Recent content by Sheezey

Interesting. I understand what you mean. That is a great way to look at it. Good advice. Thank you for your help.

Ok got it! The "some distance from b" in my moment equation is = to Ycp. Which requires me to find the moment of inertia. Ixx = (1/12)(8)(hcsc(60))3 Ycp = (Ixx(8)(hcsc(60))3sin(60))/((h/2)(8hcsc(60)) ∑M = -150000+18339.6+287.8h2(Ycp) = 0 solve for h ∴ h = 10.59 ft Thanks for your help guys!

Ok so the submerged area of the plate is going to be (8ft)*(the submerged length of A to B)... If I create a triangle on the drawing and call the unknown submerged length 'x' I get x=h/sin(60). Now I can plug that into the original equation I was asking about, hydrostatic force. So I now have...

No it wouldn't. It does vary and will be become greater the deeper you go.

Thinking more about it there would be a force due to the water acting on to the steel plate. I believe this force would also act in the center of the plate. I think this is what I am looking for to insert for the area in my original equation.

Correction (7.85)(62.3lb/ft3)(15ft)(8ft)(1/12ft)(7.5ftcos(60)) = 18339.6lbft

Right. You must mean the weight force of the steel plate which would act along the center of the plate. So we would have... (7.85)(62.3lb/ft3)(15ft)(8ft)(1/12ft)(7.5cos(60)) = 18339.6lb Does that look right?

Hi billy_joule, Thank you for responding. The torque about B due to the self weight would be 10,000lb* 15ft = 150,000lbft

Homework Statement Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at 20°C. The gate is 1-in-thick steel, SG =7.85. Compute the water level h for which the gate will start to fail. Homework Equations F = δhcgA Ycp =...