Location of a Centroid

Apr 2010
3
0
I'm not really sure where to begin with this one. Any help would be greatly appreciated.

 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I'm not really sure where to begin with this one. Any help would be greatly appreciated.

\(\displaystyle \bar{y} = \frac{y}{2}\)

\(\displaystyle dA = y \, dx\)

\(\displaystyle y_c = \frac{\frac{1}{2} \int_0^b y^2 \, dx}{\int_0^b y \, dx}\)

where \(\displaystyle y = H\sin\left(\frac{\pi x}{b}\right) \)
 
Apr 2010
3
0
Thank you, that's exactly what I needed. It makes much more sense now