Monday, December 11, 2017

The Physics of DNA

I have always been interested in biology, hence why I am a molecular biology major, but taking this intro physics class has made me wonder more about the physics behind biology. I have just accepted that many of the biological processes that occur, and have never really thought about what drives them: spoiler, it’s physics. So I decided to do some digging into the physics behind biology, specifically DNA.
Researchers from the Pratt School of Engineering at Duke University investigated the double helix structure of DNA. One of the hallmarks of DNA is its stability, which is largely due to its double helix structure. The double helix structure seems to be the result of a balance between many forces: “stacking forces” occur between bases across the length of the DNA strand, and “pairing forces” are observed between complementary base pairs. These physicists devised a system that allowed them to focus on stacking forces, while previous studies struggled to separate stacking and complementary forces.
Different bases have different bonding properties: adenine forms strong and regular stacking bonds, while thymine forms the weakest and more disorganized stacking bonds. Strands of just adenine bases showed complex elasticity. I wonder if these bonds were studied/approximated as springs. They stretched the bonds with increasingly greater force and made a surprising discovery: the measurements plateaued off in not one but two places (23 and 113 pico-Newtons). The leveling off indicates breaking of bonds and that the helix unfolds. Thymine strands show no plateauing and do not resist stretching. I am really curious what it is about the structure of the bonds that causes them to show different elasticities and resistance to forces; this is an area that is being studied more now.
A physicist's view of DNA describes DNA through a physics based lens. The DNA double helix is much like a spring, in the sense that is can be stretched but has a resistance to stretching. Proteins input energy/force and “walk along” the DNA; this “melts” the DNA by supercoiling and bending it. These forces cause the DNA to denature and melt. When placed in a fluid, DNA tends to  fold in on itself, while it resists being pulled. DNA has intrinsic forces which resist stretching and deviation from a double helix. There is something about the structure of a double helix which allows for balance between pairing forces, stacking forces, and other forces that DNA comes into contact with in biological systems. I’m sure that this is an oversimplified view of the physics of DNA, but I think it’s pretty interesting that DNA has adopted a complex structure based on physics that allows it to best balance all of the forces which it experiences.

References:
Duke University: Pratt School of Engineering. (2007). Unraveling the physics of DNA's double

Mashaghi, A., & Katan, A. (2013). A physicist's view of DNA. De Physicus, De Physicus,

(24e,3), 59-61.

Sunday, December 10, 2017

Physics of the Titanic: Could Jack Have Survived?

Very recently, I was scrolling on Facebook trying to procrastinate on some of the work I have to do this week. As I was scrolling through the endless stream of news, cat videos, fail videos, etc., I saw a post talking about whether Jack could have survived at the end of the Titanic and whether or not to condemn Rose as a cold-hearted person for not letting Jack join her on top of the floating door. Since we have recently covered the topic of buoyancy, I decided I should give a crack at it and see whether or not Jack could have survived.

In order to start this problem, we need to use the equation for the force of buoyancy:

Fb = Density (fluid) x Volume (fluid displaced) x g

We know that the acceleration due to gravity is 9.8 m/s^2. For the density of the fluid, or freezing ocean water in this case, we can estimate the density to be about 1070 kg/m^3 (https://www.windows2universe.org/earth/Water/density.html). The only thing missing is the volume of the ocean water displaced. For simplicity's sake, we will assume that the door on which Rose was just completely submerged. Thus, if we find the volume of door, we can use this in calculating our buoyancy force. The dimensions of the door can be estimated to be 2.00 m by 1.00 m by 0.125 m, giving a volume of 0.250 m^3 (http://physicsbuzz.physicscentral.com/2012/09/sorry-girls-titanic-doors-were-made-of.html). Now that we have this information, we can calculate the buoyancy force acting on the floating door:

Fb = 1070(0.250)(9.8) = ~2620 N

In order to determine whether both Rose and Jack could have gotten on the floating door and lived, we need to figure out the force of gravity acting against the buoyancy force. The force of gravity is given by

Fg = (m [door] + m [Rose] + m [Jack])g

By simply googling the weights of Kate Winslet and Leonardo DiCaprio, we can use 63 kg and 75 kg for the masses of Rose and Jack respectively. We now only need the mass of the door. One of the woods used to make the doors of the Titanic was oak, which has a density of 770 kg/m^3 (https://www.engineeringtoolbox.com/wood-density-d_40.html). By multiplying this density by the volume of the door, we find the mass of the door to be 192.5 kg. With this information in hand, we are ready to calculate the force of gravity, which is found to be:

Fg = (192.5+63+75)(9.8) = ~3240 N

We see that Fg > Fb if both Jack and Rose are on the door at the same time. This means that the door would have sank if both Jack and Rose were on it, ultimately meaning they would have died. To confirm that the door wouldn't sink with just Rose on it and confirm that the movie did not have bad physics (at least in this instance), we can calculate the force of gravity if it were just Rose on the door.

