Question A.3: Referring to the area A of Concept Application A.2, consider......

Solution. Divide the area A′ into its components A_1 \text { and } A_3 (Fig. A.10b). Recall from Concept Application A.2 that C is located 46 mm above the lower edge of A. The ordinates \bar{y}_1^{\prime} \text { and } \bar{y}_3^{\prime} of A_1 \text { and } A_3 and the first moment Q_{x^{\prime}}^{\prime} of A′ with respect to x′ are

\begin{aligned}Q_{x^{\prime}}^{\prime} & =A_1 \bar{y}_1^{\prime}+A_3 \bar{y}_3^{\prime} \\& =(20 \times 80)(24)+(14 \times 40)(7)=42.3 \times 10^3 mm ^3\end{aligned}

Alternative Solution. Since the centroid C of A is located on the x′ axis, the first moment Q_{x^{\prime}} of the entire area A with respect to that axis is zero:

Q_{x^{\prime}}=A \bar{y}^{\prime}=A(0)=0

Using A” as the portion of A located below the x′ axis and Q_{x^{\prime}}^{\prime \prime} as its first moment with respect to that axis,

Q_{x^{\prime}}=Q_{x^{\prime}}^{\prime}+Q_{x^{\prime}}^{\prime \prime}=0 \quad \text { or } \quad Q_{x^{\prime}}^{\prime}=-Q_{x^{\prime}}^{\prime \prime}

This shows that the first moments of A′ and A” have the same magnitude and opposite signs. Referring to Fig. A.10c, write

Q_{x^{\prime}}^{\prime \prime}=A_4 \bar{y}_4^{\prime}=(40 \times 46)(-23)=-42.3 \times 10^3 \ mm ^3

and

Q_{x^{\prime}}^{\prime}=-Q_{x^{\prime}}^{\prime \prime}=+42.3 \times 10^3 \ mm ^3