The face of a dam is shaped like an isosceles trapezoid with a lower base of 44 meters, an upper base of 108 meters, and a height of 114 meters. Find the force on the face of the dam when the water level is 35 meters below the top of the dam.

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creamydhaka creamydhaka · Answer 1 · 2020-03-12T08:32:48+01:00

Answer:

[tex]F=4047408940.35\ N[/tex]

Explanation:

Given:

length of upper base of trapezoid, [tex]u=108\ m[/tex]

length of lower base of trapezoid, [tex]l=44\ m[/tex]

height of the trapezoid, [tex]y=114\ m[/tex]

water level from the top of the dam, [tex]h'=35\ m[/tex]

The depth of water from the base of the dam wall:

[tex]d=y-h'[/tex]

[tex]d=114-35[/tex]

[tex]d=79\ m[/tex]

Now the pressure on the dam wall due to water upto the given level:

[tex]P=\rho.g.d[/tex]

[tex]P=1000\times 9.8\times 79[/tex]

[tex]P=774200\ Pa[/tex]

Now refer the schematic for the area on which the water pressure acts.

Now we have the area as:

[tex]A=\frac{1}{2}\time (sum\ of\ parallel\ sides)\times height[/tex]

[tex]A=0.5(44+ 44.3509+44)\times 79[/tex]

[tex]A=5227.8596\ m^2[/tex]

Now the force on the dam wall can be given as:

[tex]F=P.A[/tex]

[tex]F=774200\times 5227.8596[/tex]

[tex]F=4047408940.35\ N[/tex]