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Buoyant Force and Volume

by bud102
Tags: buoyant, force, volume
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bud102
#1
Oct7-05, 12:57 PM
P: 3
I've already determined the first part of the question, which is regarding the buoyant force. An object weights 5N and displaces 3.5 N of water when placed in the beaker. The buoyant force is = to the displaced water which is 3.5N. Now, here's where I'm stuck. The volume doubles, and weight stays the same. What is the buoyant force? My guess is that since the volume is doubling, the amount displaced will double so it will be 7N and the object will float because the amount displaced is greater than the weight. Is that accurate?
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Fermat
#2
Oct7-05, 01:56 PM
HW Helper
P: 876
Not accurate, I'm afraid.

The volume of the object may double but it will only displace double the original volume of water if it it completely immersed.

If you think about it ...

The downward force, the weight of the object, is 5N and if the upwards force, the buoyant force, is 7N, then the object is going to accelerate upwards!

When the volume doubles, try to figure out if the object floats or will it be completely immersed.
mezarashi
#3
Oct7-05, 02:10 PM
HW Helper
P: 660
Quote Quote by bud102
The buoyant force is = to the displaced water which is 3.5N.
I agree completely.

Quote Quote by bud102
My guess is that since the volume is doubling, the amount displaced will double so it will be 7N and the object will float because the amount displaced is greater than the weight.
I also agree. If suddenly the volume doubles, it will now displace double the volume of water, which will create 7N of bouyant force. The body will indeed accelerate upwards until it surfaces. Now part of the object will be above water and partly underwater. Take a guess, yes, the part underwater will displace just enough water to support its weight of 5N. Thus the object floats. Check your notes for conditions for floating ;)

bud102
#4
Oct7-05, 02:17 PM
P: 3
Buoyant Force and Volume

Then the object would float then because the deeper the object is immersed, the greated the buoyant focce and once the weight of the disperased water is equal to the weight of the object, it will float. Since the volume is doubled, the bouyant force will never really double because once it equals the weight it will float. Density plays a role because the more dense an item is, the less it has to be immersed in order to displace the amount of water to float?
mezarashi
#5
Oct7-05, 02:52 PM
HW Helper
P: 660
Quote Quote by bud102
Then the object would float then because the deeper the object is immersed, the greated the buoyant focce and once the weight of the disperased water is equal to the weight of the object, it will float. Since the volume is doubled, the bouyant force will never really double because once it equals the weight it will float.
Bingo! In this case, the bouyant force will max at 5N.

Quote Quote by bud102
Density plays a role because the more dense an item is, the less it has to be immersed in order to displace the amount of water to float?
You can look at it from a couple of views. Here's one: mass = density x volume. The more dense an object is, the heavier it will be for any given volume. Volume is proportional to how much water it can displace which is proportional to the max possible bouyant force on it.
bud102
#6
Oct7-05, 03:30 PM
P: 3
Thank You All For Your Help! :)


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