HFBoards mathematics behind Newton's laws
 Register FAQ Members List Search Today's Posts Mark Forums Read
 Mobile Hockey's Future Become a Sponsor Site Rules Support Forum vBookie Page 2
 Notices Please do not post or solicit links to illegal game streams.

 Sciences A place to discuss natural, applied & social sciences, along with any other academically-oriented topics of interest to membership.

# mathematics behind Newton's laws

 10-01-2015, 08:36 PM #1 KrisLetAngry #teamKanye     Join Date: Dec 2013 Location: Saskatchewan Country: Posts: 12,719 vCash: 50 mathematics behind Newton's laws Remove if not allowed. I'm relearning in college after a few years in away from school and the professor essentially went fast. I just need to know the mathematics behind it. If any of yu know a source online I'd love to have it. Note: I know how to do math
 10-01-2015, 09:34 PM #2 Dave Registered User   Join Date: Oct 2009 Posts: 4,512 vCash: 500 can you be more specific?
 10-02-2015, 11:49 AM #3 Fugu Administrator HFBoards     Join Date: Nov 2004 Location: ϶(°o°)ϵ Posts: 36,706 vCash: 500 https://en.wikibooks.org/wiki/Classi...tonian_Physics
 10-02-2015, 12:18 PM #4 Hippasus 1,9,45,165,495,1287,     Join Date: Feb 2008 Location: Bridgeview Posts: 4,758 vCash: 500 Conservation of momentum Conservation of linear momentum: work over time = power A nailgun jams and is placed on its side. The potential energy of the temporary friction is released propelling the 8 g nail into a piece of 9 kg 2 x 4. This law can reveal the initial velocity of the nail by means of algebraic backcalculation. This is a simple model where we assume neither ground nor air resistance, no significant gravitational impact, and no movement of the wood directly prior to impact. mass = inertia = m (9.008 kg)(.40 m/s) = (.008 kg)v 3.60320 = .008v v = 450.4 m/s If the resultant external force on a system of objects is null, the sum of the vectors will remain constant. Since energy cannot be destroyed, but only transformed, mass can be regarded as one form of the superset of energy. The conservation between mass and energy is predicted by relativity theory as energy holds constant in isolated and bound systems. Conservation of angular momentum: Using the Earth as an example: mass = m = 5.9736 x 10^24 kg mean radius = r = 6.371 x 10^6 m mean orbital radius = R = 1.496 x 10^11 m day = 86,400 s year = 365.256366 d = 3.155815002 x 10^7 s Angular momentum of the Earth = L-(total) = L-(sub 1) + L-(sub 2) Orbital momentum = L-(sub 1) = Iw-(sub 1) I-(sub 1) = moment of inertia for a point mass around an orbit = mR^2 = (5.9736 x 10^24)(1.496 x 10^11)(1.496 x 10^11) = 13.36901238 x 10^46 = 1.336901238 x 10^47 w-(sub 1) = angular speed for a point mass around an orbit = (2(Pi)) / year = (2(Pi)) / (3.155815002 x 10^7) = 1.990986577 x 10^-7 Iw-(sub 1) = (1.336901238 x 10^47)(1.990986577 x 10^-7) = L-(sub 1) = 2.661752419 x 10^40 kg*m^2*s^-1 Axial momentum = L-(sub 2) = Iw-(sub 2) I-(sub 2) = moment of inertia for a sphere about its axis = (2/5)mr^2 = (.4)(5.9736 x 10^24)(6.371 x 10^6)(6.371 x 10^6) = 96.98651179 x 10^36 = 9.698651179 x 10^37 w-(sub 2) = angular speed for a sphere about its axis = (2(Pi)) / day = 6.28318 / 86,400 = 7.2 x 10^-5 Iw-(sub 2) = (9.698651179 x 10^37)(7.2 x 10^-5) = 69.83028849 x 10^32 = L-(sub 2) = 6.983028849 x 10^33 = .0000006983028849 x 10^40 kg*m^2*s^-1 L-(total) = (2.661752419 x 10^40 kg*m^2*s^-1) + (.0000006983028849 x 10^40 kg*m^2*s^-1) = 2.6617531173028849 x 10^40 = 2.6618 x 10^40 kg*m^2*s^-1 http://geophysics.ou.edu/solid_earth...ts.html#Moment http://www.uwgb.edu/dutchs/PLATETEC/ChangeRotn0.HTM The law of conservation of angular momentum for the Earth [simplified model with nothing more than algebra. This is a quick introduction to the basic idea of conservation mathematics]: moment of inertia = I I of uniform sphere = (2/5)mr^2 linear momentum = mv angular speed = w angular momentum = Iw 6.28318 r/s = 60 revolutions per minute .00069444 revolutions per minute = .000727219 radian/s v = d/t [(1)] angular momentum for a circular orbit = mvr [(1)] angular momentum for a circular orbit = mass x velocity x rotation [(1)] mean distance to the Sun = 1.496 x 10^8 km [Wikipedia] mean distance to the Sun from the Earth’s orbit = 149.6 x 106 km [(1)] 149.6 x 10^9 m [(1)] 5.9736 x 10^24 kg = mass of Earth [Wikipedia] 5.9742 x 10^24 kg [(1)] corrected now I = (.4)(5.9736 x 10^24 kg)(6,371,000 m)^2 = 2.38944 x 6,371,000 = (15,223,122.24 x 10^24 kg)m^2 Iw = ((15,223,122.24 x 10^24 kg)m^2)(.000727219 radian/s)^2 = 11,070.54373 kgm^2 v = d/t = ((2)(Pi)r)/t = ((2)(Pi)r)/(365 x 24 x 3600 s) = (9.39964522 x 10^8 km) Kinetic energy of rotation <-> KE-sub-r <-> (1/2)Iw^2 KE-sub-r = (1/2)Iw^2 = (.5)((15,223,122.24 x 10^24 kg)m^2)(.000727219 radian/s)^2 This is a more accurate measure of this problem from: (1) http://www.livephysics.com/problems-...-momentum.html v = d/t = ((2)(Pi)r)/t = ((2)(Pi)r)/(365 x 24 x 3600 s) (2) http://answers.yahoo.com/question/in...5210640AA0tKmi radian conversion from RPM: http://www.onlineconversion.com/foru...1127224365.htm joule: http://en.wikipedia.org/wiki/Joule scientific notation: http://en.wikipedia.org/wiki/Scientific_notation
10-03-2015, 10:22 AM
#5
shortshorts
The OG Kesler Hater

