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diff --git a/src/pages/sta/frames_and_machines.astro b/src/pages/sta/frames_and_machines.astro
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--- a/src/pages/sta/frames_and_machines.astro
+++ b/src/pages/sta/frames_and_machines.astro
@@ -57,13 +57,75 @@ Frame and machine analysis involves the following steps:
 <Enumerate>
 	<Item> Identify any two-force members and apply the two force member assumptions (see the Truss reference pages for more details)
     </Item>
-	<Item> Identify any multi force members and make free body diagrams for each multip force member. Remember that the same forces in two different free body diagrams will be equal and opposite!
-    </Item>
+	<Item>
+		Identify any multi force members and make free body diagrams for each multi force member.
+		When two connected members share a common force at their connection point, that force is equal in magnitude, but acts in the direction opposite of the direction it acts on the other member.
+		This comes from Newton's third law, as this force is internal to the overall system.
+	</Item>
 	<Item> Use equations of equilibrium to solve for the unknown forces in the free body diagrams. 
     </Item>
 </Enumerate>
 
 </p>
+	<Example title = "Ladder" solution = "True" id = "ladder">
+		<Image src = "/sta/frames_and_machines/ladder.jpg" width = "3">Image from <a target = "_blank" rel = "noopener noreferrer" href = "https://pxhere.com/en/photo/710585">PxHere.</a> CC0</Image>
+		<p>
+			A ladder is a common example of an A-frame.
+			An A-frame is a basic structure designed to bear loads in a lightweight economical manner.
+			The simplest form of an A-frame is formed by joining two similarly sized beams at an angle of 45 degrees or less and additionally connecting them with a third horizontal beam, so that the structure looks like that of an uppercase letter "A."
+		</p>
+		<p>
+			Consider the simply supported A-frame shown below, which has a pin support at point <InlineEquation equation = "A,"/> a roller support at point <InlineEquation equation = "E,"/> and is supporting a vertical load <InlineEquation equation = "P"/> at the midpoint of line <InlineEquation equation = "CD."/>
+		</p>
+		<Image src = "/sta/frames_and_machines/a_frame.png" width = "3"/>
+		<p>
+			Free body diagrams for each member would like this.
+		</p>
+		<Image src = "/sta/frames_and_machines/a_frame_exploded.png" width = "3"/>
+		<p>
+			Take note that if a common force is shared between two free body diagrams, the force will be equal in magnitude, but act in the direction opposite of the direction it acts on the other member—this comes from Newton's third law.
+			Equilibrium equations can then be written for each member to solve for the unknowns.
+		</p>
+		<p>
+			For this example, we are asked to find the internal force at point <InlineEquation equation = "B"/> on member <InlineEquation equation = "ABC."/>
+			Assume that point <InlineEquation equation = "B"/> is at the midpoint of member <InlineEquation equation = "ABC"/> and that point <InlineEquation equation = "D"/> is at the midpoint of member <InlineEquation equation = "CDE."/>
+			Assume that the structure is symmetrical.
+		</p>
+		<div slot = "solution">
+			<p>
+				First, analyze the free body diagram of the entire system. Take the moment about pin <InlineEquation equation = "A"/> and solve for the reaction force at roller <InlineEquation equation = "E."/>
+			</p>
+			<DisplayEquation equation = `\\begin{align} \\sum M_A = 0 &= -\\frac{5 w}{8} P + w F_E \\\\
+				\\rightarrow F_E &= \\frac{5}{8} P \\end{align}`/>
+			<p>
+				Next, analyze the free body diagram for member <InlineEquation equation = "CDE."/> Take the moment about point <InlineEquation equation = "C"/> on member <InlineEquation equation = "CDE"/> and solve for the internal force at point <InlineEquation equation = "D."/>
+			</p>
+			<DisplayEquation equation = `\\begin{align} \\sum M_{C, CDE} = 0 &= -\\frac{w}{8} P - \\frac{h}{2} F_D + \\frac{w}{2} F_E \\\\
+				&= -\\frac{w}{8} P - \\frac{h}{2} F_D + \\frac{w}{2} \\cdot \\frac{5}{8} P \\\\
+				\\rightarrow F_D &= \\frac{3 w}{8 h} P \\end{align}`/>
+			<p>
+				Since the magnitude of the force <InlineEquation equation = "F_D"/> was found to be positive, the assumption that the force <InlineEquation equation = "F_D"/> is in the direction shown on free body diagram of member <InlineEquation equation = "CDE"/> is correct.
+			</p>
+			<p>
+				Therefore, on member <InlineEquation equation = "CDE,"/>
+			</p>
+			<DisplayEquation equation = "F_D = -\\frac{3 w}{8 h} P"/>
+			<p>
+				Next, Analyze the free body diagram for member <InlineEquation equation = "BD."/> Analyze the forces along the x-direction and solve for the internal force <InlineEquation equation = "F_B."/>
+				Remember, on member <InlineEquation equation = "BD,"/> the force <InlineEquation equation = "F_D"/> acts in the direction opposite of the direction it acts on member <InlineEquation equation = "CDE."/>
+			</p>
+			<DisplayEquation equation = `\\begin{align} \\sum F_{x, BD} = 0 &= F_B + F_D \\\\
+				0 &= F_B + \\frac{3 w}{8 h} P \\\\
+				\\rightarrow F_B &= -\\frac{3 w}{8 h} P \\end{align}`/>
+			<p>
+				On member <InlineEquation equation = "ABC,"/> the force <InlineEquation equation = "F_B"/> acts in the direction opposite of the direction it acts on member <InlineEquation equation = "BD."/>
+			</p>
+			<p>
+				Therefore, on member <InlineEquation equation = "ABC,"/>
+			</p>
+			<DisplayEquation equation = "F_B = \\frac{3 w}{8 h} P"/>
+		</div>
+	</Example>
 <!--
 <Example id="machine-example" title="Machine Example" solution="True">
 <p>   This example is from <em>Engineering Statics</em> by Daniel Baker and William Haynes. </p>