FIGURE 5.1 below shows a shaped lamina - NSC Civil Technology Civil Services - Question 5 - 2017 - Paper 1

Next, we compute the centroid (X) for each of the areas:

• Centroid of Part 1 ( X1):

  $X_1 = \frac{(b_1 + b_2)}{2} = \frac{(30 + 15)}{2} = 22.5 \ mm$

• Centroid of Part 2 ( X2):

  Since this is a rectangle, the centroid is at half the width, hence:

  $X_2 = 30 \ mm + \frac{60}{2} = 60 \ mm$

Using the formula for the centroid:

$X = \frac{ \sum (A \cdot X)}{ \sum A}$

Where:

• Total Area $= A_1 + A_2 = 1012.5 + 1500 = 2512.5 \ mm^2$
• Contribution from Part 1:$= A_1 \cdot X_1 = 1012.5 \cdot 22.5 = 22781.25 \ mm^3$
• Contribution from Part 2:$= A_2 \cdot X_2 = 1500 \cdot 60 = 90000 \ mm^3$

Now we can substitute these values into the centroid formula:

$X_{total} = \frac{22781.25 + 90000}{2512.5} \approx 41.15 \ mm$

Finally, round off to two decimal places to get the final answer: 41.15 mm.