Gael (unknown to me) took a photo of Professor Tim Reston explaining 2nd year degree level seismic surveys (in other words seismic surveys for beginners) to me. I thought my own students would find it amusing to see me concentrating to try to understand a concept – I need to brush up on my maths and physics!!
Juan (Ships Systems) has kindly written the rest of tonight’s blog. It is an excellent explanation; can you answer the questions at the end? 🙂
Why does a ship float?
Can you think of a good explanation of why things float? Sure we know heavy things sink in water and light things float, but is that the whole story? After all, the James Cook has a mass of 5,401 tonnes (=5.4 million kilograms), and yet that floats…
A few more seconds to think about it?
OK you must have guessed (I’m sure you did) that ol’ Newton might have something to do with it and that we’ll have to think about forces. If the ship is not accelerating downwards (we call that sinking), then there must be a force equal to the ship’s weight opposing it’s downward motion, or: the sum of all vertical forces must equal zero. The vertical components of forces acting on the ship are in equilibrium. What are these forces?
So we have some forces to calculate! If the James Cook has a mass of 5,401 tonnes, how much does it weigh? I’ll give you a moment to think about it.
Got it yet?
mass x (acceleration due to gravity) = weight
5,401,000kg x 9.81m/s2 = 52,983,810N
Or about 53MN (meganewtons)
Yes, it’s that old chestnut: the force = mass x acceleration equation. An old favourite of mine. So we’ve managed to calculate the force (weight) that the ship exerts onto the sea. And, to repeat, if the ship is not sinking, then the force that the sea exerts on the ship must exactly counteract the weight of the ship. That’s right. The sea is exerting 53MN on the ship and we call that opposing force buoyancy.
Does the sea exert 53MN on you?
“Hang on!” I hear you shout. “Forces don’t just come from nowhere ! Why is the sea exerting 53MN on the ship? Where does buoyancy come from?!” Hold your horses. I was just getting to that.
If you jumped into the sea, would it exert 53MN on you?
If you said yes to that question, leave the room in shame.
I can imagine no circumstances in which you could ever exert 53MN on the sea. (well… unless you were travelling very, very fast and hit the water, coming to a full stop in a very short amount of time. I’m interested – can you make some assumptions and work out how fast you’d have to hit the water for 53MN to be exerted?).
So if the sea doesn’t exert 53MN on you, what’s different between you and the James Cook? What if you were the same size as usual but had the mass of the James Cook? So you’d be really, really dense. (Is there a joke in there somewhere?) Your kg/m3 would be very high. Would the sea exert 53MN on you? Or would you simply sink beneath the waves like a stone?
- We’ll forget the little inconvenience of the strong and weak nuclear forces…
- The average volume of a human body is 0.0664m3.
Sinking like a stone
You’d sink beneath the waves like a stone. Because? The sum of vertical forces must no longer be equal to zero! The buoyancy force exerted on you by the sea was not enough to keep you afloat. You stop accelerating downwards when you reach your terminal velocity in water, but only stop moving when the sum of the buoyancy force and the reaction provided by the seabed counteract your very great weight and vertical forces reach equilibrium. The sea did not come to your rescue with 53MN. In fact, the sea probably exerted a measly 670 N of buoyancy (why??)
The magic property
So the difference? Volume. Yes well done. The only other property we haven’t discussed yet that might be different between you and the James Cook is volume – the space inside a shape, measured in metres cubed (m3). Why is this important to buoyancy? Unfortunately, here, I must digress (again) to a story about a man in a bathtub…
A man in a bathtub
There was once a man called Archimedes who lived in the city of Syracuse. One evening, he got into the bath after a hard day’s work, and suddenly noticed that the water level went up as he climbed in. (Have you ever noticed this? It’s simply amazing!! Try it next time you take a bath!).
Archimedes realised that this could only mean one thing – that the volume of the displacement of water in the bath tub was equal to the volume of his body. He was so excited by this discovery he shouted “Eureka!” and, jumping out of the bath, he ran through the streets of Syracuse (naked – I should have warned you that this story was Intended For Audiences of 18 Years or Above) proclaiming his discovery.
So why does the sea exert 53MN on the James Cook?
Do you remember earlier, we calculated that the James Cook exerts 53MN on the sea? And, as the ship is not sinking then the sea must be exerting 53MN back on the James Cook?
Now we know from Archimedes that an object immersed in fluid (a fancy way for saying ‘a thing in liquid or gas’) displaces a volume of fluid equal to the volume of the object. Shall we check how much force the sea must be exerting?
The James Cook has a volume of 5430.5m3. The same amount of water must be displaced (pushed out of the way) by the James Cook sitting on it. Here’s everything you need to calculate the force exerted by the sea:
Volume of seawater displaced = 5430.5m3
Density of seawater = 1025kg/m3 (fresh water is 1000kg/m3)
Acceleration due to gravity = 9.81m/s2
The force exerted by the sea
Hopefully, you managed this calculation:
(Mass of seawater)kg = (density of seawater)kg/m3 x (volume displaced)m3
Then, now you’ve got the mass of seawater displaced, you could work out the force using Netwon’s Second Law:
(Force)N = (mass of seawater)kg x (acceleration due to gravity)m/s2
And the answer you got should be within 3% of 53MN, or 53,000,000N, which is around the weight we calculated for the ship!
The conclusion: Archimedes’ Principle
So in the end, we have discovered that the weight of displaced fluid is equal to the buoyancy force exerted on an immersed object. This is called Archimedes’ principle – named after the man in the bathtub himself.
If that immersed object floats, then the buoyancy force exerted on that object must equal that object’s weight.
If that immersed object sinks, then the buoyancy force exerted on that object must be less than that object’s weight.
(Buoyancy force)N = (Weight of displaced fluid)N
(Weight of displaced fluid)N = (Density of fluid)kg/m3 x (Volume displaced)m3 x (Acceleration due to gravity)m/s2
Sinking happens when:
(Buoyancy force)N < (Mass of object)kg x (Acceleration due to gravity)m/s2
Testing your understanding
Now that’s over with (phew), you are almost an expert in buoyancy. Have a go at these crackers to see how much you’ve understood…
- Show that when you jump into the sea, you experience a buoyancy force of about 670N. What assumptions did you make?
- Will the James Cook sink slightly or lift slightly when going from seawater to freshwater?
- If a bathtub is an exact cuboid 1.5m long, 0.6m wide and containing water 0.4m deep, show that the water level would go up by about 7cm when you climb in.
A little physics homework for you all 🙂
Thank you Juan :). Watch out for Juan’s career profile in the next day or so.
The egret is still alive, it’s eaten some fish but is not drinking anything.