Completed teeter

Author: AiStudent

Book/build.py

@@ -137,11 +137,11 @@ def build_index(sources):
     ]),
     ("Newtonian Mechanics", [
         ("Single particle mechanics", "Physics/src/NewtonianMechanics/SingleParticle.lhs")
+    ]),
+    ("Examples", [
+        ("Teeter", "Physics/src/Examples/Teeter.lhs"),
+        ("Box on an incline", "Physics/src/Examples/Box_incline.lhs"),
     ])
-    # ("Examples", [
-    #     ("Gungbräda", "Physics/src/Examples/Gungbraeda.lhs"),
-    #     ("krafter på lådor", "Physics/src/Examples/krafter_pa_lador.lhs"),
-    # ])
 ]
 
 if not os.path.exists("build"):

Physics/src/Examples/Teeter.lhs

@@ -6,50 +6,72 @@ Exam excercise 3, 2017-01-13
 > import Dimensions.Quantity
 > import Prelude hiding (length)
 
-> balk_M = 1.0 # mass
+
+Two boxes, m1 and m2, rests on a beam in balance.
+
+Known values:
+
+> beam_M = 1.0 # mass
 > m1 = 2.0 # mass
 > m2 = 5.0 # mass
 > d = 0.75 # length
-> balk_L = 5.0 # length
+> beam_L = 5.0 # length
 > two = 2.0 # one
 
 ![Teeter](teeter.png){.float-img-left}
 
 Direct implication:
 
-$$ f(x) = \frac{a}{b}$$
+> beam_left_L = (beam_L /# two) +# d
+> beam_right_L = beam_L -# beam_left_L
 
-> balk_left_L = (balk_L /# two) +# d
-> balk_right_L = balk_L -# balk_left_L
+We want to be able to represenet the torques.
 
-We want to be able to represenet the angular momentums.
-Here are some propositions.
+A torque (sv. vridmoment) is defined as:
 
-> m1_vrid = m1 *# balk_left_L 
+$$ \tau = distance\ from\ turning\ point \cdot force $$
 
-> balk_L_vrid = ((balk_left_L /# balk_L) *# balk_M) *# (balk_left_L /# two)
+Since all force values will be composited of a mass and the gravitation, we can ignore the gravitation.
 
+$$ \tau = distance\ from\ turning\ point \cdot mass $$
 
-> balk_H_vrid = ((balk_right_L /# balk_L) *# balk_M) *# (balk_right_L /# two)
 
+> m1_torq = m1 *# beam_left_L 
 
-m2_vrid = m2 * x
+To get the beams torque on one side, we need to divide by 2 because the beam's torque is spread out linearly (the density of the beam is equal everywhere), which means the left parts mass centrum is \emph{half the distance} of the left parts total length.
 
-VL = HL
+$$ beamL_{\tau} = beamL_{M} \cdot \frac{distance}{2} $$
 
-m1_vrid + balk_L_vrid = m2_vrid + balk_H_vrid
+$$ beamL_{\tau} = \frac{beam\ left\ length}{beam\ length} \cdot beam_M \cdot \frac{beam\ left\ length}{2} $$
 
-m1_vrid + balk_L_vrid - balk_H_vrid  = m2_vrid
+> beamL_torq = ((beam_left_L /# beam_L) *# beam_M) *# (beam_left_L /# two)
 
-(m1_vrid + balk_L_vrid - balk_H_vrid) / m2  = x
+> beamR_torq = ((beam_right_L /# beam_L) *# beam_M) *# (beam_right_L /# two)
 
-> x = (m1_vrid +# balk_L_vrid -# balk_H_vrid) /# m2
+We make an expression for $m2_{\tau}$, which involves our unknown distance x.
 
-Security check:
+$$ m2_{\tau} = m2 \cdot x $$
+
+For the teeter to be in balance, both sides torques should be equal.
+
+$$ Left\ side\ torque = Right\ side\ angular\ torque $$
+
+We try to break out $m2_{\tau}$ and then x.
 
-> m2_vrid = m2 *# x
+$$ m1_{\tau} + beamL_{\tau} = m2_{\tau} + beamR_{\tau} $$
+
+$$ m1_{\tau} + beamL_{\tau} - beamR_{\tau} = m2_{\tau} $$
+
+$$ \frac{m1_{\tau} + beamL_{\tau} - beamR_{\tau}}{m2}  = x $$
+
+Our solution:
+
+> x = (m1_torq +# beamL_torq -# beamR_torq) /# m2
+
+Security check:
 
-> vL = m1_vrid +# balk_L_vrid
-> hL = m2_vrid +# balk_H_vrid
+> m2_torq = m2 *# x
 
+> left_side_torque = m1_torq +# beamL_torq
+> right_side_torque = m2_torq +# beamR_torq