## Chapter 2

## Q. 2.3

## Step-by-Step

## Verified Solution

The system shown in Figure 2.17 is in equilibrium. The nature of internal force developed in AB is essentially tensile. Taking moment of forces about the hinge C and equating to zero, we get

\begin{gathered} \sum M_c=0 \\ P \times 2.4=F_{ AB } \times 1.5 \\ P=\frac{F_{ AB } \times 1.5}{2.4}=\frac{40 \times 550 \times 1.5}{2.4}=13750 N =13.75 kN \end{gathered}