1. **Problem Statement:** Determine the greatest load $P$ that can be applied to the triangular truss so that no member experiences tension exceeding 2 kN or compression exceeding 1.5 kN. 2. **Setup and Assumptions:** The truss is equilateral with side length $4$ m and angles of $60^\circ$. Supports at $A$ (pin) and $B$ (roller) hold the structure. Load $P$ is applied downward at vertex $C$. 3. **Method:** Use static equilibrium and method of joints to find forces in members $AC$, $BC$, and $AB$ in terms of $P$. 4. **Equilibrium at Joint C:** The load $P$ acts downward. Members $AC$ and $BC$ are symmetrical, so forces in $AC$ and $BC$ are equal in magnitude, call this force $F$. 5. **Geometry:** Each member forms $60^\circ$ angles. The vertical component of $F$ from each member balances $P$: $$2F \sin 60^\circ = P \implies F = \frac{P}{2 \sin 60^\circ} = \frac{P}{2 \times \frac{\sqrt{3}}{2}} = \frac{P}{\sqrt{3}}$$ 6. **Force Type:** Since $P$ pulls down, members $AC$ and $BC$ are in tension (pulling away from joint $C$). 7. **Force in Member AB:** At joint $A$, sum of forces must be zero. Member $AB$ force $F_{AB}$ is horizontal. The horizontal component of $F$ in $AC$ is $F \cos 60^\circ = F \times \frac{1}{2} = \frac{P}{\sqrt{3}} \times \frac{1}{2} = \frac{P}{2\sqrt{3}}$. Since $A$ is a pin support, reaction forces balance this. Member $AB$ force must balance horizontal components of $AC$ and $BC$ (which are equal and opposite), so $F_{AB}$ is compression. 8. **Calculate $F_{AB}$:** The force in $AB$ equals the horizontal component of $AC$ (or $BC$): $$F_{AB} = F \cos 60^\circ = \frac{P}{2\sqrt{3}}$$ 9. **Constraints:** - Tension limit: $F \leq 2$ kN for $AC$ and $BC$ - Compression limit: $F_{AB} \leq 1.5$ kN 10. **Apply limits:** - From tension limit: $$F = \frac{P}{\sqrt{3}} \leq 2 \implies P \leq 2 \sqrt{3} \approx 3.464$$ - From compression limit: $$F_{AB} = \frac{P}{2\sqrt{3}} \leq 1.5 \implies P \leq 1.5 \times 2 \sqrt{3} = 3 \sqrt{3} \approx 5.196$$ 11. **Final Load $P$:** The greatest load $P$ is limited by the smaller value: $$P_{max} = 3.464$$ 12. **Answer:** The maximum load $P$ that can be applied without exceeding member force limits is approximately **3.46** (units consistent with kN).