GATE 2023 CE (CE2) – Question 57

Correct answer: 17.91 to 18.09

Explanation

**Step 1: free body of the horizontal member PQ.** PQ is pinned at P to the vertical member RS, is supported by the brace TU at U, and carries 10 N downward at Q. The distances along PQ are PU = 1.2 m and UQ = 0.6 m, so PQ = 1.8 m.

**Step 2: take moments about P** (the pin, whose force has no moment):
$$F_{brace,y}\times1.2=10\times1.8\;\Rightarrow\;F_{brace,y}=15\text{ N}.$$

**Step 3: the brace direction.** The brace goes from T (1 m below P on the vertical member) to U (1.2 m to the right of P), so its rise-to-run ratio is $1:1.2$. The force in the brace acts along the brace, so its horizontal component is related to the vertical one by
$$F_{brace,x}=F_{brace,y}\times\frac{1.2}{1}=15\times1.2=18\text{ N}.$$

**Step 4: horizontal equilibrium of PQ.** The horizontal force from the brace on PQ must be balanced by the horizontal force from the pin at P. So the horizontal component of the force transferred at P has magnitude
$$\mathbf{18.0\ N}.$$