Fg = (192.5+63)(9.8) = ~2500 N

We see that Fg < Fb in the case where Rose is the only one on the door, confirming that she could be on top of the door without it sinking. Thus, despite the cries and calls of all those fans of Jack who claim he could have been saved, we see that there would have been no way for both Jack and Rose to be on top of the door without it sinking. One of them had to die in the end for the other to have a chance to survive.

Thanks for reading.

Physics of Pressure Cookers

Pressure cookers are a great solution to cooking foods quickly while simulating the effects of a long braising process.  It works roughly according to the ideal gas law, PV=nRT.  Although the ideal gas law has limitations in regard to application in the real world, the principles behind it are applicable to the pressure cooker.  The machine works as follows, a liquid inside is brought to a boil.  This boiling liquid changes states of matter and becomes a gas.  Because the gas is limited by the volume of the pressure cooker, V remains constant.  The trapped gas increases the internal pressure of the container and without adding or subtracting any particles, n remains constant as well.  R is constant at 8.314J/molK, so for the equation to remain true the increase in pressure results in an increase in temperature greater than that of an open container in which the volume the gas can occupy is much greater.  This increase in temperature allows for the food inside to cook much faster.

            Interestingly, cooking at high altitudes can often be challenging since the lower atmospheric pressure makes cooking time longer.  A pressure cooker can account for this since it compensates for some of this loss of pressure.  It is still under the physics of pressure, but can give you a couple more minutes back when you are cooking a nice warm dinner after a night of skiing in the mountains.

Snow Boots

This morning when I woke up, there was snow on the ground. My first thought was I should have gone to school in the south, then my second thought was questioning what I should wear, given the fact that it is snowy, cold, and most likely slippery outside. And then, because physics has to do with everything in our lives, I thought about the physics behind wearing good shoes in the snow.

I chose my winter boots because I know form experience that I am less likely to slip and fall when I have them on. But why is that? In order to not slip I would have to have enough force of static friction with the ground. The material used to make the sole of the shoe would determine its coefficient of static friction, and some materials would result in more resistance to slipping. Therefore, one could assume that the material my winter boots are made out of have a higher coefficient of static friction then my leather booties or keds.


This is something I should have thought about more in high school. When I was a senior, anxiously awaiting my college acceptance letters in late March, I would check the mail after school every day. Just a few days before being accepted to Colgate, I put on my Steve Madden tall leather boots and headed out to the mailbox. Next thing I remember was standing up next to my mailbox, I do not remember walking out there or falling. I got the mail and on my way back inside, realized my tooth was broken. When I went into the bathroom inside my house to look at it, I realized my scarf was completely soaked in blood. Two layers of 8 stitches and a concussion diagnosis later, I now realize that although stylish, tall leather boots are not what I should wear in March in Minnesota – there isn’t enough friction between them and the ground.

Saturday, December 9, 2017

Venice and the Importance of Buoyancy

In class on Friday we talked about thermal expansion and how the earth's rising temperature leads to the rising sea level. The increase in temperature is of course due to human emissions of greenhouse gases, such as carbon dioxide. Greenhouse gases trap heat near the earth's surface and are a vital part of our atmosphere, but as more of these gases are released into the atmosphere, more heat is trapped. This poses serious threats to the environment and humans alike.

In Venice, which is built on a series of islands off the coast of northeastern Italy, flooding has historically been a problem. Venice is only about a meter above sea level, so when there is a high enough tide, parts of the city are submerged in water. The rising sea level makes this problem even more serious because it increases the frequency and severity of the floods. 

In 1987, the Venice Water Authority began to take steps to protect Venice and the other populated areas around the lagoon in which Venice is located. One of the steps was to start planning the implementation of gates at each of the three inlets from the Adriatic Sea into the lagoon. In 2003, after years of planning, the project began. 



The project is know as "MOSE", which stands for "MOdulo Sperimentale Elettromeccanico" (Experimental Electromechanical Model". When the project is finished, there will be a total of 78 gates (41 at one inlet, 18 at another, and 19 at the other). The gates are made of steel and mostly hollow. The smallest gate is 18.5 by 20 by 3.6 meters and the largest one is 29.5 by 20 by 4.5 meters. Each gate is housed in a structure on the seafloor and attached to two hinges. When "resting", the gate is filled with seawater. When there is a high tide, some of the seawater is expelled by pumping air into the gate, causing it to rise through buoyancy.  


