4. Experiment 1 Member Forces

Using the Method of Joints and an applied load of $W=500 \mathrm{~N}$, calculate the theoretical forces in the three members, stating whether tension or compression. [Hint: By inspection, state the force in member AB , and then analyse the force balance at joint C .]
$$
\begin{array}{ll}
\sum F_{y}=0: & F_{A C} \sin 45^{\circ}-W=0 \
\sum F_{x}=0: & -F_{B C}-F_{A C} \cos 45^{\circ}=0
\end{array}
$$

Show that $F_{A C}=\sqrt{2} W$ tension and $F_{B C}=W$ compression.

Question

4. Experiment 1 Member Forces

Using the Method of Joints and an applied load of $W=500 \mathrm{~N}$, calculate the theoretical forces in the three members, stating whether tension or compression. [Hint: By inspection, state the force in member AB , and then analyse the force balance at joint C .]
$$
\begin{array}{ll}
\sum F_{y}=0: & F_{A C} \sin 45^{\circ}-W=0 \
\sum F_{x}=0: & -F_{B C}-F_{A C} \cos 45^{\circ}=0
\end{array}
$$

Show that $F_{A C}=\sqrt{2} W$ tension and $F_{B C}=W$ compression.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To analyze the vertical forces at joint C, we set up the equation: $ F_{AC} \sin(45^{\circ}) - W = 0 $ ‖ ‖ step 2 ⋮ Rearranging gives us: $ F_{AC} \sin(45^{\circ}) = W $ ‖ ‖ step 3 ⋮ Since $ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $, we can substitute: $ F_{AC} \frac{\sqrt{2}}{2} = 500 $ ‖ ‖ step 4 ⋮ Solving for $ F_{AC} $ yields: $ F_{AC} = 500 \cdot \frac{2}{\sqrt{2}} = 500\sqrt{2} $ ‖ ‖ step 5 ⋮ Therefore, $ F_{AC} = \sqrt{2} W $ is in tension. ‖ ‖ step 6 ⋮ Now, analyzing the horizontal forces at joint C, we set up the equation: $ -F_{BC} - F_{AC} \cos(45^{\circ}) = 0 $ ‖ ‖ step 7 ⋮ Rearranging gives us: $ F_{BC} = -F_{AC} \cos(45^{\circ}) $ ‖ ‖ step 8 ⋮ Substituting $ F_{AC} = 500\sqrt{2} $ and $ \cos(45^{\circ}) = \frac{\sqrt{2}}{2} $: $ F_{BC} = -500\sqrt{2} \cdot \frac{\sqrt{2}}{2} $ ‖ ‖ step 9 ⋮ Simplifying gives us: $ F_{BC} = -500 $ ‖ ‖ step 10 ⋮ Therefore, $ F_{BC} = W $ is in compression. ‖ ∻Answer∻ ⚹ $ F_{AC} = \sqrt{2} W $ (tension) and $ F_{BC} = W $ (compression) ⚹ ∻Key Concept∻ ⚹Force balance at a joint in a truss system⚹ ∻Explanation∻ ⚹The forces acting on a joint must balance out to maintain equilibrium, allowing us to determine the forces in the members connected to that joint. In this case, we found that $ F_{AC} $ is in tension and $ F_{BC} $ is in compression.⚹◊What is the value of $F_{AC}$ when the applied load $W$ is 500 N?⍭ Generate me a similar question◊

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