A triangular plate with base 6 m and height 2 m is submerged vertically in water such that the highest vertex of the plate is 4 meters below the surface and the base is horizontal to the surface.4 m6 m2 mExpress the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1,000 kg/m3.)

The hydrostatic force is the force due to the effect of pressure acting on an area of a submerged body's surface

The expression for the hydrostatic force acting against one side of the plate and the magnitude of the force are;

The reason the above values are correct are as follows:

The known parameters of the plate are;

Base length of the triangular plate, l = 6 m

Height of the triangular plate, h = 2 m

The depth of the highest vertex below the surface of the water = 4 m

Required:

To express the hydrostatic force against a side of the plate as an integral

Solution:

The area of small slit of the plate, dA = l × dh

The ratio of the length to height of the plate is given as follows;

Therefore;

l = 3·h

Which gives;

dA = l × dh = 3·h × dh

dA = 3·h × dh

The force, dF acting on a small element dA, is given as follows;

dF = ρ·g·(4 + h)·dA

Therefore;

dF = ρ·g·(4 + h)·3·h × dh

dF = 3·ρ·g·(4·h + h²) × dh

Given that the height is 2 meters, the value of h ranges from 0 to 2, which gives;

Evaluating the integral gives;

Therefore, the hydrostatic force against one side of the plate as an integral is given as follows;

The hydrostatic force, against a side of the plate,  F ≈ 313,600 N

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