What I found so fascinating about these gates is that, despite all of the complicated engineering that goes into their construction and implementation, as well as the importance that they serve, they are raised simply through buoyancy. This is not only simple, but probably more efficient than using something like hydraulics to lift something that weighs about 300 tons. Buoyancy is so simple, in fact, that I was able to calculate how much seawater, and how much air, is present within one of the gates when it is raised and approximately in static equilibrium. I was able to do this only using what we learned in class about torque, force, and buoyancy (I also used a bunch of triangles and basic trigonometry). 

The gate I decided to do my calculations for was one of the gates at the Malamocco inlet. This inlet is both the deepest and has the largest gates. The dimensions of the gate in question are 29.5 x 20 x 4.5 and the inlet is about 14 m deep. The gates are designed to stop a tide of up to 3 meters, so if there was no high tide, the gate would be 17 m above the seafloor. This was the situation I considered because I did not have to take into account the different heights of water on each side of the gate and how this would effect the static equilibrium. 

First I calculated the volume of air and volume of steel in the gate when it was empty. To do this, I used the density of air (1.225 kg/m^3) and steel (8050 kg/m^3) and the approximation of the mass of the gate of 300 tons and the total volume calculated from the dimensions.. The conversion from tons to kg is 1 ton=907.185 kg. The calculations for this are shown below:

300 tons * (907.185 kg/1 ton) = 2.72 * 10^5 kg

V-total = 29.5 m * 20 m *4.5 m = 2655 m^3

1.225 kg/m^3 * V-air + 8050 kg/m^3 * (2655 m^3 - V-air) = 2.72 * 10^5 kg

V-air = 2622 m^3
V-steel = 33.4 m^3

Then I figured out what forces and torques act on the gate in static equilibrium. A free-body diagram showing the torques, forces, angles of each torque with respect to the object, axis of rotation, and the distance the torques are from the axis of rotation are shown below:




The torque and force from gravity and buoyancy are shown acting on the center of mass of the gate, which I treated as a rectangular prism, even though the actual shape is slightly different. The axis of rotation is the hinge, for which there is no torque because the force acts at a radius of 0 with respect to the axis of rotation (note that the force from the hinge accounts for both hinges so the force for each hinge is half of what I calculated). 𝜃-1 = 31.7° and 𝜃-2 = 148.3°. The equations that show the sum of all the forces and all the torques are shown below:

ΣF = 0 N = F-H +⍴Vg - mg where V is the seawater displace by the gate and ⍴ is the density of the seawater
Σ𝝉 = 0 N*m = 𝝉-B + 𝝉-g = ⍴Vg*sin(31.7°)*10 m - mg*sin(148.3°)*10 m

For the buoyancy force, I had to figure out how much seawater was displaced because the gate is not completely submerged. I did so by using similar triangles and trigonometry. The diagram below shows the dimensions of the gate and the triangles I used to find the area of the gate that is out of the water:



I split the part of the gate that is out of the water into a rectangle and a triangle. The rectangle has sides of 3.53 m and 4.5 m while the triangle has a base of 2.78 m and a height of 4.5 m. I then calculated the area and multiplied by the width of the gate to find the volume that is out of the water:

(3.53 m * 4.5 m + 0.5*2.78 m*4.5 m) * 29.5 m  = 654.9 m^3

Subtracting this value from the total volume of the gate gave me the volume of the gate that is underwater and therefore how much seawater is displaced:

2655 m^3 - 654.9 m^3 = 2000.1 m^3 = V of seawater displaced

Plugging this back in the equation for buoyancy force and using the density of seawater (1029 kg/m^3), I got:

F-B = ⍴Vg = 1029 kg/m^3 * 2000.1 m^3 *9.8 m/s^2 = 2.02 * 10^7 N

With the buoyancy force, I then calculated the buoyancy torque:

𝝉-B = (2.02*10^7 N) *10 m * sin(31.7°) = 1.06 *10^8 N*m

Then I used the fact that the sum of the torques is 0 N*m to solve for the volume of air and seawater in the gate. To do so, I rewrote the mass of the gate as follows:

m = ⍴-steel*V-steel + ⍴-air*V-air + ⍴-seawater*V-seawater

and because the original V-air (2622 m^3)= V-air +V-seawater:

m = ⍴-steel*V-steel + ⍴-air (2622 m^3 -V-seawater) + ⍴-seawater*V-seawater

Now the sum of the torques can be written as follows:

