Added Sunny's solutions

src/Advanced/Geometry of Masses/parts/0 balance 1d.tex

@@ -5,6 +5,7 @@ Consider a mass $m_1$ on top of a pin. \par
 Due to gravity, the mass exerts a force on the pin at the point of contact. \par
 For simplicity, we'll say that the magnitude of this force is equal the mass of the object---
 that is, $m_1$.
+
 \begin{center}
 	\begin{tikzpicture}[scale=2]
 		\fill[color = black] (0, 0.1) circle[radius=0.1];
@@ -12,9 +13,10 @@ that is, $m_1$.
 
 		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
 
-
 		\draw[color = black, opacity = 0.5] (1, 0.1) circle[radius=0.1];
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1, 0) -- (0.85, -0.3) -- (1.15, -0.3) -- cycle;
+
+		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5]
+			(1, 0) -- (0.85, -0.3) -- (1.15, -0.3) -- cycle;
 
 		\draw[->, line width = 0.5mm] (1, 0) -- (1, -0.5) node[below] {$m_1$};
 		%\draw[->, line width = 0.5mm, dashed] (1, 0) -- (1, 0.5) node[above] {$-m_1$};
@@ -44,8 +46,8 @@ As you might expect, it is not stable: the rod pivots and falls down.
 		\draw[color = black, opacity = 0.5] (1.2, 0.0) circle[radius=0.1];
 		\draw[-, line width = 0.5mm, opacity = 0.5] (0.7, 0) -- (1.9, 0);
 
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1.5, 0) -- (1.35, -0.3) -- (1.65, -0.3) -- cycle;
-
+		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5]
+			(1.5, 0) -- (1.35, -0.3) -- (1.65, -0.3) -- cycle;
 
 		\draw[->, line width = 0.5mm] (1.2, 0) -- (1.2, -0.5) node[below] {$m_1$};
 		%\draw[->, line width = 0.5mm, dashed] (1.5, 0) -- (1.5, 0.5) node[above] {$f_p$};
@@ -57,6 +59,7 @@ It therefore exerts a \textit{torque} on the mass-rod system, causing it to rota
 \pagebreak
 
 
+
 \definition{Torque}
 Consider a rod on a single pivot point.
 If a force with magnitude $m_1$ is applied at an offset $d$ from the pivot point,
@@ -110,7 +113,6 @@ Find the position of this system's center of mass. \par
 
 		\draw[-, line width = 0.5mm] (-0.5, 0) -- (1.5, 0);
 
-
 		\fill[color = black] (-0.5, 0) circle[radius=0.1];
 		\node[above] at (-0.5, 0.2) {$3$};
 
@@ -139,9 +141,24 @@ Do the same for the following system, where $m_1$ and $m_2$ are arbitrary masses
 	\end{tikzpicture}
 \end{center}
 
+\begin{solution}
+	The CoM will be such that the rod is split into $d_1+d_2=1$
+	according to the relation $m_1 d_1 = m_2 d_2$.
+	This is sufficient, but if you want to solve for one of the $d$,
+	you get $d_1 = \frac{m_2}{m_1+m_2}$.
+
+	\vspace{2mm}
+
+	This should be intuitive; the distance of each mass from the CoM
+	is proportional to the other mass's share of the total mass.
+\end{solution}
+
 \vfill
 \pagebreak
 
+
+
+
 \definition{}
 Consider a massless, horizontal rod of infinite length. \par
 Affix a finite number of point masses to this rod. \par
@@ -172,5 +189,12 @@ Consider a one-dimensional system of masses consisting of $n$ masses $m_1, m_2,
 with each $m_i$ positioned at $x_i$. Show that the resulting system always has a unique center of mass. \par
 \hint{Prove this by construction: find the point!}
 
+\begin{solution}
+	\begin{equation*}
+		x_0 = \frac{1}{M}\sum_{i=1}^n m_i x_i
+	\end{equation*}
+	where $M = \sum_{i=1}^n m_i$
+\end{solution}
+
 \vfill
 \pagebreak