Watch lecture 1.5, 1.6
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- maths
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- statics
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- dynamics
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title: MMME1028 // Statics and Dynamics
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title: MMME1028 // Statics
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---
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# Lecture L1.1, L1.2
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@ -38,10 +38,14 @@ Can be found here [here](./lecture_exercises/mmme1028_l1.2_exercises_2021-10-04.
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- Body is in equilibrium if sum of all forces and moments acting on
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body are 0
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<details>
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<summary>
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### Example
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Determine force $F$ and $x$ so that the body is in equilibrium.
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</summary>
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1. Check horizontal equilibrium
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@ -62,6 +66,8 @@ Determine force $F$ and $x$ so that the body is in equilibrium.
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$x = 6$
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</details>
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## Free Body Diagrams
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A free body diagram is a diagram of a single (free) body which shows all
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@ -138,3 +144,304 @@ This principle can be useful in determining moments.
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5. Solve everything symbolically (algebraicly) until the end
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6. Check your answers make sense
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7. Don't forget the units
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# Lecture L1.4
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## Tension and Compression
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- The convention in standard mechanical engineering problems is that positive values are for
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tension and negative values for compression
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- Members in tension can be replaces by cables, which can support tension but not compression
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- Resisting compression is harder as members in compression can buckle
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## What is a Pin Joint?
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- Pin jointed structures are structures where joints are pinned (free to rotate)
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- Pin joints are represented by a circle (pin) about which members are free to rotate:
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- A pin joint transmits force but cannot carry a moment
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## What is a Truss?
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- Trusses are an assembly of many bars, which are pin jointed in design but do not rotate due to
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the geometry of the design. A pylon is a good example of this
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- Trusses are used in engineering to transfer forces through a structure
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- When pin jointed trusses are loaded at the pins, the bars are subjected to pure tensile or
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compressive forces.
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These bars are two force members
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## Equilibrium at the Joints
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### Forces at A
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$$\sum F_y(A) = \frac P 2 + T_{AB}\sin{\frac \pi 4} = 0 \rightarrow T_{AB} = -\frac P 2$$
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\begin{align*}
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\sum F_x(A) &= T_{AB}\cos{\frac pi 4} + T_{AC} = 0 \\
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T_AC &= -\frac{-P} 2 \times \frac{\sqrt{2}} 2 = \frac P 2
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\end{align*}
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### Forces at B
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Add the information we just obtained from calculating forces at A:
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And draw a free body diagram for the forces at B:
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$$
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\sum F_y(B) = -\frac{-P} 2 \sin{\frac \pi 4} - T_{BC} = 0 \rightarrow
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T_{BC} = \frac P {\sqrt 2} \times \frac {\sqrt 2} 2 = \frac P 2
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$$
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\begin{align*}
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\sum F_x(B) &= -\frac{-P}{\sqrt2}\cos{\frac \pi 4} + T_{BD} = 0 \\
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T_{BD} &= -\frac P 2
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\end{align*}
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## Symmetry in Stuctures
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Symmetry of bar forces in a pin jointed frame depends on to aspects:
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1. Symmetry of the stucture
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2. Symmetry of the loading (forces applied)
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Both conditions must be met to exploit symmetry.
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# Lecture L1.5, L1.6
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## Method of Sections
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The method of sections is very useful to find a few forces inside a complex structure.
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If an entire section is in equilibrium, so are discrete parts of the same structure.
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This means we an isolate substructures and draws free body diagrams for them.
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We must add all the forces acting on the substructure.
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Then we make a virtua cut through some of the members, replacing them with forces.
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Then we can write 3 equilibrium equations for the substructure:
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1. 1 Horizontal, 1 vertical, and 1 moment equation
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2. Either horizonal or vertical and 2 moment equations
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3. 3 moment equations
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<details>
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<summary>
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### Example 1
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</summary>
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Draw a virtual cut through the structure, making sure to cut through all the bars whose forces
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you are trying to find:
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Draw the free body diagram, substituting cut bars by forces:
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As there are three unknown forces, we need three equilibrium equations.
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#### First Equation: Moments about E
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\begin{align*}
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\sum M(E) &= \frac P 2 \times 2L + T_{DF}L = 0 \\
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T_{DF} &= -P
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\end{align*}
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#### Second Equation: Vertical Equilibrium
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\begin{align*}
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\sum F_y &= \frac P 2 + T_{EF}\sin{\frac \pi 4} = 0 \\
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T_{EF} &= -\frac P {\sqrt2}
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\end{align*}
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#### Third Equation: Horizontal Equilibrium
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\begin{align*}
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\sum F_x = T_{DF} + T_{EF}\cos{\frac \pi 4} + T_{EG} = 0 \\
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T_{EG} = \frac {3P} 2
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\end{align*}
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#### Taking Moments from Outside the Structure
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If we only needed EG, we could have taken moments about point F, outside our substructure:
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\begin{align*}
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\sum M(F) &= \frac P 2 \times 3L -T_{EG}L = 0 \\
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T_{EG} &= \frac {3P} 2
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\end{align*}
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</details>
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## Zero-Force Members
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Consider the free body diagram for the joint at G:
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$$\sum F_y(G) = T_{FG} = 0$$
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$$\sum F_x(G) = -T_{EG} + T_{GJ} = 0 \rightarrow T_{EG} = T_{GJ}$$
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Meaning that the structure is effecively the same as this one:
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Why was it there?
