This Thanksgiving Day my family went to watch the Thanksgiving Day Parade in New York City. It was particularly windy and cold at 32 degrees Fahrenheit. They said that the workers had to hold the parade balloons particularly low. This made me wonder how much force the workers must apply to the balloons to hold them down. If the balloon is not moving at all in the Y direction, with up being positive, the sum of the forces equals the buoyant force, the force of the balloon and the force of the people holding it down.
Fy=pair*g*Vairdisplaced + phelium*g*Vballoon - Fpeople=0 N
The buoyant force and the force of the balloon are acting in the positive direction because the buoyant force acts up and the force of the ballon is the force of the helium, whose density is less than that of air so it pulls the balloon upward. The force of the people is then the only force in the downward direction.
The largest balloon ever was a Shamu balloon with 18000ft^3 or 509.7m^3. How much force was needed to hold this balloon down?
Fy=(1.269kg/m^3)(509.7m^3)(9.8m/s^2)+(1.786E-4kg/m^3)(509.7m^3)(9.8m/s^2)-Fpeople
Fpeople(total) = 6340 N
If there are approximately 40 people holding down each balloon then each person is applying about 159N of force. This is only the force in the Y direction. With the wind there were likely other forces in the X direction which would have put more work on the balloon holders to hold it in place.
Source: http://www.thedailybeast.com/articles/2011/11/25/macy-s-thanksgiving-day-parade-floats-balloons-more-statistics.html
Monday, December 9, 2013
Sunday, December 8, 2013
How does the length of a bridge change with the seasons?
Harnessing my inner 10 year old, I think bridges are
completely rad. The fact that we can build such huge structures across great
distances safely is pretty awesome
So, given my fascination with bridges, I
figured I’d take a closer look at the physics that bridge builders have to
consider when planning and putting together these remarkable pieces of
engineering. For the sake of ease of calculations, I’m considering only the
effects of temperature, not any other weather factors. I’m also assuming that
the bridge will respond in a linear fashion to the changes in temperature.
So, given my fascination with bridges, I
figured I’d take a closer look at the physics that bridge builders have to
consider when planning and putting together these remarkable pieces of
engineering. For the sake of ease of calculations, I’m considering only the
effects of temperature, not any other weather factors. I’m also assuming that
the bridge will respond in a linear fashion to the changes in temperature.
The bridge I’m considering is essentially a solid slab of steel
spanning a river. I’m going to take a look both at winter and summer effects of
temperature on the bridge.
SUMMER
ΔT = +30oC
α = 12x10-6 oC-1
Lo = 200m
ΔL = ?
ΔL = αLoΔT
ΔL = (12x10-6 oC-1)(200m)(30oC)
ΔL = 0.072m = 7.2cm
WINTER
ΔT = -20 oC
α = 12x10-6 oC-1
Lo = 200m
ΔL = ?
ΔL = αLoΔT
ΔL = (12x10-6 oC-1)(200m)(-20oC)
ΔL = -0.048m = -4.8cm
As a means of comparison, what if it was an aluminum bridge?
SUMMER
ΔT = +30 oC
α = 25x10-6 oC-1
Lo = 200m
ΔL = ?
ΔL = αLoΔT
ΔL = (25x10-6 oC-1)(200m)(30oC)
ΔL = 0.15m = 15cm
WINTER
ΔT = -20 oC
α = 25x10-6 oC-1
Lo = 200m
ΔL = ?
ΔL = αLoΔT
ΔL = (25x10-6 oC-1)(200m)(-20oC)
ΔL = -0.10m = -10cm
Guess it’s a good thing we make bridges out of steel and not aluminum!
