger compartment.
  8. Jul 9, 2008 #7

    Okay ... thanks. So for the problem, does this mean that before the car sinks, the bouyant force will equal to mg? So if x cubic meter of the passenger's compartment is full of water, does this mean that x*(density of water = 998)*g + 1800*g = bouyant force of the car? So since water can't get into the wheels, does this mean that there is an upward force from .75+.8 volumes of air *density of air * mg? And they must be equal?

  9. Jul 9, 2008 #8


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    This will work out fine if you are consistent with the buoyant force on the passenger compartment.

    This formula for the buoyant force is not right. The buoyant force on a submerged tire would be the same no matter what's inside it. The buoyant force on a tire filled with lead would be the same as the buoyant force on a tire filled with air (if the volumes were the same). So here, the mass of the air in the tires in included in the 1800 kg mass of the car, and the weight of the air is helping to pull the car down.

    So you should not have (density of air) in the water's buoyant force expression. (Also, those volumes are more than just the tires--it's the entire car except for the passenger compartment.)

    Also, the total buoyant force on the car arises from the total volume of water displaced by the car. You have the volume of the front of the car (.75 m^3), and you have the volume of the back of the car (0.8 m^3). What about the passenger compartment?
  10. Jul 9, 2008 #9
    Oh okay .. thanks!
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