Assignment 1 - =12640.6N (2) lets suppose the vertical...

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Assignment 1 07000414x Dai Yifeng Fluid Mechanics According to fluid mechanics Douglas( second edition ) 2.7 A mass of 50kg acts on a piston of area 100cm2. What is the intensity of pressure on the water in contact with the underside of the piston, if the piston is in equilibrium. Solution: since the force exerted by the mass of 50kg is F=mg , and the piston is in equilibrium, so F= A that is to say A=mg the intensity of pressure P=mg/A =50*9.81/0.01=49050N / 3.1 A circular lamina 125 cm in diameter is immersed in water so that the distance of its edge measured vertically below the free surface varies from 60 cm to 150 cm. Find the total force due to the water acting on one side of the lamina, and the vertical distance of the centre of pressure below the surface. Solution: (1)dF=pdA=ρghdA=ρg·sinα·y·dA so the total force F=∫dF=ρg·sinα∫y·dA=ρg·sinα·y·A since h= sinα·y is the depth of centroid so h=(60+150)/2=105cm=1.05m so F=ρg·h·A=ρg·h·π·(D/2) ·(D/2) =1000*9.81*1.05*π*(1.25/2) *(1.25/2)
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Unformatted text preview: =12640.6N (2) lets suppose the vertical distance of the centre of pressure is h, then h/ sinF=ydF that is to sayg hA(x)= gsinh/ sinI(x) so h/ sin= I(x)/ A(x)= I(x)/(yA) since I(x)=I+y*y*A so h/ sin=y+I/(yA) I=(1/4)* *(D/2) *(D/2) *(D/2) *(D/2)=(1/4) **(1.25/2) *(1.25/2) *(1.25/2) *(1.25/2)=0.1198 sin=60/x=150/(125+x), then x=250/3, so sin=18/25 h= [y+I/(yA)] sin={[1.05/(18/25)]+0.1198/[1.05/(18/25)**(1.25/2) *(1.25/2)]}* (18/25) =1.098m 3.15 A pair of lock gates, each 3 m wide, have their lower hinges at the bottom of the gates and their upper hinges 5 m from the bottom. The width of the lock is 5.5m. Find the reaction between the gates when the water level is 4.5 m above the bottom on one side and 1.5 m on the other. Assuming that this force acts at the same height as the resultant force due to the water pressure find the reaction forces on the hinges....
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Assignment 1 - =12640.6N (2) lets suppose the vertical...

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