The profile of the inner face of a dam takes the form of a parabola with the equation 18y=x2, where y m is the height above the base and x m is the horizontal distance of the face from the vertical reference plane. The water level is 27 m above the base. take g=9.8 m/s2 (Density of water is 1000 kg/m3)

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the horizontal thrust on the dam (per metre width) due to the water pressure is 3.57 MN⋅m−1 in three significant figures.

the vertical thrust on the dam (per metre width) due to the water pressure is 3.89 MN⋅m−1 in three significant figures.

the point where the line of action of horizontal thrust force pass through is 18 m below the free surface of water

the point where the line of action of horizontal thrust force pass through is 9 m below the free surface of water

answer is A.

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Detailed Solution

Horizontal component H = Thrust on verticalProjection AC divided by width =121000×9.8×272 N⋅m−1 =3.57 MN⋅m−1Vertical component V = weight of water ABC area ABC=∫027xdy=18∫027y1/2dy=2318273/2 m2=396.8 m2Vertical component =1000×9.8×396.8 N⋅m−1=3.89 MN⋅m−1

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