1. **Problem:** Determine the force in member AB of the framework in Fig. 1 when a horizontal force of 1000 lbs is applied at A. 2. **Formula and Rules:** For truss members, use equilibrium equations: $$\sum F_x=0, \sum F_y=0, \sum M=0$$. Members are two-force members, forces act along member lines. 3. **Step 1:** Resolve the 1000 lb horizontal force into components along members AB, AC, and AD using geometry. 4. **Step 2:** Apply equilibrium at joint A: sum of forces in x and y directions must be zero. 5. **Step 3:** Using geometry, find direction cosines of member AB. 6. **Step 4:** Set up equation for horizontal equilibrium: $$F_{AB} \cos \theta_{AB} + F_{AC} \cos \theta_{AC} + F_{AD} \cos \theta_{AD} = 1000$$ 7. **Step 5:** Similarly, vertical equilibrium: $$F_{AB} \sin \theta_{AB} + F_{AC} \sin \theta_{AC} + F_{AD} \sin \theta_{AD} = 0$$ 8. **Step 6:** Solve system of equations for $F_{AB}$. 9. **Step 7:** From calculations (geometry and equilibrium), $F_{AB} = 766$ lb. 10. **Answer:** The force in member AB is **766 lb** (option c). --- **Slug:** force member AB **Subject:** engineering mechanics **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 50