The plane truss shown in the figure is hinge-supported at E and F. The truss is subjected to vertical downward force at R and horizontal force at G. (Figure not to scale) The force (in kN) along with its nature in member JF is
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Find the support reactions first (F_y=25 kN, from moments about E), then use a vertical section through the last panel and vertical equilibrium to isolate member JF.
Step 1: Get the reactions the same way, then switch to joint-by-joint work instead of a section cut.
As before, taking moments about E (a horizontal reaction at F contributes nothing about E, since both hinges sit on the same base line) gives $F_y = 25$ kN upward, and $E_y=-5$ kN from vertical equilibrium of the whole truss. This part does not depend on the internal bracing pattern at all, so it stays the anchor point for both solution routes.
Step 2: Walk the zero-force members joint by joint, starting from the far top-left corner.
At L, the two members meeting there (to M and to G) are the only ones present, with no applied load, so equilibrium in two perpendicular directions forces both to zero. Moving to M, the same style of argument (one already-zero member, two unknowns, no load at the joint) forces the next pair to zero, and this repeats again at N. So the whole run of top-chord and connecting diagonal members from L up to just short of R turns out to carry no force at all, and the top chord only gets loaded once we reach the loaded joint R itself.
Step 3: Handle the middle chord under the horizontal load separately.
At G, the 40 kN horizontal load has nowhere else to go once the members found in Step 2 are known to be zero, so it pushes straight along the middle chord: G-H works out to 40 kN compression, and this same 40 kN carries forward, member by member, along H-I and I-J, since none of the intermediate joints has its own applied load to divert it.
Step 4: Bring the 20 kN load into the picture at joint R.
At R, the top-chord members are known to be zero from Step 2, leaving the 20 kN load to be carried by the members that actually reach down into the truss body from this point; equilibrium at R sends this load down as a 20 kN compressive force in the vertical member below R.
Step 5: Isolate joint F and use both of its equilibrium equations together.
At F, three members meet (the bottom chord coming in from the left, a vertical from above, and the diagonal J-F), together with the support reaction $F_y=25$ kN found in Step 1. Since the vertical member above F and the incoming bottom-chord member trace back, through the zero-force chain identified above, to carry no force, the only member left to balance the 25 kN reaction in the vertical direction is J-F itself, inclined at 45 degrees:
\[
T_{JF}\sin45^\circ = -F_y \;\Rightarrow\; T_{JF} = -F_y\sqrt2 = -25\sqrt2\text{ kN, i.e. }25\sqrt2\text{ kN compression}
\]
This matches the section-cut answer from the other solution, using an entirely separate, joint-by-joint route.
Step 6: Same honest flag as the other solution.
Both independent methods land on the same number, $25\sqrt2$ kN compression, which is option C here, not the answer key's stated option A ($10\sqrt2$ kN compression). Since the exact interior bracing near R, J and V could not be confirmed with full certainty from the figure, this is reported as a genuine mismatch to flag for review rather than adjusted to fit the key.
\[ \boxed{T_{JF} = 25\sqrt2\text{ kN compression by this working; does not match the stated key of }10\sqrt2\text{ kN compression.}} \]
