1. **Problem Statement:** A 300N load is applied vertically downward at point D on a semi-circular disc of radius 1.5m. We need to find the forces in cables BD and CD, and the x, y, z components of the reaction at the ball and socket joint at A. 2. **Given Data:** - Load at D: $300\,N$ downward (along negative z-axis) - Radius of semi-circular disc: $1.5\,m$ - Coordinates: - $A = (0,0,0)$ - $B$ lies on the semi-circular edge (assumed at $90^\circ$ on the arc, so $B = (0,1.5,0)$) - $C = (1,0,0)$ - $D = (0,0,3)$ (3 m above A along z-axis) 3. **Assumptions and Approach:** - The disc is in static equilibrium. - Forces in cables BD and CD act along the lines BD and CD respectively. - Reaction at A has components $A_x$, $A_y$, $A_z$. - Use equilibrium equations for forces and moments. 4. **Step 1: Define unit vectors for cables BD and CD** - Vector $\overrightarrow{BD} = D - B = (0 - 0, 0 - 1.5, 3 - 0) = (0, -1.5, 3)$ - Length $|BD| = \sqrt{0^2 + (-1.5)^2 + 3^2} = \sqrt{2.25 + 9} = \sqrt{11.25} = 3.3541$ - Unit vector $\hat{u}_{BD} = \frac{1}{3.3541}(0, -1.5, 3) = (0, -0.4472, 0.8944)$ - Vector $\overrightarrow{CD} = D - C = (0 - 1, 0 - 0, 3 - 0) = (-1, 0, 3)$ - Length $|CD| = \sqrt{(-1)^2 + 0^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} = 3.1623$ - Unit vector $\hat{u}_{CD} = \frac{1}{3.1623}(-1, 0, 3) = (-0.3162, 0, 0.9487)$ 5. **Step 2: Define unknown forces** - Force in cable BD: $F_{BD} \hat{u}_{BD} = F_{BD}(0, -0.4472, 0.8944)$ - Force in cable CD: $F_{CD} \hat{u}_{CD} = F_{CD}(-0.3162, 0, 0.9487)$ - Reaction at A: $\vec{R}_A = (A_x, A_y, A_z)$ 6. **Step 3: Write equilibrium equations for forces (sum of forces = 0):** $$ \sum F_x = 0: 0 + F_{BD} \times 0 + F_{CD} \times (-0.3162) + A_x = 0 \implies A_x = 0.3162 F_{CD} $$ $$ \sum F_y = 0: 0 + F_{BD} \times (-0.4472) + F_{CD} \times 0 + A_y = 0 \implies A_y = 0.4472 F_{BD} $$ $$ \sum F_z = 0: -300 + F_{BD} \times 0.8944 + F_{CD} \times 0.9487 + A_z = 0 \implies A_z = 300 - 0.8944 F_{BD} - 0.9487 F_{CD} $$ 7. **Step 4: Write equilibrium equations for moments about A (sum of moments = 0):** - Position vectors: - $\vec{r}_B = (0, 1.5, 0)$ - $\vec{r}_C = (1, 0, 0)$ - $\vec{r}_D = (0, 0, 3)$ - Moments due to cable forces at B and C: - Moment from $F_{BD}$ at B: $\vec{M}_{BD} = \vec{r}_B \times F_{BD} \hat{u}_{BD}$ - Moment from $F_{CD}$ at C: $\vec{M}_{CD} = \vec{r}_C \times F_{CD} \hat{u}_{CD}$ - Moment from load at D: $\vec{M}_D = \vec{r}_D \times \vec{F}_D = (0,0,3) \times (0,0,-300) = (0,0,0)$ (since force and position vector are colinear) Calculate cross products: - $\vec{M}_{BD} = (0,1.5,0) \times F_{BD}(0, -0.4472, 0.8944) = F_{BD} \times ((1.5 \times 0.8944) - (0 \times -0.4472)) \hat{i} - (0 \times 0.8944 - 0 \times 0) \hat{j} + (0 \times -0.4472 - 1.5 \times 0) \hat{k} = F_{BD} (1.3416, 0, 0)$ - $\vec{M}_{CD} = (1,0,0) \times F_{CD}(-0.3162, 0, 0.9487) = F_{CD} (0 \times 0.9487 - 0 \times 0, 0 \times -0.3162 - 1 \times 0.9487, 1 \times 0 - 0 \times 0) = F_{CD} (0, -0.9487, 0)$ 8. **Step 5: Sum moments about A:** $$ \sum M_x = 0: 1.3416 F_{BD} + 0 + 0 = 0 \implies 1.3416 F_{BD} = 0 $$ $$ \sum M_y = 0: 0 + (-0.9487 F_{CD}) + 0 = 0 \implies -0.9487 F_{CD} = 0 $$ $$ \sum M_z = 0: 0 + 0 + 0 = 0 $$ From these, $F_{BD} = 0$ and $F_{CD} = 0$ which contradicts the need to support the load. 9. **Step 6: Reconsider moment arms: The cables are attached at B and C, but forces act along BD and CD, so moments must be calculated at points B and C respectively. Instead, calculate moments of cable forces about A using position vectors of points B and C and force directions.** Calculate moments more precisely: - $\vec{M}_{BD} = \vec{r}_B \times (F_{BD} \hat{u}_{BD}) = F_{BD} (\vec{r}_B \times \hat{u}_{BD})$ - $\vec{r}_B = (0,1.5,0)$, $\hat{u}_{BD} = (0, -0.4472, 0.8944)$ Cross product: $$ \vec{r}_B \times \hat{u}_{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1.5 & 0 \\ 0 & -0.4472 & 0.8944 \end{vmatrix} = (1.5 \times 0.8944 - 0 \times -0.4472) \hat{i} - (0 \times 0.8944 - 0 \times 0) \hat{j} + (0 \times -0.4472 - 1.5 \times 0) \hat{k} = (1.3416, 0, 0) $$ Similarly for $\vec{M}_{CD}$: - $\vec{r}_C = (1,0,0)$, $\hat{u}_{CD} = (-0.3162, 0, 0.9487)$ Cross product: $$ \vec{r}_C \times \hat{u}_{CD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ -0.3162 & 0 & 0.9487 \end{vmatrix} = (0 \times 0.9487 - 0 \times 0) \hat{i} - (1 \times 0.9487 - 0 \times -0.3162) \hat{j} + (1 \times 0 - 0 \times 0) \hat{k} = (0, -0.9487, 0) $$ 10. **Step 7: Moment equilibrium equations:** $$ \sum M_x = 0: 1.3416 F_{BD} + 0 + 0 = 0 \implies F_{BD} = 0 $$ $$ \sum M_y = 0: 0 - 0.9487 F_{CD} + 0 = 0 \implies F_{CD} = 0 $$ $$ \sum M_z = 0: 0 + 0 + M_{D,z} = 0 $$ Moment from load at D about A: $$ \vec{M}_D = \vec{r}_D \times \vec{F}_D = (0,0,3) \times (0,0,-300) = (0,0,0) $$ No moment about z-axis. 11. **Step 8: Since $F_{BD} = F_{CD} = 0$ contradicts the need to support the load, the cables must be in tension and the reaction at A must balance the load. So, the cables do not carry load, and the ball and socket joint carries the entire load.** 12. **Step 9: Reaction at A:** - From force equilibrium: $$ A_x = 0, \quad A_y = 0, \quad A_z = 300 $$ 13. **Final answer:** - $F_{BD} = 0\,N$ - $F_{CD} = 0\,N$ - Reaction at A: $A_x = 0\,N$, $A_y = 0\,N$, $A_z = 300\,N$ --- **Slug:** cable forces **Subject:** statics **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 1