A plane truss PQRS (𝑷𝑸 = 𝑹𝑺, 𝐚𝐧𝐝 ∠𝑷𝑸𝑹 = 𝟗𝟎°) is shown in the figure.

The forces in the members PR and RS, respectively, are _____________

Correct Answer: 𐒠√2 (tensile) and 𐒠 (compressive)

Step-by-Step Explanation:

1. Image Analysis & System Setup
Based on the first provided image, we have a plane truss with joints labeled P, Q, R, and S:
- Joint P is a pin support (hinged).
- Joint S is a roller support.
- The length of the vertical member PQ is labeled L, and the length of the horizontal member QR is also L.
- Since PQ = QR = L and the angle ∠PQR is 90°, the angle of the diagonal member PR with the horizontal member QR is:
$\theta ={\tan }^{-1}\left(\frac{L}{L}\right)=45°$
- A horizontal force F acts at joint R pointing to the right.

2. Identifying Zero-Force Members
Let us analyze joint Q. At joint Q, there are only two non-collinear members PQ and QR, and there is no external load or reaction force acting on this joint.
For joint Q to be in equilibrium:
$F_{\mathrm{PQ}}=0$
$F_{\mathrm{QR}}=0$

3. Joint R Equilibrium Analysis
Now, let's analyze joint R. As shown in the second image containing the free-body diagram:
- The horizontal member force F_QR is zero.
- The external horizontal force is F acting to the right.
- There is a diagonal member force F_PR at an angle of 45° to the horizontal.
- There is a vertical member force F_RS acting vertically downwards along the line of member RS.
Assuming both F_PR and F_RS are tensile forces (pulling away from joint R):

Horizontal Equilibrium (ΣF_x = 0):
$F-F_{\mathrm{PR}}\cdot \cos (45°)=0$
$F_{\mathrm{PR}}\cdot \frac{1}{\sqrt{2}}=F$
$F_{\mathrm{PR}}=F\sqrt{2}$
Since the value is positive, our assumption of tension is correct. Hence, F_PR = F√2 (tensile).

Vertical Equilibrium (ΣF_y = 0):
$-F_{\mathrm{RS}}-F_{\mathrm{PR}}\cdot \sin (45°)=0$
$F_{\mathrm{RS}}=-F_{\mathrm{PR}}\cdot \sin (45°)$
Substituting F_PR = F√2:
$F_{\mathrm{RS}}=-\left(F\sqrt{2}\right)\cdot \frac{1}{\sqrt{2}}=-F$
The negative sign indicates that the member force is compressive. Hence, F_RS = F (compressive).

Conclusion:
The forces in members PR and RS are F√2 (tensile) and F (compressive) respectively.