1. **Problem statement:** Determine the forces in cables AC and AB needed to hold a 20-kg ball D in equilibrium with given parameters $F = 300$ N and $d \approx 1$ m. Then, for a 20-kg ball D with $F = 100$ N applied horizontally at A, find the largest dimension $d$ so the force in cable AC is zero. 2. **Given data:** - Mass of ball D: $m = 20$ kg - Gravitational acceleration: $g = 9.81$ m/s$^2$ - Force at A: $F = 300$ N (first), $F = 100$ N (second) - Vertical distance BC = 1.5 m - Horizontal distance from wall to A = 2 m - Vertical distance from bottom of wall to C = $d$ m --- ### FIRST PROBLEM 3. Calculate weight of ball D: $$W = mg = 20 \times 9.81 = 196.2\, \text{N}$$ 4. Coordinates for points relative to bottom of wall: - C: $(0, d)$ - B: $(0, d + 1.5)$ - A: $(2, d)$ 5. Vector for cable AB from A to B: $$\overrightarrow{AB} = (0 - 2, (d + 1.5) - d) = (-2, 1.5)$$ Length: $$|AB| = \sqrt{(-2)^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5$$ Unit vector: $$\hat{u}_{AB} = \left(\frac{-2}{2.5}, \frac{1.5}{2.5}\right) = (-0.8, 0.6)$$ 6. Vector for cable AC from A to C: $$\overrightarrow{AC} = (0 - 2, d - d) = (-2, 0)$$ Length: $$|AC| = 2$$ Unit vector: $$\hat{u}_{AC} = (-1, 0)$$ 7. Forces on point A in equilibrium: sum of forces = 0 Forces: - $F$: horizontal of magnitude 300 N to right: $(300, 0)$ - Weight $W$: vertical down: $(0, -196.2)$ - Tensions in cables: - $T_{AB} \hat{u}_{AB}$ - $T_{AC} \hat{u}_{AC}$ Equilibrium equations: Horizontal: $$300 - 0.8 T_{AB} - T_{AC} = 0$$ Vertical: $$0 + 0.6 T_{AB} + 0 = 196.2$$ (weight acts down so force upward is positive tension component) 8. Solve vertical: $$0.6 T_{AB} = 196.2 \implies T_{AB} = \frac{196.2}{0.6} = 327$$ 9. Solve horizontal: $$300 - 0.8 \times 327 - T_{AC} = 0 \implies T_{AC} = 300 - 261.6 = 38.4$$ --- ### SECOND PROBLEM 10. Given $F = 100$ N, find $d$ such that $T_{AC}=0$. From equilibrium: Horizontal: $$100 - 0.8 T_{AB} - 0 = 0 \implies 0.8 T_{AB} = 100 \implies T_{AB} = \frac{100}{0.8} =125$$ Vertical: $$0.6 T_{AB} = 196.2 \implies 0.6 \times 125 = 75 \neq 196.2$$ This contradiction means the angles must change to maintain equilibrium (i.e., $d$ changes). 11. Let unit vector of AB depend on $d$: $$\overrightarrow{AB} = (-2,1.5)$$ but vertical component changes to $d +1.5 - d = 1.5$ fixed; horizontal fixed at -2. Length: $$|AB| = \sqrt{(-2)^2 + (1.5)^2} = 2.5$$ Unit vector vertical component remains $\frac{1.5}{2.5} = 0.6$ regardless of $d$. So to have $T_{AC}=0$, cable AC must bear no force, so cable AB bears all weight, but given $F$ only 100 N, tension in AB insufficient. To change this, re-examine geometry of AC cable vertical coordinate: AC vector from A at $(2,d)$ to C at $(0,d)$ is horizontal at any $d$, meaning tension in AC is horizontal only. Force vertical component only from AB, so vertical force balance: $$T_{AB} \times 0.6 = 196.2 \implies T_{AB} = \frac{196.2}{0.6} = 327$$ Horizontal balance: $$100 - 0.8 \times 327 - T_{AC} = 0$$ If $T_{AC}=0$, $$100 - 261.6 = -161.6 \neq 0$$ Conclusion: force $F = 100$ insufficient to satisfy equilibrium with $T_{AC}=0$ at original geometry. 12. Actually, the problem wants largest $d$ so $T_{AC}=0$. To make $T_{AC} = 0$, the cable AB must be vertical (no horizontal component), i.e., horizontal distance between A and B is zero: Distance between A and B horizontally is 2 m, fixed. So change $d$ so vector AB is vertical: $$x_B - x_A = 0 \implies 0 - 2 = -2 \neq 0$$ Not possible. Alternatively, apply force horizontally $F = 100$ at A, change $d$ to reduce vertical component of AB cable. But since height difference is fixed (1.5 m), only $d$ changes vertical position of C and B, but B is always 1.5 m above C: Thus, vertical change of $d$ shifts both B and C up/down, so ratio changes. To have $T_{AC} = 0$, horizontal force must equal horizontal component of tension AB: $$F = T_{AB} \cos \theta$$ Vertical balance: $$T_{AB} \sin \theta = W$$ From triangle with horizontal 2 m, vertical $1.5$ m: $$\tan \theta = \frac{1.5}{2} = 0.75$$ So $\sin \theta = \frac{1.5}{2.5} = 0.6$, $\cos \theta = 0.8$ Force: $$F = T_{AB} \cos \theta = \frac{W}{\sin \theta} \cos \theta = W \frac{\cos \theta}{\sin \theta} = W \cot \theta$$ Given $F=100$, solve for $\cot \theta$: $$100 = 196.2 \times \cot \theta \implies \cot \theta = \frac{100}{196.2} = 0.51$$ Then $$\theta = \arccot(0.51) = \arctan \left( \frac{1}{0.51} \right) = \arctan(1.96) \approx 63.6^\circ$$ From $\theta = \arctan(\frac{1.5}{2})$, now want $$\tan \theta = \frac{1.5}{2} = 0.75$$ originally, need new $d$ so $$\tan \theta = \frac{h}{2} = \frac{h}{2} = \tan 63.6^\circ = 2$$ So vertical distance between A and B is $$h = 2 \times \tan 63.6^\circ = 2 \times 2 = 4$$ Since vertical distance between B and C is $1.5$ m, and B is located at $d + 1.5$, then $$h = (d + 1.5) - d = 1.5$$ originally, fixed. Contradiction appears because geometry assumptions conflict. 13. Actually, $d$ is the vertical distance from bottom to C. Increasing $d$ raises B and C up together, so vertical distance between A and B increases by $d$. Therefore, $$h = (d + 1.5) - d_A$$ Where $d_A$ is vertical position of A, originally $d$. Since A has vertical coordinate $d$, B has $d + 1.5$. Distance AB vertical component is $$\Delta y= (d + 1.5) - d =1.5$$ fixed. Therefore, changing $d$ does not change vertical height difference between A and B. Thus, no change to angle $\theta$, meaning force in AC can never be zero unless force $F$ changes otherwise. Hence, issue is in problem statement or assumptions. --- **Final answers:** - For first problem: $$T_{AB} = 327\, \text{N}, \quad T_{AC} = 38.4\, \text{N}$$ - For second problem: given $F=100$ N, force in cable AC cannot be zero by changing $d$ alone because vertical difference between B and A remains 1.5 m independent of $d$. ---