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--- 183_notes:examples:walking_in_a_boat [2014/10/01 05:43] – pwirving
+++ 183_notes:examples:walking_in_a_boat [2014/10/01 05:51] (current) – pwirving
@@ Line 83: / Line 83: @@
  $x_{cm,f} = \dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m}$
+As indicated the center of the mass of the system as not changed position therefore:
  $x_{cm,f} = x_{cm,i}$
+So we can relate the equations for initial and final to each other:
  $\dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$
-Same denominator
+Both of these equations have the same denominator and so we can cancel it out:
  ${M(x+D+\dfrac{L}{2}) + m(x+D)} = {M(D+\dfrac{L}{2}) + m(D+L)}$
-Solve for x,
+Multiply out to solve for x:
-$M_x + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$
+$Mx + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$
-Cancel like terms
+Cancel like terms we get:
  $Mx + mx = mL$
@@ Line 104: / Line 106: @@
  $(M+m)x = mL$
+Therefore:
  $x = (\dfrac{m}{M+m})L$
@@ Line 110: / Line 114: @@
 Does this make sense?
+If you want to check whether something makes sense a good start is to check the units:
 Units  $(x) = m$    $(\dfrac{m}{M+m})$  = unitless
-If M is really big then  $x \equiv 0$ , think oil thanker
+Therefore m is the remaining unit.
+Another check is that we know that if M is really big then  $x \equiv 0$ , think of a similar scenario to one just discussed occurring on a oil thanker.
  $x = (\dfrac{m}{M+m})L  \equiv \dfrac{m}{M}L \equiv 0$  when  $M>>m$
-If M = 0 then x  $\equiv L$ , no boat limit
+Another check would be to check what happens when M = 0 then x  $\equiv L$ , i.e. if there was no boat.
  $x = (\dfrac{m}{M+m})L  \equiv \dfrac{m}{M}L \equiv L$  when  $m>>M$
 So the motion of the center of mass of a system is dictated by the net external force.