Σ𝝉 = 0 N*m = 𝝉-B + 𝝉-g = 𝝉-B - (⍴-steel*V-steel + ⍴-air (2622 m^3 -V-seawater) + ⍴-seawater*V-seawater)g*sin(148.3°)*10 m

-1.06 * 10^8 = -(8050 kg/m^3*33.4 m^3 + 1.225 kg/m^3(2622 m^3 - V-seawater) + 1029 kg/m^3*V-seawater)g*sin(148.3°)*10 m

V-seawater = 1736 m^3
V-air = 885.6 m^3

And just for fun, I calculated the average density of the gate:

⍴-avg = (m-steel + m-air + m-seawater)/(V-air + V-seawater + V-steel) = 780 kg/m^3

Even though 885.6 m^3 is a lot of air, relative to the size of the gate, it is surprisingly small and only takes up about a third of the space in the gate. This shows how significant a difference of a factor of almost 1000 in densities can be and how a very simple concept can be used as a solution to a very large scale problem.


Sources:

https://en.wikipedia.org/wiki/MOSE_Project#Lido_inlet
https://www.mosevenezia.eu/mose/?lang=en
http://www.mega-project.eu/assets/exp/resources/The_MOSE_project.pdf






Keystone Pipeline

On November 16, 2017 the Keystone pipeline had a leak in Amherst, South Dakota. According to an NPR article, 210,000 gallons of oil spilled out of the pipeline before it was shut down.  The spill covers approximately a 100 yard radius of flat grassland. According to a CNN article, the pipeline is 36 inches (.91m) in diameter and can carry 830,000 barrels of oil per day. There are 42 gallons in a barrel and one gallon is approximately0.004m^3. Thus 1.32x10^5 m^3 of oil flow each day, at a rate of 1.53m/s.

When the oil is flowing through the pipeline, it moves with laminar flow because the pipeline has the same height throughout its entirety.

A lot of oil spills are caused by corrosion at the bottom of the pipe to the point where a hole is created. I was curious what the pressure was at the bottom of the pipeline:

Pressure = (rho)(g)(h) + Patm
Density of oil = 900kg/m^3
Pressure = (900kg/m^3)(9.8m/s^2)(.91m) +1.013x10^5Pa = 1.09x10^5 Pa

If corrosion did occur all the way through and a hole developed on the bottom of the pipe, I wanted to know what the velocity of the oil would be when it exited:

P1 + rhogh1 + 1/2rhov1^2 = P2 + rhogh2 + 1/2rhov2^2
1.013x10^5Pa + (900kg/m3)(9.8m/s^2)(.91m) + 0 =  1.013x10^5Pa + 0 + .5(900kg.m^3)(v2)^2
1.09x10^5 = 1.013x10^5 + 450v2^2
8.03x10^3 = 450v2^2
v2^2 = 17.8m/s
v2 = 4.2m/s

The current crossectional area of the pipeline is (.91m/2)^2π = .65m^2. Periodically throughout the pipeline there are places where the line can be closed off to adjust the flow rate or in case of emergency. The pipeline company can control when this happens. If for example they close it half way, the new crossectional area would be .65m^2 / 2, or .325m^2. What would the resulting velocity of the oil be?

A1V1 = A2V2
(.65m^2)(1.53m/s) = .325m/s^2(V2)
V2 = 3.05m/s


Sources: 
https://www.npr.org/sections/thetwo-way/2017/11/16/564705368/keystone-pipeline-oil-spill-reported-in-south-dakota 
http://www.cnn.com/2017/11/16/us/keystone-pipeline-leak/index.html











Thermal Expansion and Jars

As I was opening a jar of jam yesterday at lunch, I thought about the different ways people use to open stuck jarsgetting someone stronger than you, putting duct tape on the lid and yanking, running the jar under hot water, etc. 
So why does the hot water method work? It comes down to thermal expansion. Using the equation for volume,

ΔV=βVΔT

we can begin to see how it would work. Since the hot water is the same temperature for both the metal lid of the jar and the glass body, that stays constant. The original volume of the jar and the lid at the point where they screw together interlock, so should be very similar in volume, so that should not affect the change in volume either. What does vary between the lid and the jar is the volumetric coefficient, which for glass is 25.5 x 10 K¹ (at 20°C)and for steel (the most common metal jar lids are made of) is anywhere from 33.0 to 39.0 x 10 K¹ (again at 20°C). 
So if we can disregard the change in temperature and the original volume as the same for both the glass and the lid, calling it "x", the equation becomes 

ΔV=β(x)

which by using the coefficients stated above shows that the metal will expand more than the glass will, so it will become easier to twist the lid off.