Question

The depth of the centre of pressure on a vertical rectangular gate 8 m wide and 6 m high, when the water surface coincides with the top of the gate, is

Answer 1

The depth of the centre of pressure on a vertical rectangular gate when the water surface coincides with the top of the gate, is 4m.

The depth of the centre of pressure on a vertical rectangular lamina is given as  2 × h ÷ 3, where 'h' is the height of the rectangular gate.

Given, h = 6m, from the question.

Putting h = 6m, we get depth of the centre of pressure as

(2×6÷ 3)m = 4m

Required answer = 4m.

Answer 2

The depth of the centre of pressure on a vertical rectangular gate is 4 metre.

Explanation:

We are given the value of height h= 6m ; we are given the width = 8m

The depth of of COP Of rectangular vertical laminar is = 2 $\times$ $\frac{h}{3}$

here in the formula h resembles the height of the gate.

Putting the value of h in the above formula and solving it,

                                        ⇒ $2\times \frac{6}{3}$ = 4m

Hence the centre of pressure that is required in the question is of value 4m.

To know more:

Total pressure on 1 m x 1 m gate immersed vertically at a depth of 2 m below the free water surface will be

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