There has been a factory fire! During the fire, the fire spread to a room in the back filled with chemicals. Most notable of the chemicals was petroleum distillate. This chemical is highly flammable and very dangerous.The arrival of the fire upon the chemical room resulted in the contact with the petroleum distillate which caused a huge explosion. The container split into four separate pieces each with the same shape but different volumes. The velocity of a particular piece will be calculated and that velocity will be used to determine the force of the material. That piece has a hit an employee running by the room during the fire. The force will be used to see if it would have broken a rib of the person running by.
One piece of the container that came off had a mass of 300 grams and a velocity of 4100 m/s going North. The second piece had a mass of 400 grams and a velocity of 3075 m/s going South. The third piece had a mass of 398 grams and a velocity of 4200 m/s going 20 deg South of West. The fourth piece has a mass of 456 grams with an unknown velocity and direction.
Calculations for x and y velocities of mass 3: cos(20)*4200=3946.71=x-velocity sin(20)*4200=1436.48=y-velocity
(m1i)(v1i)+(m2i)(v2i)+(m3i)(v3i)+(m4i)(v4i)=(m1f)(v1f)+(m2f)(v2f)+(m3f)(v3f)+(m4f)(v4f) 0+0+0+0=(.3)(0)+(.4)(0)+(.398)(-3946.71)+(.456)(v4xf) 0=0+0+(-1570.79)+(.456)(v4xf) 1570.79=(.456)(v4xf) v4xf=3444.72 m/s
(m1i)(v1i)+(m2i)(v2i)+(m3i)(v3i)+(m4i)(v4i)=(m1f)(v1f)+(m2f)(v2f)+(m3f)(v3f)+(m4f)(v4f) 0+0+0+0=(.3)(4100)+(.4)(-3075)+(.398)(-1436.48)+(.456)(v4yf) 0=(1230)+(-1230)+(-571.72)+(.456)(v4yf) 571.72=(.456)(v4yf) v4yf=1253.77 m/s
(v4f)²=(v4xf)²+(v4yf)² (v4f)²=(3444.72)²+(1253.77)² (v4f)²=13438012.77 v4f=3665.79 m/s
arctan(opposite/adjacent)=angle (in deg) North of East arctan(1253.77/3444.72)=20 deg North of East
The fourth piece from the exploding container of petroleum distillate will travel at a velocity of 3665.79 m/s 20 deg North of East.
The poor individual was hit by the fourth piece of the container at 3665.79 m/s and was struck right in the ribs. Luckily the person miraculously survived; however, doctors and physicists examined the piece that hit the employee to determine the force it delivered. The physicist knew, with air resistance in mind, that the force was equal to D*v². D was a constant representing air resistance.
D=(pCA/2) p is the density of the air, C is the drag constant which is based on the shape of the solid, and A is the silhouette area (area of the shape from the front) A fairly standard air density is 1.225 kg/m^3, so p=1.225 kg/m^3. The shape of the piece was a cube so the drag constant is 1.05, C=1.05. The silhouette area was .001 square meters, A=.001 m².
D=(pCA/2)=((1.225*1.05*.001)/2)=6.43E-4 kg/m F=Dv² v²=13438016.32 Dv²=(6.43E-4)(13438016.32)=8642.32 N of force
With a force of 8642.32 N exerted on his ribs, the employee had some cracked ribs. There is a 25% chance that a rib will crack when 3300 N is exerted on it, so 8642.32 N guarantees at least one rib cracking. After running an x-ray of the patient, the doctor found multiple cracked ribs, the physicist was glad he had done the math correct, but felt sorrow for the injured employee.
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