Question

The axial force in the member PQ of the plane truss shown in figure is:

(Aromatic compound with an isobutyl side chain) $\xrightarrow[KOH, Heat]{KMnO_4}$ ?

• A. $\sqrt{3}W$
• B. $\sqrt{2}W$
• C. W
• D. $\frac{\sqrt{3}}{2}W$

Hint:

In a $45^\circ$-$45^\circ$-$90^\circ$ triangle relationship, the diagonal (hypotenuse) force is always $\sqrt{2}$ times the vertical component. Since the vertical reaction is $W$, the member force must be $\sqrt{2}W$.

Answer 1

The Correct Option is B

Concept: For a symmetric truss with a symmetric load, the vertical reactions at the supports are equal to half the total vertical load. We can then use the method of joints at one of the supports to find the member forces.

Step 1: Determine the support reactions.
Total vertical load = $2W$. Due to symmetry, the vertical reaction at $Q$ ($V_Q$) and $R$ ($V_R$) is: \[ V_Q = V_R = \frac{2W}{2} = W \]

Step 2: Apply Method of Joints at Joint Q.
Let $F_{PQ}$ be the force in member $PQ$. At joint $Q$, considering vertical equilibrium ($\sum F_y = 0$): \[ V_Q + F_{PQ} \sin(45^\circ) = 0 \] (Assuming $F_{PQ}$ is compressive and acting towards the joint) \[ W = F_{PQ} \frac{1}{\sqrt{2}} \] \[ F_{PQ} = \sqrt{2}W \text{ (Compression)} \]