Question

find the centroid of I section whose top angle equals to 10 cm into 2.5 cm equals to 10 cm and 2.5 CM angle equals to 15 cm into 3.5 CM ​

Answer 1

Answer:

We are gonna use symmetry . See picture our

centroid will lie upon y axis .

We can assume that at place of all three sections there are point masses are placed on middle of these sections .

Assume that density of flakes is 1 . so

$m = \sigma \times area \\ so \\ m _{1} = 1 \times 10 \times 2.5 = 25 \\ m _{2} = 1 \times 10 \times 2.5 = 25 \\ m _{3} = 1 \times 15 \times 2.5 = 37.5$

Now y coordinates of all three point masses are 2.5 , 0 and -2.5 respectively .

So now centroid or centre if mass will be at

$y = \frac{25 \times 2.5 + 25 \times 0 + 37.5 \times( - 2.5) }{25 + 25 + 37.5} \\ \\ = - \: \frac{2.5 \times 12.5}{87.5} \\ \\ = - \: 0.357$

So centroid will lie at distance of 0.357 below middle point of mid section .