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An equilateral triangular plate with sides 4 m is submerged vertically in water so that the base is even with the surface. Express 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 1000 kg/m3.)

Answer :

Answer:

Hydrostatic force = 7800N

Step-by-step explanation:

The hydrostatic force is the force on a body when there is change its in hydrostatic pressure on the surface of the body when its equilibrium position shifts.

Let us first look at the formula for hydrostatic force

hydrostatic force = the pressure at the center x the area

Pressure = density x gravity x height

hydrostatic force = [sqrt(4)/2 m][1000 kg/m^3 * 9.8m/s^2] x [(1/2)(4)(4sqrt(4)/2) m^2]

Calculating the above data we get;

Hydrostatic force = 7800N

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Answer:

The Hydrostatic force is =7800N

Step-by-step explanation:

Let recall the following,

The hydrostatic force is a force that occurs when a force resulting from pressure of a body or fluid at rest

The hydrostatic force is the pressure at the center x the area

The Pressure is = density x gravity x height

The acceleration due to gravity = 9.8m^2

The weight of the water (density) = 1000 kg/m3

So find the mass at the center which is,

√(4)/2 that is below the water level

The hydrostatic force is = (√(4)/2m) (1000 kg/m^3 x 9.8m/s) ((1/2)(4)(4√(4)/2)m^2)

Therefore, in expressing the hydrostatic force against one side of the plate is.

(√(4)/2m) (1000 kg/m^3 x 9.8m/s) ((1/2)(4)(4√(4)/2)m^2) =7800N

The Hydrostatic force =7800N

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