Question:
For the truss shown in the figure, the magnitude of force in the member PR is ______ kN.
For the truss shown in the figure, the magnitude of force in the member PR is ______ kN.
Show Hint
Use symmetry to get the support reactions first, then resolve the two members meeting at joint P.
Updated On: Jul 28, 2026
- \( 10 \)
- \( 0 \)
- \( \dfrac{10}{\sqrt{3}} \)
- \( \dfrac{20}{\sqrt{3}} \)
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The Correct Option is C
Solution and Explanation
Step 1: Find the support reactions.
The truss is made of two equilateral triangles, PQR and RST, each of side 6 m, joined at R, with a pin at P, a roller at T, and a 20 kN load acting straight down at R. The whole truss and the load are symmetric about the vertical line through R, so each support carries half the load: \(R_P = R_T = 10\) kN vertically, and the horizontal reaction at the pin P is zero since there is no horizontal load anywhere.
Step 2: Set up joint P.
Only two members meet at P: PQ, which rises at 60 degrees to the horizontal since angle QPR is 60 degrees, and PR, which runs horizontally to R. Taking tension as positive in each member, the vertical reaction \(R_P = 10\) kN acts upward at this joint.
Step 3: Resolve forces at joint P.
Vertical equilibrium: \(F_{PQ} \sin 60^{\circ} + 10 = 0\), so \(F_{PQ} = -10 / \sin 60^{\circ} = -20/\sqrt{3}\) kN, meaning PQ is in compression.
Horizontal equilibrium: \(F_{PQ}\cos 60^{\circ} + F_{PR} = 0\), so \(F_{PR} = -F_{PQ}\cos 60^{\circ} = -(-20/\sqrt{3})(0.5) = 10/\sqrt{3}\) kN.
Final Answer:
The positive sign shows PR is in tension, and its magnitude works out cleanly. \[ \boxed{|F_{PR}| = \dfrac{10}{\sqrt{3}} \approx 5.77 \text{ kN}} \]
The truss is made of two equilateral triangles, PQR and RST, each of side 6 m, joined at R, with a pin at P, a roller at T, and a 20 kN load acting straight down at R. The whole truss and the load are symmetric about the vertical line through R, so each support carries half the load: \(R_P = R_T = 10\) kN vertically, and the horizontal reaction at the pin P is zero since there is no horizontal load anywhere.
Step 2: Set up joint P.
Only two members meet at P: PQ, which rises at 60 degrees to the horizontal since angle QPR is 60 degrees, and PR, which runs horizontally to R. Taking tension as positive in each member, the vertical reaction \(R_P = 10\) kN acts upward at this joint.
Step 3: Resolve forces at joint P.
Vertical equilibrium: \(F_{PQ} \sin 60^{\circ} + 10 = 0\), so \(F_{PQ} = -10 / \sin 60^{\circ} = -20/\sqrt{3}\) kN, meaning PQ is in compression.
Horizontal equilibrium: \(F_{PQ}\cos 60^{\circ} + F_{PR} = 0\), so \(F_{PR} = -F_{PQ}\cos 60^{\circ} = -(-20/\sqrt{3})(0.5) = 10/\sqrt{3}\) kN.
Final Answer:
The positive sign shows PR is in tension, and its magnitude works out cleanly. \[ \boxed{|F_{PR}| = \dfrac{10}{\sqrt{3}} \approx 5.77 \text{ kN}} \]
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