HW 12a help and hints:
#1b: Inside the boat with the person are several blocks of wood. Suppose the person now throws the
blocks of wood into the pool. (a) Does the boat float higher, lower, or at the same level relative to the
water? (b) Does the water level in the pool increase, decrease, or stay the same?
For question 1, please think about: in order to support the weight, how much water volume needs to be
displaced. In part b, when the wood is out of the boat, does it float or sink? does it displace water? does it
displace more water or less water than the amount it displaced when it was in the boat?
This is a very tricky question.
When it is in the boat, the wood's weight needs to be supported by buoyancy force, hence, it contribute to
displace some water volume. Let's say the wood's weight contribute to displace more volume 1 (V1) of
water.
When it is out the boat, it floats and it still displace some amount of water Let's say V2.
How is V1 related to V2?
When the wood is inside or out side of a boat, which way it will displace more water?
Or V1 is = V2?
If it is not wood, but metal, which sinks after taking out of the boat.
How much water does it displace when it is in the boat? and How much water does it displace when it is
out of the boat and sink to the bottom?
If it is a piece of metal, you get different answers.
Read the new lecture notes posted online can be helpful. In the lab, you will be asked with a similar
question but you would throw a piece of metal out of the "boat" in your lab. The answer is different there.
Think about why.
#2 An airtight box has a removable (massless) lid of area 3•10-2 m2. A lid is placed on it while it is on top
of a mountain (where Patm=0.83•105 Pa). It is then taken to sea level, where Patm = 1.013•105 Pa. How
much force will be required to remove the lid?
(How would you do this if the lid had mass, and this value was given?)
Recall what we learned in the 5th chapter:
1, Always identify your object of interest. (The lid)
2, Draw diagram to show all the forces acting on it.
3, Solve force equqations. Ftotal =ma , Ftotal=0 or
Fupward – Fdownward=0 , etc...
In the past, in step 2, we never worry about forces come from air pressure. Because air pressure is
everywhere, and is equal in all directions. When you put a box or any object on a desk, there is always air
with the same air pressure between the box and the desk, unless special thing was done to get rid of the air,
like when you squeeze an air-tight hanger. Normally, when you put something on the desk, air pressure
presses object from the top and also pushes the object from the bottom. So the total effect from the air
pressure is canceled. (There is always air not vacuum between the object and the desk).
Now, we need to worry about air pressure, especially in this problem, since the air inside the airtight box
was sealed in advance, after you move the box to other places the air pressure inside did not change, but
the air pressure outside can be different. If the outside has a higher pressure than inside, the lid is pressed
down tightly and an extra force from you is needed to lift the lid.
Hence when we draw diagram of all forces on the lid, we have to include the pressure on top of the
lid(outside P * Area) and the push on the bottom of the lid (inside P * Area) . Notice that they are in
opposite directions and they are not the same size.
Now, in order to lift the lid, how much more forces you need to apply upward? The total force pushing
upward needs to be more than the total force pressing downward. Find the force difference from inside
and outside pressure.(also notice that we didn't include the mg of the lid, because as you will calculate
forces due to pressure difference are so much larger than common mg of a lid. Mg is too small comparing
to P*A and we ignored it. Another reason we didn't worry about air pressure before is that air pressure
normally do not change within local area, inside a room, or within 10 or 20 meters of height. Only when
you go to high mountains, air pressure drops noticeably.
#3 In this problem, please make sure to use the density of seawater = 1025 kg/m3.
The bottom of the Mariana Trench, located in the Pacific Ocean, is about 11,000 m below the ocean's
surface, making it the deepest ocean depth on Earth. If an underwater vehicle were exploring the region,
at a depth of 10500 m, a.) What force would the water exert on the vehicle's circular observation window,
which has a radius of 0.1 m? b.) The mass of a jetliner is about 1.2·105 kg. Please fill in the following
relationship: The force on the window is as much as the weight of
these jetliners.
Student A: Find the force exerted on the window.
Pressure from the water = e(density)*g*h(depth)+Patm
F=pressure/area(of the window); therefor F=p*a
The density of water as stated in the problem is 1025kg/m^3 so I did 1025*g*depth=P then P*a to find the
force on the window but this came up wrong. Am I using the density right? My answer was very large.
Man, Weining:
Use SI unit to avoid mistakes.
Yes, 1 Newton is a very small amount of force. 1 kg (about 2 pounds) object weights 9.8 Newtons.
100gram. Like a a small apple weights one newton. And a 70kg person, weights about 700 newtons.
Under the deepest place in the ocean, you are supporse to get huge pressue and large force in the unit of
newtons. The force=pressure*area down there needs support all the weight of the 10000 meter deep sea
water on top of it.
Student B: im having trouble getting the first part. how did you go about it?
Man, Weining: Don't think too complicated things. This is not a tricky problem.
In part a, they ask you to find the force on the small window. You know the area of the window, and the
force comes from the liquid pressure down to the deep sea. How large is the pressure down there? (the
deeper you go, the higher the pressure is. How is P related to depth h?
You should find that pressure out and then the total force on that window
Part b is pretty straightforward. You got the total fore from part a,
They just want you to get a sense of how large that force is.
So they ask you to compare the total force you got in partA with weight of jet (mg), all you need to do is
to find the force in part a is how many times of the heavy jet's mg?
Student B: my only problem here is that im not sure which height is the depth that i need to use. i'm given
2 heights and i know p=Patm + egh to find the pressure and then pressure=f/a to solve for force. I
remember you talked about in lecture about distinguishing which is the correct 'h" to use for depth.
