Question

Explain the ways in which the hydrostatic force is approximated against one side of the plate by a Riemann sum, also express the force and evaluate it.

Short answer

The integral for the hydrostatic force is\({\rm{ }}\int_0^4 3 75(2 + y)dy\)and is evaluated as\(6000lb\).

Step-by-step solution

Assume the horizontal axis for explanation of solution.

It is given that the vertical plate has a rectangular dimension of \(6ft\)by\(4ft\).

The plate is \(2ft\)under the water.

Assume the horizontal axis at a depth of \(2ft\)under the water on the side of the plate.

Let the vertical axis passes through the middle of the plate.

Consider the \({\rm{ith}}\) strip of width \(\Delta y\) be at the distance \(y_i^*\) below the horizontal axis such that the area is\(2 \times 3 \times \Delta y = 6\Delta y\).

The pressure of the \({\rm{ith}}\)is\(\rho {d_i} = 62.5\left( {2 + y_i^*} \right)\).

The force is computed as

\(\begin{aligned}{}{F_i} &= {P_i}{A_i}\\ = \rho {d_i}{A_i}\\ &= 62.5\left( {2 + y_i^*} \right)6\Delta y\\ &= 375\left( {2 + y_i^*} \right)\Delta y\end{aligned}\)

If there are\(n\) strips the sum of forces becomes\(\sum\limits_{i = 1}^n {{F_i}} = \sum\limits_{i = 1}^n 3 75\left( {2 + y_i^*} \right)\Delta y\).

Apply limit\(n\)tends to\(\infty \)in order to obtain the integral.

Apply limit\(n\) tends to\(\infty \)in order to obtain the integral as follows.

\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{F_i}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n 3 75\left( {2 + y_i^*} \right)\Delta y = \int_0^4 3 75(2 + y)dy\)

Compute the integral:

\(\begin{aligned}{}\int_0^4 3 75(2 + y)dy\\ &= 375\left( {2y + \frac{{{y^2}}}{2}} \right)_0^4\\ &= 375\left( {2(4) + \frac{{{{(4)}^2}}}{2}} \right)\\ &= 375(8 + 8) &= 6000{\rm{lb}}\end{aligned}\)

Therefore, the integral for the hydrostatic force is\({\rm{ }}\int_0^4 3 75(2 + y)dy\)and is evaluated as\(6000lb\).