1. **State the problem:** We have a system with three springs connected at point A: spring AC, spring AB, and spring AD. The block at D is in equilibrium. We need to find the mass of the block at D. 2. **Given data:** - Unstretched length of spring AB: $L_{AB0} = 3$ m - Length of spring AB in equilibrium: $L_{AB} = 4$ m (horizontal distance) - Length of spring AC in equilibrium: $L_{AC} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ m - Length of spring AD in equilibrium: unknown, but vertical from A to D - Spring constants: $k_{AC} = 20$ N/m, $k_{AB} = 30$ N/m, $k_{AD} = 40$ N/m 3. **Calculate the extension of each spring:** - Extension of AB: $x_{AB} = L_{AB} - L_{AB0} = 4 - 3 = 1$ m - Extension of AC: unstretched length is not given, but since AC is connected from A to C with 3 m vertical and 3 m horizontal, assume unstretched length $L_{AC0} = 3\sqrt{2}$ m (same as equilibrium length), so extension $x_{AC} = 0$ m - Extension of AD: unstretched length unknown, but since AD hangs vertically and the block is in equilibrium, the extension $x_{AD}$ will be the vertical displacement of the block from the unstretched length. 4. **Determine forces in the springs:** - Force in AB: $F_{AB} = k_{AB} x_{AB} = 30 \times 1 = 30$ N - Force in AC: $F_{AC} = k_{AC} x_{AC} = 20 \times 0 = 0$ N - Force in AD: $F_{AD} = k_{AD} x_{AD} = 40 x_{AD}$ N 5. **Analyze equilibrium at point A:** - The block at D is held in equilibrium, so the sum of forces at A is zero. - Forces from springs AB and AC act horizontally and diagonally, while AD acts vertically downward. 6. **Resolve forces into components:** - Force in AB acts horizontally to the right: $F_{ABx} = 30$ N, $F_{ABy} = 0$ - Force in AC acts along the line AC at 45° (since 3 m vertical and 3 m horizontal): - $F_{ACx} = F_{AC} \times \frac{3}{3\sqrt{2}} = 0$ - $F_{ACy} = F_{AC} \times \frac{3}{3\sqrt{2}} = 0$ - Force in AD acts vertically downward: $F_{ADx} = 0$, $F_{ADy} = -40 x_{AD}$ 7. **Sum of forces in horizontal direction:** $$ \sum F_x = F_{ABx} + F_{ACx} + F_{ADx} = 30 + 0 + 0 = 30 \text{ N} $$ Since the block is in equilibrium, horizontal forces must balance, so there must be an equal and opposite force, but since no other forces are given, assume the system is fixed to balance this. 8. **Sum of forces in vertical direction:** $$ \sum F_y = F_{ABy} + F_{ACy} + F_{ADy} - mg = 0 $$ Since $F_{ABy} = 0$, $F_{ACy} = 0$, and $F_{ADy} = -40 x_{AD}$, and the weight of the block is $mg$, we have: $$ -40 x_{AD} - mg = 0 \implies mg = 40 x_{AD} $$ 9. **Find $x_{AD}$:** - The unstretched length of AD is not given, but since the block is at equilibrium, the extension $x_{AD}$ equals the vertical displacement of the block from the unstretched length. - Assume the unstretched length of AD is zero (or negligible), so $x_{AD}$ equals the length of AD. 10. **Calculate the length of AD:** - Since the block is at D below A, and the system is in equilibrium, the vertical displacement is equal to the extension. - From the geometry, the vertical length of AD is unknown, but since the problem does not provide it, assume the extension $x_{AD}$ equals the length of AD. 11. **Calculate mass $m$:** - Using $mg = 40 x_{AD}$ and $g = 9.8$ m/s², - Without the length of AD, we cannot find $m$ numerically. **Conclusion:** - The problem lacks the length of spring AD or its extension to find the mass. - If the length of AD or its extension is given, mass can be calculated by $m = \frac{40 x_{AD}}{9.8}$. **Slug:** spring mass **Subject:** physics