Let's break the problem down **step by step**: --- ## **Given Data** - **Tension in AC:** \( T_{AC} = 100 \) lbs - **AE (coaxial cable):** Applies force only in **x-direction** (let's call this \( F_{AE} \)) - **AB:** Pole resists only a compressive force **along its axis** (\( F_{AB} \)) - **Find:** Forces in **AB, AD, AE**. --- ## **Step 1: Draw a Free-Body Diagram (FBD) at A** At point A, the forces are: - \( \vec{F}_{AB} \) (along AB) - \( \vec{F}_{AC} \) (along AC, known magnitude) - \( \vec{F}_{AD} \) (along AD) - \( \vec{F}_{AE} \) (along x-axis only) --- ## **Step 2: Get Unit Vectors for Each Cable** Let's define the direction cosines for each member using the angles given. ### **a) AC** - 60° down from y-axis (in yz-plane), 20° from x-axis (in xy-plane) - Let's use the geometry. Let’s assume the coordinates for A are at the origin (for simplicity), and use the angles to define directions. #### **Direction Cosines:** For AC: - **From the figure:** - 60° from y-axis (down to C) - 20° from x-axis (in xy-plane) - Unit vector for AC: \[ \vec{u}_{AC} = \langle \cos(60^\circ)\cos(20^\circ),\; \sin(60^\circ),\; \cos(60^\circ)\sin(20^\circ) \rangle \] - x-component: \( \cos(60^\circ)\cos(20^\circ) \) - y-component: \( \sin(60^\circ) \) - z-component: \( \cos(60^\circ)\sin(20^\circ) \) ### **b) AD** - 60° down from y-axis, 48° from x-axis (in the negative z direction) - Unit vector for AD: \[ \vec{u}_{AD} = \langle \cos(60^\circ)\cos(48^\circ),\; \sin(60^\circ),\; -\cos(60^\circ)\sin(48^\circ) \rangle \] ### **c) AB** - Along y-axis only (vertical): \( \vec{u}_{AB} = \langle 0, 1, 0 \rangle \) ### **d) AE** - Along x-axis: \( \vec{u}_{AE} = \langle 1, 0, 0 \rangle \) --- ## **Step 3: Write Equilibrium Equations at A** \[ \sum \vec{F} = 0 \] Let the unknown forces be \( F_{AB} \), \( F_{AD} \), \( F_{AE} \). \[ F_{AB} \vec{u}_{AB} + F_{AD} \vec{u}_{AD} + 100 \vec{u}_{AC} + F_{AE} \vec{u}_{AE} = 0 \] Write components for x, y, z: ### **x-component:** \[ 0 \cdot F_{AB} + F_{AD} \cdot u_{AD,x} + 100 \cdot u_{AC,x} + F_{AE} \cdot 1 = 0 \] ### **y-component:** \[ F_{AB} \cdot 1 + F_{AD} \cdot u_{AD,y} + 100 \cdot u_{AC,y} + 0 = 0 \] ### **z-component:** \[ 0 \cdot F_{AB} + F_{AD} \cdot u_{AD,z} + 100 \cdot u_{AC,z} + 0 = 0 \] --- ## **Step 4: Plug in Numerical Values** Let's calculate the direction cosines numerically: - \( \cos(60^\circ) = 0.5 \) - \( \sin(60^\circ) = 0.866 \) - \( \cos(20^\circ) = 0.94 \) - \( \sin(20^\circ) = 0.342 \) - \( \cos(48^\circ) = 0.669 \) - \( \sin(48^\circ) = 0.743 \) #### **For AC:** - x: \( 0.5 \times 0.94 = 0.47 \) - y: \( 0.866 \) - z: \( 0.5 \times 0.342 = 0.171 \) #### **For AD:** - x: \( 0.5 \times 0.669 = 0.335 \) - y: \( 0.866 \) - z: \( -0.5 \times 0.743 = -0.372 \) #### **For AB:** - x: \( 0 \), y: \( 1 \), z: \( 0 \) #### **For AE:** - x: \( 1 \), y: \( 0 \), z: \( 0 \) --- ## **Step 5: Write Out the System** ### **x:** \[ 0.335 F_{AD} + 47 + F_{AE} = 0 \] ### **y:** \[ F_{AB} + 0.866 F_{AD} + 86.6 = 0 \] ### **z:** \[ -0.372 F_{AD} + 17.1 = 0 \] --- ## **Step 6: Solve for Unknowns** **From z-component:** \[ -0.372 F_{AD} + 17.1 = 0 \implies F_{AD} = \frac{17.1}{0.372} \approx 46.0~\text{lbs} \] **From y-component:** \[ F_{AB} + 0.866 \times 46.0 + 86.6 = 0 \] \[ F_{AB} + 39.836 + 86.6 = 0 \] \[ F_{AB} = -126.4~\text{lbs} \] (Negative means it's in compression, as expected.) **From x-component:** \[ 0.335 \times 46.0 + 47 + F_{AE} = 0 \] \[ 15.41 + 47 + F_{AE} = 0 \] \[ F_{AE} = -62.41~\text{lbs} \] (Negative means the force is in the negative x-direction, i.e., tension pulling left.) --- ## **Step 7: Final Answers** ### **Forces in Each Member** - **AB:** \( \boxed{126.4~\text{lbs (compression)}} \) - **AD:** \( \boxed{46.0~\text{lbs (tension)}} \) - **AE:** \( \boxed{62.4~\text{lbs (tension, negative x-direction)}} \) --- ## **Summary Table** | Member | Force (lbs) | Nature | |--------|-------------|---------------| | AB | 126.4 | Compression | | AD | 46.0 | Tension | | AE | 62.4 | Tension (−x) | --- Let me know if you'd like a more detailed explanation of any step or a diagram!