Ycp=bh3/(12)(h/2)b + (h/2)
The Attempt at a Solution
I found the angle of the gate from the horizontal to be 53.1 degrees. Using this, I found y to be 2.5m and H (vertical distance to centroid) to be 6 m (2.5sin(53.1)+4).
I then found Fw=9810*6*(4*5)=1177200N, and Ycp=(4*53)/(12*20*2.5) +2.5=3.33m.
By taking the moment at the ground, I found Fw*(5-3.33)-5P=0, so P= 393kN. My book says the answer should be 523 kN. Help please.