Man, Weining: Apr 29, 10:44 PM
While the underwater vehicle were at a depth of 10750 m,. Hence, to calculate the pressure on the vehicle,
you would need the depth of this location.
In the class I mentioned, the depth is the distance from the water surface to the actual location.
That's the difference from depth and height.
Height increases when you go upward.
Depth increases when you go deeper (downward)
Also you will find under such deep water, egh is way larger than Patm, Patm almost make no contribution
to the calculation.
Man Weining:
Area of the window is not the radius or diameter of the window.
also pay attention to dimensions, Pressure = force / area
Force = pressure * area.
Force is not = pressure * length.
0.3m is obviously a length only.
What's the area of a circular window with radius of 0.3m?
Also units:
egh=P 1025kg/m^3 * 9.8m/s^2 * 10850m = 108988250kg/ms^2 = 108988250 Pa
108988250kg/ms^2 * 0.3m = 32696475 kg/s^2
kg/s^2 is not Newton,
One newton is going to give 1 kg object 1m/s^2 acceleration.
Hence, 1 N= 1 kgm/s^2
1N =1 kg/ms^2 * 1m^2 = 1Pa*1m^2
#4 A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 25.0 N;
when it is completely immersed in water the scale reads 21.8 N. (a) What is the volume of the block?m3
(b) What is the density of the block?
Someone asked:
I don't understand what I'm doing wrong with part B of number 4. It says that I am within 10% correct of the
answer. I calculate the buoyancy force and then to find the volume of the object in part A. I then took the Mg
which is 20 N and divided it by gravity to get mass. Then I took the mass and divided it by volume of the object.
Can you help me please?
My answer:
You can notice that you only kept 2 significant figures for the answer in part A.
Your answer is 0.00027 m^3, but it can actually be 0.000274 or 0.000265...
0.00027 has only TWO significant figures.
The difference you introduced by throwing out the 3rd non-zero digits can be around 0.000005. The ratio between
what you throw away and the actual number is about 5/270, about 2 % of error.
When you use this number 0.27 to calculate part B, it will also give you 2% or error.
(you realize those 3 digits of zeros in front of 27 doesn't help you, 0.00027, 2.7 or 27 all only have 2 significant
figures. The round of error percentage are all about 0.5/27 = 5/270 about 2%.
To avoid this kind of rounding off error, you would want to keep one more significant digits before your final
answer. You may find the density actually to be 7400 or 7600 kg/m^3,...
Webassign likes numbers within 1% of accuracy. That means it is good to keep THREE significant figures. For
example, make it clear that it is 0.000274, but not 0.00027....will help.
It is not necessary to keep 4 significant digits, though.
NOTICE that significant figures has nothing to do with where the decimal points.
0.00027 m^3 = 0.27 Litter = 270 ml = 270 cm^3. They all only have 2 significant figures.
While 0.000274, 0.274, and 274 Have three significant figures . To keep 3 siginificant figures will keep your
uncertainty less than 1%, because you know it is 0.00274, not 0.00273.
Three solid objects (A, B, and C) are placed in water. The objects each have constant densities of 190, 260,
and 740 kg/m3, respectively.
a.) How do you immediately know all three objects will float?
b.) Find the fraction of the volume submerged for each object.
c.) If object C has the dimensions (length=6 m, width=8 m, height=2 m), how far does this object stick out
of the water when floating (in this orientation)?
Student A: I m able to find out how much volume is sticking out,but I'm not sure how I would calculate
this length for part C?
Man, Weining: good question.
If you find out how much volume is inside the water,
Vsub/(area of the box bottom) = height inside the water.
(area of the box bottom)= L*W
Total height - height inside the water = height sticking out.
Or If you find out how much volume is out the water,
This volume/ /(area of the box) = height inside the water.
Also you know since it has a uniform box shape,
The area from the top to the bottom is the same.
The percentage of (volume out of the waver) / (total volume) should be = (height out of the waver) / (total
height)
Student A: I found the amt out of water is equal to .27
I multiplied .27 times the length of box, 7 m
my answer was 1.89 is wrong. Why?
Man, Weining: You are asked to find the HEIGHT out of water.
Why you used .27 times the LENGTH of box.
If you found the percentage of (volume out of the waver) / (total volume) = 0.27, you could use 0.27 times
height to find HEIGHT out of water.
#6 What is the acceleration of a rising hot-air balloon if the ratio of the air density outside the balloon to
that inside is 1.29? Neglect the mass of the balloon fabric and the basket.
Man, Weining:
Again, identify your object, balloon and its inside air.
Identify the forces on the object:
mg of the inside air and
buoyancy force from out side air.
Ignore mg of the balloon fabric.
The total force would be? which minus which?
The acceleration would be? Ftotal / mass of your object (inside air)
When you try to write Ftotal or mass , there are unknown things like volume V, Density_inside,
Density_outside...But don't worry, Volume cancels at the end. Density_outside=1.29* density_inside....
See Lecture notes. Read the new uploaded notes about Chapter 15 before this HW. Important Equation:
buoyancy force = density of liquid *g * volume of displaced liquid. Also notice that, volume of displaced
liquid = volume UNDER water. (Volume under or immersed in the outside fluid.) For question 6, You can
first express gravity of the hot air, and the buoyancy force from the outside air with their density, volume,
and g. (ignore the mass of the balloon skin. ) Then you find the total force on the hot air balloon . Then
you divide that by the mass of the hot air to get acceleration. (notice that volume of the balloon can be
canceled at the end...)