Physics of Fly systems
So I was working in a theater over the past week setting up lighting equipment for a show, and I needed to raise what is called a batten up using what is called a loft block. A batten is a 80 gauge steel pipe that weighs around 111.97 kg and a loft block is a winch that has a radius of 150mm or .150m. In addition to the wight of the pipe itself, there were 10 lighting instruments on the batten. There were four "source 4"s three 4" fresnels and 3 8" fresnels which weigh 6.4kg, 3.3kg, and 6.5kg respectively. This brings the weight of the batten to 166.97kg in total which represents a force of 1636.306 N pulling on an aircraft cable that suspends the pipe. So I wanted to find out what torque was required to lift the batten:
166.97kg* 9.8 m/s^2 = 1636.306 N
Fg = Ft which means that the force due to gravity is the same as the force of the torque required to move the batten
This means that the torque required to move the batten is:
T= rF (because the sin() is going to equal 1)
T = .150 m * (1636.306 N) = 245.4 N/m
In addition, I wanted to know how much energy would be required to move the batten using the loft block. Assuming that only one cable was attached to batten, I found that the kinetic energy equaled:
I = 1/3 ( 166.97kg) (9.144m)^2 = 4653.6 kgm^2
KE = 1/2 (I) (w^2)
w = v/r assuming that the batten was raised at a speed of 1 m/s
w = 1.0 m/s / .150m = 6.67 rads/s
KE = .5 * 4653.6 kgm^2 * (6.67)^2
KE = 207033.8 J = 207.033 kJ
Though this may seem like a large amount of energy needed to raise a batten, there are multiple lift lines on the pipe which would lessen the amount of energy needed by lessening the moment of inertia. But even without multiple lift lines, it is possible to lift a batten using a hand crank. However, most theaters use a mechanic crank that pulls much more effectively with little to no effort from a stage hand.
Sources:
http://www.etcconnect.com/DOCS/DOCS_DOWNLOADS/DATASHTS/7060L1009_SOURCE_FOUR_36_SPEC_SHT_VG.PDF
http://www.altmanltg.com/altman-lighting-fresnels/75Q%20&%20175Q/fresnel-75Q.PDF
http://www.bhphotovideo.com/c/product/173850-REG/Arri_531602_650W_Fresnel_120_240V_AC.html
http://en.wikipedia.org/wiki/Fly_system
http://www.engineeringtoolbox.com/steel-pipes-weights-d_774.html
166.97kg* 9.8 m/s^2 = 1636.306 N
Fg = Ft which means that the force due to gravity is the same as the force of the torque required to move the batten
This means that the torque required to move the batten is:
T= rF (because the sin() is going to equal 1)
T = .150 m * (1636.306 N) = 245.4 N/m
In addition, I wanted to know how much energy would be required to move the batten using the loft block. Assuming that only one cable was attached to batten, I found that the kinetic energy equaled:
I = 1/3 ( 166.97kg) (9.144m)^2 = 4653.6 kgm^2
KE = 1/2 (I) (w^2)
w = v/r assuming that the batten was raised at a speed of 1 m/s
w = 1.0 m/s / .150m = 6.67 rads/s
KE = .5 * 4653.6 kgm^2 * (6.67)^2
KE = 207033.8 J = 207.033 kJ
Though this may seem like a large amount of energy needed to raise a batten, there are multiple lift lines on the pipe which would lessen the amount of energy needed by lessening the moment of inertia. But even without multiple lift lines, it is possible to lift a batten using a hand crank. However, most theaters use a mechanic crank that pulls much more effectively with little to no effort from a stage hand.
Sources:
http://www.etcconnect.com/DOCS/DOCS_DOWNLOADS/DATASHTS/7060L1009_SOURCE_FOUR_36_SPEC_SHT_VG.PDF
http://www.altmanltg.com/altman-lighting-fresnels/75Q%20&%20175Q/fresnel-75Q.PDF
http://www.bhphotovideo.com/c/product/173850-REG/Arri_531602_650W_Fresnel_120_240V_AC.html
http://en.wikipedia.org/wiki/Fly_system
http://www.engineeringtoolbox.com/steel-pipes-weights-d_774.html
World Cup Physics
In light of the recent World Cup drawing, I wanted to look
at the physics behind arguably one of the best players in the world, Cristiano
Ronaldo. He has been recorded to kick at
soccer ball at a rate of 60 miles per hour which 96.5 km/h. 96.5 km/h is
approximately 26.8 meters/second. The
weight of a standard FIFA soccer ball is anywhere between 420-445 g. The radius of a standard soccer ball is 11cm
which is 0.11m.
