Discussion
A plane truss is simply supported at \(P\) and \(R\) as shown. A downward force \(F\) is applied at hinge \(Q\). The axial force developed in member \(PS\) is
Answer: B
The force in \(PS\) is \(\frac{\sqrt{5}}{2}F\) Compressive.
The load \(F\) at \(Q\) sits midway between the supports, so each support carries \(F/2\) upward. At joint \(P\) only two members meet: \(PQ\) (horizontal) and \(PS\), which rises \(L/2\) over a run of \(L\), so its length is \(\frac{\sqrt{5}}{2}L\) and \(\sin\alpha = \frac{1}{\sqrt{5}}\).
Vertical equilibrium of joint \(P\), taking member forces as tensile:
\[ \begin{aligned} \tfrac{F}{2} + F_{PS}\sin\alpha &= 0 \\ F_{PS} &= -\tfrac{F}{2}\sqrt{5} = -\tfrac{\sqrt{5}}{2}F \end{aligned} \]
The negative sign means compression: the inclined member pushes down on the support to balance its upward reaction. \(\sqrt{5}F\) comes from using the full load \(F\) instead of the reaction \(F/2\), and the tensile options have the sign reversed.
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