Join Date: Oct 2008
Country:
Posts: 12,347
vCash: 457
Quote:
 Originally Posted by Fugu https://en.wikibooks.org/wiki/Classi...tonian_Physics
To dumb it down even more...

f = ma

left = right
up = down

clockwise = counterclockwise

do algebra to solve what you're looking for.

 10-04-2015, 10:14 PM #6 KrisLetAngry #teamKanye     Join Date: Dec 2013 Location: Saskatchewan Country: Posts: 12,719 vCash: 50 Hello. Mostly the sum of forces = O If there is no acceleration And when there is angles attached to the proble Got some help from fellow classmates this weekend and hopefully some more before the test.
 10-04-2015, 10:39 PM #7 Dave Registered User   Join Date: Oct 2009 Posts: 4,512 vCash: 500 For that it would be trigonometry (sin cos tan). Establish a coordinates system, I'll assume for the level you're at it will almost always be X-Y coordinates. Then resolve each force into it's component (X-Y) forces using trig. Once everything has been broken down into components you can do a SUM all X forces, and SUM all Y forces. Now you'll have two resultant forces acting on the object which can be combined into one single force with an appropriate angle.
10-06-2015, 11:47 PM
#8
KrisLetAngry
#teamKanye

Join Date: Dec 2013
Country:
Posts: 12,719
vCash: 50
Quote:
 Originally Posted by Dave For that it would be trigonometry (sin cos tan). Establish a coordinates system, I'll assume for the level you're at it will almost always be X-Y coordinates. Then resolve each force into it's component (X-Y) forces using trig. Once everything has been broken down into components you can do a SUM all X forces, and SUM all Y forces. Now you'll have two resultant forces acting on the object which can be combined into one single force with an appropriate angle.
Yep. Basically

Just spent a lot of time practicing and actually getting answers right.

Hopefully I can get as good of a mark as my math test, but that ain't happening

 10-18-2015, 01:57 AM #9 KrisLetAngry #teamKanye     Join Date: Dec 2013 Location: Saskatchewan Country: Posts: 12,719 vCash: 50 Thought I'd mention I got a 70% got every question on Newtons while doing the math perfect except for losing 1 mark for a rounding error. Multiple choice killed me though. Regardless got 70 and that's 10 more then I needed.

Forum Jump