The rotational kinetic energy that a soccer ball has when
kicked by Ronaldo:
The angular velocity is w=v/r
w=
26.8 m/s / 0.11 m = 243.63 radians/sec
Minimum KErot = ½ I w2
= ½ (0.4*0.42 kg*(0.11m)2)
* (243.63 rad/s)2
=
60.32 J
Maximum KErot = ½ I w2
= ½ (0.4*0.445 kg*(0.11m)2)
* (243.63 rad/s)2
=
63.92 J
Jack Could Have Lived
I analyzed the scene in the Titanic where Rose is on a piece of wood and jack is in the water. I wanted to determine whether Jack and Rose could both fit onto the piece of wood because I think Rose was being selfish. I estimated that the piece of wood to be a rectangle with a length of 1.69 m and a width of .75m connected to a triangle with the same height and width. I estimated the thickness of the board to be .2 meters.
Calculations:
Volume of Board
V=1.90125 m2 X 0.2 m
V=.38025 m3
Jack is 73 kg
Rose is 63 kg
Buoyant Force of the
Wood
Fb=PVG
Fb=(1025 kg/m3)(.38025 m3)(9.8
m/s2)
Fb=3819.61125 N
ΣF=Fb+Mgwood+
Mg (rose+jack)
ΣF=(3819.61125
N) - 228 kg(9.8m/s2) - (73+63kg)(9.8m/s2)
ΣF=252.41125
N
The total force is still pointing upward so Rose and Jack
could have both fit on the board and been saved.
Saturday, December 7, 2013
The Tension Force in Spider Man’s Web
By Kody Lyng
This past week, a new trailer for the upcoming sequel to The Amazing Spider Man series was released.
In one of the clips, Spider-Man falls from the top of the Empire State Building
in order to pursue a criminal. I would like to take a closer look at the
tension force in Spider-Man’s web when he is at the bottom of his fall. We must
first make a few assumptions:
Spider-Man shoots his web at the top of the Empire State
Building
At the bottom of his fall, he is 1 meter above the ground
His web does not stretch at all
Spider-Man weighs 62 kg
Ignore air resistance
In order to find his velocity at the bottom of his swing, we
must set his kinetic energy at the bottom of the swing equal to his initial
potential energy.
1/2mv2 = mgh
v = sqrt(2gh) = sqrt[2*9.8 m/s2*(443m-1m)] = 93.1
m/s
In order to find the tension force of the web, we must set
the net force equal to the tension force subtracted by the force of gravity.
The net force will be the centripetal force or
Fc = m*v2/r
So…
m*v2/r = FT – mg or FT
= mv2/r + mg
FT = (62 kg)*(93.1 m/s)2/(443 m-1 m) +
(62 kg)*(9.8 m/s2)
FT = 1820 Newtons
Home Alone Paint Cans
I saw a commercial for the classic holiday movie "Home Alone" and wondered about the physics that I saw. The scene I'm referring to is the one where Macaulay Culkin throws paint cans tied to a rope off of the banister that swing and hit burglars in the head.
http://www.youtube.com/watch?v=2Lb92tL6R4A
I was interested in the force with which the paint cans hit the robbers.
http://www.youtube.com/watch?v=2Lb92tL6R4A
I was interested in the force with which the paint cans hit the robbers.
- I estimated the length of the rope and the mass of the paint can
- L=3m
- mass of paint can=4.5kg
- I found this equation online for the velocity of a pendulum at the bottom of it's swing (aka the point of contact with the robbers' heads)
- v = √{2gL[1-cos(a)]}
- I assumed that the rope didn't get taught until ~45degrees
- v=(2)(9.8m/s^2)(3m)[1-cos(45)]=5.3m/s
- a(radial)=v^2/r
- a(radial)=5.3^2/3m=9.3m/s^2
- F=ma
- F=(4.5kg)(9.3m/s^2)=42N
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