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- The structure may be designed for other loading patterns
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- The bar may prevent the struture from becoming a mechanism
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- A zero force member may also be there to prevent buckling
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## Externally Applied Moments
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Externally applied moments are dealt with in the same way as external forces, but they only
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contribute to moment equations and not force equilibrium equations.
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## Distributed Load
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A distriuted load is applied uniformly to a bar or section of a bar.
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It can be represented by a single force through the midpoint its midpoint.
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<details>
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<summary>
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### Example 1
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</summary>
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Is equivalent to:
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</details>
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## Equivalent Loads
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When loads are applied within a bar, as far as support reactions and bar forces in *other* bars
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are concered, we can determine *equivalent node forces* using equilibrium
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# Lecture L1.6
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<details>
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<summary>
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### Example 1
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The figure shows a light roof truss oaded by a force $F = 90$ kN at 45\textdegree to the horizontal
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at point $B$.
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</summary>
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a. Find the reaction forces at A and D using equilibrium applied to the whole structure.
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> 1. Add unknown quantities to the diagram
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>
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> 
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>
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> 2. Consider the number of unknowns --- there are 3 therefore 3 equations are needed
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> 3. Decide which equilibrium equation to start with
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>
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> Horizontal equilibrium:
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>
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> \begin{align*}
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\sum F_x &= R_{Ax} - F\cos45 = 0 \\
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R_{Ax} &= 63.6\text{ kN}
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> \end{align*}
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>
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> Vertical equilibrium:
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>
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> \begin{align*}
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\sum F_y &= R_{Ay} + R_{Dy} - F\sin = 0 \\
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R_{Ay} + R_{Dy} &= \frac{\sqrt2 F} 2
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> \end{align*}
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>
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> Moment equation:
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>
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> \begin{align*}
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> \sum M_{xy}(B) &= 4.5R_{Ay} - 4.5R_{Dy} - L_{BC}R_{Ax} = 0 \\
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> \frac{L_{BC}}{4.5} &= \tan30 = \frac 1 {\sqrt3} \rightarrow L_{BC} = 2.6 \\
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> \end{align*}
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>
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> 4. Solve for $R_{Ay}$
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>
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> \begin{align*}
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> 4.5R_{Ay} &= 4.5R_{Dy} + 2.6\times\frac{\sqrt2 F}{2} & R_{Dy} = \frac{\sqrt2 F}{2} - R_{Ay} \\
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> 4.5R_{Ay} &= 4.5\left(\frac{\sqrt2 F}{2} - R_{Ay}\right) + 2.6\times\frac{\sqrt2 F}{2} \\
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> R_{Ay} &= 0.56F = 50.2\text{ kN}
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> \end{align*}
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>
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> 5. Substitute to find $R_{Dy}$
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>
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> \begin{align*}
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> R_{Dy} = \frac{\sqrt2 F}{2} - R_{Ay} = (0.71-0.56)F = 13.3\text{ kN}\\
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> \end{align*}
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b. Use the graphical/trigonometric method o check your answer.
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> Write the reaction at A as a single force with unknown direction:
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>
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> 
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>
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> When three forces act on an object in equilibrium, they must:
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>
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> 1. Make a triangle of forces
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> 2. Go through a single point
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>
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> So we can figure out the angle of $R_{A}$ by drawing it such that all the lines of action of
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> all forces go through the same point:
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>
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> 
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>
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> \begin{align*}
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L_{DE} &= L_{BC} = 2.6 \\
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L_{EG} &= \tan45\times L_{BE} = 4.5 \\
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L_{DE} &= 2.6+4.5 = 7.1 \\
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\\
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\tan\theta &= \frac{L_{DG}}{L_{AD}} = 0.79 \\
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\theta &= 38.27
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> \end{align*}
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>
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> Now draw the force triangle:
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>
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> 
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>
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> Using the sine rule we find out $R_A$ and $R_{Dy}$, which are $81.1$ kN and $13.4$ kN,
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> respectively.
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>
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> Now we can check our answers in part (a) and (b) are the same:
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>
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> - $R_{Dy} = 13.4$
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> - $R_{Ax} = 81.1\cos38.27 = 63.6$
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> - $R_{Ay} = 81.1\sin.27 = 50.2$
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>
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> The methods agree.
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</details>
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